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University Physics Volume 2
SENIOR CONTRIBUTING AUTHORS
SAMUEL J. LING, TRUMAN STATE UNIVERSITY JEFF SANNY, LOYOLA MARYMOUNT UNIVERSITY WILLIAM MOEBS, FORMERLY OF LOYOLA MARYMOUNT UNIVERSITY

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Contents

Preface 1

UNIT 1 THERMODYNAMICS

CHAPTER 1
Temperature and Heat 5
Introduction 5 1.1 Temperature and Thermal Equilibrium 6 1.2 Thermometers and Temperature Scales 7 1.3 Thermal Expansion 10 1.4 Heat Transfer, Specific Heat, and Calorimetry 17 1.5 Phase Changes 25 1.6 Mechanisms of Heat Transfer 33 Chapter Review 51

CHAPTER 2
The Kinetic Theory of Gases 67
Introduction 67 2.1 Molecular Model of an Ideal Gas 68 2.2 Pressure, Temperature, and RMS Speed 78 2.3 Heat Capacity and Equipartition of Energy 88 2.4 Distribution of Molecular Speeds 93 Chapter Review 98

CHAPTER 3
The First Law of Thermodynamics 109
Introduction 109 3.1 Thermodynamic Systems 110 3.2 Work, Heat, and Internal Energy 112 3.3 First Law of Thermodynamics 116 3.4 Thermodynamic Processes 122 3.5 Heat Capacities of an Ideal Gas 125 3.6 Adiabatic Processes for an Ideal Gas 127 Chapter Review 132

CHAPTER 4
The Second Law of Thermodynamics
Introduction 143 4.1 Reversible and Irreversible Processes 144 4.2 Heat Engines 146 4.3 Refrigerators and Heat Pumps 148 4.4 Statements of the Second Law of Thermodynamics 4.5 The Carnot Cycle 152 4.6 Entropy 157

143 150

4.7 Entropy on a Microscopic Scale 163 Chapter Review 168

UNIT 2 ELECTRICITY AND MAGNETISM

CHAPTER 5
Electric Charges and Fields 179
Introduction 179 5.1 Electric Charge 180 5.2 Conductors, Insulators, and Charging by Induction 5.3 Coulomb's Law 190 5.4 Electric Field 195 5.5 Calculating Electric Fields of Charge Distributions 5.6 Electric Field Lines 210 5.7 Electric Dipoles 213 Chapter Review 216

186 202

CHAPTER 6
Gauss's Law 229
Introduction 229 6.1 Electric Flux 230 6.2 Explaining Gauss’s Law 239 6.3 Applying Gauss’s Law 245 6.4 Conductors in Electrostatic Equilibrium 258 Chapter Review 267

CHAPTER 7
Electric Potential 279
Introduction 279 7.1 Electric Potential Energy 280 7.2 Electric Potential and Potential Difference 7.3 Calculations of Electric Potential 298 7.4 Determining Field from Potential 308 7.5 Equipotential Surfaces and Conductors 7.6 Applications of Electrostatics 319 Chapter Review 324

287 311

CHAPTER 8
Capacitance 335
Introduction 335 8.1 Capacitors and Capacitance 336 8.2 Capacitors in Series and in Parallel 345 8.3 Energy Stored in a Capacitor 351 8.4 Capacitor with a Dielectric 354 8.5 Molecular Model of a Dielectric 357 Chapter Review 364

CHAPTER 9
Current and Resistance 373

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Introduction 373 9.1 Electrical Current 374 9.2 Model of Conduction in Metals 379 9.3 Resistivity and Resistance 384 9.4 Ohm's Law 393 9.5 Electrical Energy and Power 397 9.6 Superconductors 403 Chapter Review 408
CHAPTER 10
Direct-Current Circuits 419
Introduction 419 10.1 Electromotive Force 420 10.2 Resistors in Series and Parallel 427 10.3 Kirchhoff's Rules 439 10.4 Electrical Measuring Instruments 451 10.5 RC Circuits 454 10.6 Household Wiring and Electrical Safety 460 Chapter Review 465
CHAPTER 11
Magnetic Forces and Fields 479
Introduction 479 11.1 Magnetism and Its Historical Discoveries 480 11.2 Magnetic Fields and Lines 482 11.3 Motion of a Charged Particle in a Magnetic Field 486 11.4 Magnetic Force on a Current-Carrying Conductor 491 11.5 Force and Torque on a Current Loop 496 11.6 The Hall Effect 499 11.7 Applications of Magnetic Forces and Fields 502 Chapter Review 506
CHAPTER 12
Sources of Magnetic Fields 519
Introduction 519 12.1 The Biot-Savart Law 520 12.2 Magnetic Field Due to a Thin Straight Wire 523 12.3 Magnetic Force between Two Parallel Currents 527 12.4 Magnetic Field of a Current Loop 529 12.5 Ampère’s Law 532 12.6 Solenoids and Toroids 539 12.7 Magnetism in Matter 544 Chapter Review 551
CHAPTER 13
Electromagnetic Induction 565
Introduction 565 13.1 Faraday’s Law 566 13.2 Lenz's Law 570

13.3 Motional Emf 574 13.4 Induced Electric Fields 582 13.5 Eddy Currents 586 13.6 Electric Generators and Back Emf 590 13.7 Applications of Electromagnetic Induction 596 Chapter Review 598
CHAPTER 14
Inductance 611
Introduction 611 14.1 Mutual Inductance 612 14.2 Self-Inductance and Inductors 615 14.3 Energy in a Magnetic Field 620 14.4 RL Circuits 623 14.5 Oscillations in an LC Circuit 628 14.6 RLC Series Circuits 631 Chapter Review 634
CHAPTER 15
Alternating-Current Circuits 643
Introduction 643 15.1 AC Sources 644 15.2 Simple AC Circuits 645 15.3 RLC Series Circuits with AC 651 15.4 Power in an AC Circuit 656 15.5 Resonance in an AC Circuit 660 15.6 Transformers 664 Chapter Review 669
CHAPTER 16
Electromagnetic Waves 679
Introduction 679 16.1 Maxwell’s Equations and Electromagnetic Waves 680 16.2 Plane Electromagnetic Waves 686 16.3 Energy Carried by Electromagnetic Waves 693 16.4 Momentum and Radiation Pressure 697 16.5 The Electromagnetic Spectrum 702 Chapter Review 710
Appendix A Units 723
Appendix B Conversion Factors 727
Appendix C Fundamental Constants 731
Appendix D Astronomical Data 733
Appendix E Mathematical Formulas 735
Appendix F Chemistry 739
Appendix G The Greek Alphabet 741
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Answer Key 742 Index 789

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Preface

1

PREFACE

Welcome to University Physics, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
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Format
You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print.
About University Physics
University Physics is designed for the two- or threesemester calculus-based physics course. The text has been developed to meet the scope and sequence of most university physics courses and provides a foundation for a career in mathematics, science, or engineering. The book provides an important opportunity for students to learn the core concepts of physics and understand how those concepts apply to their lives and to the world around them.
Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency.
Coverage and scope
Our University Physics textbook adheres to the scope and sequence of most two- and threesemester physics courses nationwide. We have worked to make physics interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but to work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from science educators dedicated to the project.
VOLUME I
Unit 1: Mechanics

2

Preface

• Chapter 1: Units and Measurement • Chapter 2: Vectors • Chapter 3: Motion Along a Straight Line • Chapter 4: Motion in Two and Three Dimensions • Chapter 5: Newton’s Laws of Motion • Chapter 6: Applications of Newton’s Laws • Chapter 7: Work and Kinetic Energy • Chapter 8: Potential Energy and Conservation of
Energy • Chapter 9: Linear Momentum and Collisions • Chapter 10: Fixed-Axis Rotation • Chapter 11: Angular Momentum • Chapter 12: Static Equilibrium and Elasticity • Chapter 13: Gravitation • Chapter 14: Fluid Mechanics
Unit 2: Waves and Acoustics
• Chapter 15: Oscillations • Chapter 16: Waves • Chapter 17: Sound
VOLUME II
Unit 1: Thermodynamics
• Chapter 1: Temperature and Heat • Chapter 2: The Kinetic Theory of Gases • Chapter 3: The First Law of Thermodynamics • Chapter 4: The Second Law of Thermodynamics
Unit 2: Electricity and Magnetism
• Chapter 5: Electric Charges and Fields • Chapter 6: Gauss’s Law • Chapter 7: Electric Potential • Chapter 8: Capacitance • Chapter 9: Current and Resistance • Chapter 10: Direct-Current Circuits • Chapter 11: Magnetic Forces and Fields • Chapter 12: Sources of Magnetic Fields • Chapter 13: Electromagnetic Induction • Chapter 14: Inductance • Chapter 15: Alternating-Current Circuits • Chapter 16: Electromagnetic Waves
VOLUME III
Unit 1: Optics
• Chapter 1: The Nature of Light • Chapter 2: Geometric Optics and Image
Formation • Chapter 3: Interference • Chapter 4: Diffraction
Unit 2: Modern Physics
• Chapter 5: Relativity

• Chapter 6: Photons and Matter Waves • Chapter 7: Quantum Mechanics • Chapter 8: Atomic Structure • Chapter 9: Condensed Matter Physics • Chapter 10: Nuclear Physics • Chapter 11: Particle Physics and Cosmology
Pedagogical foundation
Throughout University Physics you will find derivations of concepts that present classical ideas and techniques, as well as modern applications and methods. Most chapters start with observations or experiments that place the material in a context of physical experience. Presentations and explanations rely on years of classroom experience on the part of long-time physics professors, striving for a balance of clarity and rigor that has proven successful with their students. Throughout the text, links enable students to review earlier material and then return to the present discussion, reinforcing connections between topics. Key historical figures and experiments are discussed in the main text (rather than in boxes or sidebars), maintaining a focus on the development of physical intuition. Key ideas, definitions, and equations are highlighted in the text and listed in summary form at the end of each chapter. Examples and chapter-opening images often include contemporary applications from daily life or modern science and engineering that students can relate to, from smart phones to the internet to GPS devices.
Assessments that reinforce key concepts
In-chapter Examples generally follow a three-part format of Strategy, Solution, and Significance to emphasize how to approach a problem, how to work with the equations, and how to check and generalize the result. Examples are often followed by Check Your Understanding questions and answers to help reinforce for students the important ideas of the examples. Problem-Solving Strategies in each chapter break down methods of approaching various types of problems into steps students can follow for guidance. The book also includes exercises at the end of each chapter so students can practice what they’ve learned.
• Conceptual questions do not require calculation but test student learning of the key concepts.
• Problems categorized by section test student problem-solving skills and the ability to apply ideas to practical situations.
• Additional Problems apply knowledge across the chapter, forcing students to identify what

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Preface

3

concepts and equations are appropriate for solving given problems. Randomly located throughout the problems are Unreasonable Results exercises that ask students to evaluate the answer to a problem and explain why it is not reasonable and what assumptions made might not be correct. • Challenge Problems extend text ideas to interesting but difficult situations.
Answers for selected exercises are available in an Answer Key at the end of the book.
Additional resources
Student and instructor resources
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About the authors
Senior contributing authors
Samuel J. Ling, Truman State University

Dr. Samuel Ling has taught introductory and advanced physics for over 25 years at Truman State University, where he is currently Professor of Physics and the Department Chair. Dr. Ling has two PhDs from Boston University, one in Chemistry and the other in Physics, and he was a Research Fellow at the Indian Institute of Science, Bangalore, before joining Truman. Dr. Ling is also an author of A First Course in Vibrations and Waves, published by Oxford University Press. Dr. Ling has considerable experience with research in Physics Education and has published research on collaborative learning methods in physics teaching. He was awarded a Truman Fellow and a Jepson fellow in recognition of his innovative teaching methods. Dr. Ling’s research publications have spanned Cosmology, Solid State Physics, and Nonlinear Optics.
Jeff Sanny, Loyola Marymount University Dr. Jeff Sanny earned a BS in Physics from Harvey Mudd College in 1974 and a PhD in Solid State Physics from the University of California–Los Angeles in 1980. He joined the faculty at Loyola Marymount University in the fall of 1980. During his tenure, he has served as department Chair as well as Associate Dean. Dr. Sanny enjoys teaching introductory physics in particular. He is also passionate about providing students with research experience and has directed an active undergraduate student research group in space physics for many years.
William Moebs, Formerly of Loyola Marymount University Dr. William Moebs earned a BS and PhD (1959 and 1965) from the University of Michigan. He then joined their staff as a Research Associate for one year, where he continued his doctoral research in particle physics. In 1966, he accepted an appointment to the Physics Department of Indiana Purdue Fort Wayne (IPFW), where he served as Department Chair from 1971 to 1979. In 1979, he moved to Loyola Marymount University (LMU), where he served as Chair of the Physics Department from 1979 to 1986. He retired from LMU in 2000. He has published research in particle physics, chemical kinetics, cell division, atomic physics, and physics teaching.
Contributing authors
Stephen D. Druger, Northwestern University Alice Kolakowska, University of Memphis David Anderson, Albion College Daniel Bowman, Ferrum College Dedra Demaree, Georgetown University

4

Preface

Edw. S. Ginsberg, University of Massachusetts Joseph Trout, Richard Stockton College Kevin Wheelock, Bellevue College David Smith, University of the Virgin Islands Takashi Sato, Kwantlen Polytechnic University Gerald Friedman, Santa Fe Community College Lev Gasparov, University of North Florida Lee LaRue, Paris Junior College Mark Lattery, University of Wisconsin Richard Ludlow, Daniel Webster College Patrick Motl, Indiana University Kokomo Tao Pang, University of Nevada, Las Vegas Kenneth Podolak, Plattsburgh State University
Reviewers
Salameh Ahmad, Rochester Institute of Technology–Dubai John Aiken, University of Colorado–Boulder Raymond Benge, Terrant County College Gavin Buxton, Robert Morris University Erik Christensen, South Florida State College Clifton Clark, Fort Hays State University Nelson Coates, California Maritime Academy Herve Collin, Kapi’olani Community College Carl Covatto, Arizona State University Alejandro Cozzani, Imperial Valley College Danielle Dalafave, The College of New Jersey Nicholas Darnton, Georgia Institute of Technology Ethan Deneault, University of Tampa Kenneth DeNisco, Harrisburg Area Community College Robert Edmonds, Tarrant County College William Falls, Erie Community College Stanley Forrester, Broward College Umesh Garg, University of Notre Dame Maurizio Giannotti, Barry University

Bryan Gibbs, Dallas County Community College Lynn Gillette, Pima Community College–West Campus Mark Giroux, East Tennessee State University Matthew Griffiths, University of New Haven Alfonso Hinojosa, University of Texas–Arlington Steuard Jensen, Alma College David Kagan, University of Massachusetts Sergei Katsev, University of Minnesota–Duluth Gregory Lapicki, East Carolina University Jill Leggett, Florida State College–Jacksonville Alfredo Louro, University of Calgary James Maclaren, Tulane University Ponn Maheswaranathan, Winthrop University Seth Major, Hamilton College Oleg Maksimov, Excelsior College Aristides Marcano, Delaware State University James McDonald, University of Hartford Ralph McGrew, SUNY–Broome Community College Paul Miller, West Virginia University Tamar More, University of Portland Farzaneh Najmabadi, University of Phoenix Richard Olenick, The University of Dallas Christopher Porter, Ohio State University Liza Pujji, Manakau Institute of Technology Baishali Ray, Young Harris University Andrew Robinson, Carleton University Aruvana Roy, Young Harris University Gajendra Tulsian, Daytona State College Adria Updike, Roger Williams University Clark Vangilder, Central Arizona University Steven Wolf, Texas State University Alexander Wurm, Western New England University Lei Zhang, Winston Salem State University Ulrich Zurcher, Cleveland State University

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CHAPTER 1
Temperature and Heat
Figure 1.1 These snowshoers on Mount Hood in Oregon are enjoying the heat flow and light caused by high temperature. All three mechanisms of heat transfer are relevant to this picture. The heat flowing out of the fire also turns the solid snow to liquid water and vapor. (credit: modification of work by “Mt. Hood Territory”/Flickr)
Chapter Outline
1.1 Temperature and Thermal Equilibrium 1.2 Thermometers and Temperature Scales 1.3 Thermal Expansion 1.4 Heat Transfer, Specific Heat, and Calorimetry 1.5 Phase Changes 1.6 Mechanisms of Heat Transfer INTRODUCTION Heat and temperature are important concepts for each of us, every day. How we dress in the morning depends on whether the day is hot or cold, and most of what we do requires energy that ultimately comes from the Sun. The study of heat and temperature is part of an area of physics known as thermodynamics. The laws of thermodynamics govern the flow of energy throughout the universe. They are studied in all areas of science and engineering, from chemistry to biology to environmental science. In this chapter, we explore heat and temperature. It is not always easy to distinguish these terms. Heat is the flow of energy from one object to another. This flow of energy is caused by a difference in temperature. The

6

1 • Temperature and Heat

transfer of heat can change temperature, as can work, another kind of energy transfer that is central to thermodynamics. We return to these basic ideas several times throughout the next four chapters, and you will see that they affect everything from the behavior of atoms and molecules to cooking to our weather on Earth to the life cycles of stars.
1.1 Temperature and Thermal Equilibrium
Learning Objectives By the end of this section, you will be able to:
• Define temperature and describe it qualitatively • Explain thermal equilibrium • Explain the zeroth law of thermodynamics
Heat is familiar to all of us. We can feel heat entering our bodies from the summer Sun or from hot coffee or tea after a winter stroll. We can also feel heat leaving our bodies as we feel the chill of night or the cooling effect of sweat after exercise.
What is heat? How do we define it and how is it related to temperature? What are the effects of heat and how does it flow from place to place? We will find that, in spite of the richness of the phenomena, a small set of underlying physical principles unites these subjects and ties them to other fields. We start by examining temperature and how to define and measure it.
Temperature
The concept of temperature has evolved from the common concepts of hot and cold. The scientific definition of temperature explains more than our senses of hot and cold. As you may have already learned, many physical quantities are defined solely in terms of how they are observed or measured, that is, they are defined operationally. Temperature is operationally defined as the quantity of what we measure with a thermometer. As we will see in detail in a later chapter on the kinetic theory of gases, temperature is proportional to the average kinetic energy of translation, a fact that provides a more physical definition. Differences in temperature maintain the transfer of heat, or heat transfer, throughout the universe. Heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. (You will learn more about heat transfer later in this chapter.)
Thermal Equilibrium
An important concept related to temperature is thermal equilibrium. Two objects are in thermal equilibrium if they are in close contact that allows either to gain energy from the other, but nevertheless, no net energy is transferred between them. Even when not in contact, they are in thermal equilibrium if, when they are placed in contact, no net energy is transferred between them. If two objects remain in contact for a long time, they typically come to equilibrium. In other words, two objects in thermal equilibrium do not exchange energy.
Experimentally, if object A is in equilibrium with object B, and object B is in equilibrium with object C, then (as you may have already guessed) object A is in equilibrium with object C. That statement of transitivity is called the zeroth law of thermodynamics. (The number “zeroth” was suggested by British physicist Ralph Fowler in the 1930s. The first, second, and third laws of thermodynamics were already named and numbered then. The zeroth law had seldom been stated, but it needs to be discussed before the others, so Fowler gave it a smaller number.) Consider the case where A is a thermometer. The zeroth law tells us that if A reads a certain temperature when in equilibrium with B, and it is then placed in contact with C, it will not exchange energy with C; therefore, its temperature reading will remain the same (Figure 1.2). In other words, if two objects are in thermal equilibrium, they have the same temperature.

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1.2 • Thermometers and Temperature Scales

7

Figure 1.2 If thermometer A is in thermal equilibrium with object B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. Therefore, the reading on A stays the same when A is moved over to make contact with C.
A thermometer measures its own temperature. It is through the concepts of thermal equilibrium and the zeroth law of thermodynamics that we can say that a thermometer measures the temperature of something else, and to make sense of the statement that two objects are at the same temperature.
In the rest of this chapter, we will often refer to “systems” instead of “objects.” As in the chapter on linear momentum and collisions, a system consists of one or more objects—but in thermodynamics, we require a system to be macroscopic, that is, to consist of a huge number (such as ) of molecules. Then we can say that a system is in thermal equilibrium with itself if all parts of it are at the same temperature. (We will return to the definition of a thermodynamic system in the chapter on the first law of thermodynamics.)
1.2 Thermometers and Temperature Scales
Learning Objectives By the end of this section, you will be able to:
• Describe several different types of thermometers • Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales
Any physical property that depends consistently and reproducibly on temperature can be used as the basis of a thermometer. For example, volume increases with temperature for most substances. This property is the basis for the common alcohol thermometer and the original mercury thermometers. Other properties used to measure temperature include electrical resistance, color, and the emission of infrared radiation (Figure 1.3).

8

1 • Temperature and Heat

Figure 1.3 Because many physical properties depend on temperature, the variety of thermometers is remarkable. (a) In this common

type of thermometer, the alcohol, containing a red dye, expands more rapidly than the glass encasing it. When the thermometer’s

temperature increases, the liquid from the bulb is forced into the narrow tube, producing a large change in the length of the column for a

small change in temperature. (b) Each of the six squares on this plastic (liquid crystal) thermometer contains a film of a different heat-

sensitive liquid crystal material. Below

, all six squares are black. When the plastic thermometer is exposed to a temperature of

,

the first liquid crystal square changes color. When the temperature reaches above

, the second liquid crystal square also changes

color, and so forth. (c) A firefighter uses a pyrometer to check the temperature of an aircraft carrier’s ventilation system. The pyrometer

measures infrared radiation (whose emission varies with temperature) from the vent and quickly produces a temperature readout. Infrared

thermometers are also frequently used to measure body temperature by gently placing them in the ear canal. Such thermometers are more

accurate than the alcohol thermometers placed under the tongue or in the armpit. (credit b: modification of work by Tess Watson; credit c:

modification of work by Lamel J. Hinton, U.S. Navy)

Thermometers measure temperature according to well-defined scales of measurement. The three most common temperature scales are Fahrenheit, Celsius, and Kelvin. Temperature scales are created by identifying two reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used.

On the Celsius scale, the freezing point of water is and the boiling point is

The unit of

temperature on this scale is the degree Celsius . The Fahrenheit scale (still the most frequently used for

common purposes in the United States) has the freezing point of water at

and the boiling point at

Its unit is the degree Fahrenheit ( ). You can see that 100 Celsius degrees span the same range as 180

Fahrenheit degrees. Thus, a temperature difference of one degree on the Celsius scale is 1.8 times as large as a

difference of one degree on the Fahrenheit scale, or

The definition of temperature in terms of molecular motion suggests that there should be a lowest possible temperature, where the average kinetic energy of molecules is zero (or the minimum allowed by quantum mechanics). Experiments confirm the existence of such a temperature, called absolute zero. An absolute temperature scale is one whose zero point is absolute zero. Such scales are convenient in science because several physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature.

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1.2 • Thermometers and Temperature Scales

9

The Kelvin scale is the absolute temperature scale that is commonly used in science. The SI temperature unit is the kelvin, which is abbreviated K (not accompanied by a degree sign). Thus 0 K is absolute zero. The freezing and boiling points of water are 273.15 K and 373.15 K, respectively. Therefore, temperature differences are the same in units of kelvins and degrees Celsius, or
The relationships between the three common temperature scales are shown in Figure 1.4. Temperatures on these scales can be converted using the equations in Table 1.1.

Figure 1.4 Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales are shown. The relative sizes of the scales are also shown.

To convert from…

Use this equation…

Celsius to Fahrenheit

Fahrenheit to Celsius

Celsius to Kelvin Kelvin to Celsius

Fahrenheit to Kelvin

Kelvin to Fahrenheit

Table 1.1 Temperature Conversions To convert between Fahrenheit and Kelvin, convert to Celsius as an intermediate step.

EXAMPLE 1.1

Converting between Temperature Scales: Room Temperature

“Room temperature” is generally defined in physics to be

. (a) What is room temperature in

is it in K?

? (b) What

Strategy To answer these questions, all we need to do is choose the correct conversion equations and substitute the known values.

10

1 • Temperature and Heat

Solution To convert from

to , use the equation

Substitute the known value into the equation and solve:

Similarly, we find that

.

The Kelvin scale is part of the SI system of units, so its actual definition is more complicated than the one given above. First, it is not defined in terms of the freezing and boiling points of water, but in terms of the triple point. The triple point is the unique combination of temperature and pressure at which ice, liquid water, and water vapor can coexist stably. As will be discussed in the section on phase changes, the coexistence is achieved by lowering the pressure and consequently the boiling point to reach the freezing point. The triplepoint temperature is defined as 273.16 K. This definition has the advantage that although the freezing temperature and boiling temperature of water depend on pressure, there is only one triple-point temperature.
Second, even with two points on the scale defined, different thermometers give somewhat different results for other temperatures. Therefore, a standard thermometer is required. Metrologists (experts in the science of measurement) have chosen the constant-volume gas thermometer for this purpose. A vessel of constant volume filled with gas is subjected to temperature changes, and the measured temperature is proportional to the change in pressure. Using “TP” to represent the triple point,

The results depend somewhat on the choice of gas, but the less dense the gas in the bulb, the better the results for different gases agree. If the results are extrapolated to zero density, the results agree quite well, with zero pressure corresponding to a temperature of absolute zero.
Constant-volume gas thermometers are big and come to equilibrium slowly, so they are used mostly as standards to calibrate other thermometers.
INTERACTIVE
Visit this site (https://openstax.org/l/21consvolgasth) to learn more about the constant-volume gas thermometer.
1.3 Thermal Expansion
Learning Objectives By the end of this section, you will be able to:
• Answer qualitative questions about the effects of thermal expansion • Solve problems involving thermal expansion, including those involving thermal stress
The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal expansion, which is the change in size or volume of a given system as its temperature changes. The most visible example is the expansion of hot air. When air is heated, it expands and becomes less dense than the surrounding air, which then exerts an (upward) force on the hot air and makes steam and smoke rise, hot air balloons float, and so forth. The same behavior happens in all liquids and gases, driving natural heat transfer upward in homes, oceans, and weather systems, as we will discuss in an upcoming section. Solids also undergo thermal expansion. Railroad tracks and bridges, for example, have expansion joints to allow them to freely expand and contract with temperature changes, as shown in Figure 1.5.

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1.3 • Thermal Expansion

11

Figure 1.5 (a) Thermal expansion joints like these in the (b) Auckland Harbour Bridge in New Zealand allow bridges to change length without buckling. (credit: modification of works by “ŠJů”/Wikimedia Commons)
What is the underlying cause of thermal expansion? As previously mentioned, an increase in temperature means an increase in the kinetic energy of individual atoms. In a solid, unlike in a gas, the molecules are held in place by forces from neighboring molecules; as we saw in Oscillations, the forces can be modeled as in harmonic springs described by the Lennard-Jones potential. Energy in Simple Harmonic Motion shows that such potentials are asymmetrical in that the potential energy increases more steeply when the molecules get closer to each other than when they get farther away. Thus, at a given kinetic energy, the distance moved is greater when neighbors move away from each other than when they move toward each other. The result is that increased kinetic energy (increased temperature) increases the average distance between molecules—the substance expands.
For most substances under ordinary conditions, it is an excellent approximation that there is no preferred direction (that is, the solid is “isotropic”), and an increase in temperature increases the solid’s size by a certain fraction in each dimension. Therefore, if the solid is free to expand or contract, its proportions stay the same; only its overall size changes.
Linear Thermal Expansion
According to experiments, the dependence of thermal expansion on temperature, substance, and original initial length is summarized in the equation
1.1
where is the instantaneous change in length per temperature, L is the length, and is the coefficient of linear expansion, a material property that varies slightly with temperature. As is nearly constant and also very small, for practical purposes, we use the linear approximation:
1.2 where is the change in length and is the change in temperature.

Table 1.2 lists representative values of the coefficient of linear expansion. As noted earlier, is the same

whether it is expressed in units of degrees Celsius or kelvins; thus, may have units of

or 1/K with the

same value in either case. Approximating as a constant is quite accurate for small changes in temperature

12

1 • Temperature and Heat

and sufficient for most practical purposes, even for large changes in temperature. We examine this approximation more closely in the next example.

Material

Coefficient of Linear Expansion

Coefficient of Volume Expansion

Solids

Aluminum

Brass

Copper

Gold

Iron or steel

Invar (nickel-iron alloy)

Lead

Silver

Glass (ordinary)

Glass (Pyrex®)

Quartz

Concrete, brick

Marble (average)

Liquids

Ether

Ethyl alcohol

Gasoline

Glycerin

Mercury

Water

Gases

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1.3 • Thermal Expansion

13

Material

Coefficient of Linear Expansion

Coefficient of Volume Expansion

Air and most other gases at atmospheric pressure

Table 1.2 Thermal Expansion Coefficients

Thermal expansion is exploited in the bimetallic strip (Figure 1.6). This device can be used as a thermometer if the curving strip is attached to a pointer on a scale. It can also be used to automatically close or open a switch at a certain temperature, as in older or analog thermostats.

Figure 1.6 The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting temperature, where its two components have the same length. (b) At a higher temperature, this strip bends to the right, because the metal on the left has expanded more than the metal on the right. At a lower temperature, the strip would bend to the left.

EXAMPLE 1.2

Calculating Linear Thermal Expansion

The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to

temperatures ranging from

to

. What is its change in length between these temperatures?

Assume that the bridge is made entirely of steel.

Strategy

Use the equation for linear thermal expansion

to calculate the change in length, . Use the

coefficient of linear expansion for steel from Table 1.2, and note that the change in temperature is

Solution Substitute all of the known values into the equation to solve for :

Significance Although not large compared with the length of the bridge, this change in length is observable. It is generally spread over many expansion joints so that the expansion at each joint is small.
Thermal Expansion in Two and Three Dimensions
Unconstrained objects expand in all dimensions, as illustrated in Figure 1.7. That is, their areas and volumes, as well as their lengths, increase with temperature. Because the proportions stay the same, holes and

14

1 • Temperature and Heat

container volumes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the piece you removed were still in place. The piece would get bigger, so the hole must get bigger too.

Thermal Expansion in Two Dimensions

For small temperature changes, the change in area is given by

1.3

where is the change in area

is the change in temperature, and is the coefficient of linear

expansion, which varies slightly with temperature. (The derivation of this equation is analogous to that of

the more important equation for three dimensions, below.)

Figure 1.7 In general, objects expand in all directions as temperature increases. In these drawings, the original boundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area increases because both length and width increase. The area of a circular plug also increases. (b) If the plug is removed, the hole it leaves becomes larger with increasing temperature, just as if the expanding plug were still in place. (c) Volume also increases, because all three dimensions increase.

Thermal Expansion in Three Dimensions

The relationship between volume and temperature is given by

, where is the coefficient of

volume expansion. As you can show in Exercise 1.60,

. This equation is usually written as

1.4

Note that the values of in Table 1.2 are equal to except for rounding.

Volume expansion is defined for liquids, but linear and area expansion are not, as a liquid’s changes in linear dimensions and area depend on the shape of its container. Thus, Table 1.2 shows liquids’ values of but not .

In general, objects expand with increasing temperature. Water is the most important exception to this rule.

Water does expand with increasing temperature (its density decreases) at temperatures greater than

. However, it is densest at

and expands with decreasing temperature between

and

(

), as shown in Figure 1.8. A striking effect of this phenomenon is the freezing of water in a

pond. When water near the surface cools down to

it is denser than the remaining water and thus sinks to

the bottom. This “turnover” leaves a layer of warmer water near the surface, which is then cooled. However, if

the temperature in the surface layer drops below , that water is less dense than the water below, and thus

stays near the top. As a result, the pond surface can freeze over. The layer of ice insulates the liquid water

below it from low air temperatures. Fish and other aquatic life can survive in water beneath ice, due to

this unusual characteristic of water.

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1.3 • Thermal Expansion

15

Figure 1.8 This curve shows the density of water as a function of temperature. Note that the thermal expansion at low temperatures is

very small. The maximum density at is only

greater than the density at , and

greater than that at . The

decrease of density below occurs because the liquid water approachs the solid crystal form of ice, which contains more empty space

than the liquid.

EXAMPLE 1.3

Calculating Thermal Expansion

Suppose your 60.0-L

-gal) steel gasoline tank is full of gas that is cool because it has just been

pumped from an underground reservoir. Now, both the tank and the gasoline have a temperature of

How much gasoline has spilled by the time they warm to

?

Strategy
The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank. (The gasoline tank can be treated as solid steel.)

Solution

1. Use the equation for volume expansion to calculate the increase in volume of the steel tank:

2. The increase in volume of the gasoline is given by this equation:

3. Find the difference in volume to determine the amount spilled as

Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.)

Significance This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed later in this chapter.

16

1 • Temperature and Heat

If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist compression with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them.

CHECK YOUR UNDERSTANDING 1.1
Does a given reading on a gasoline gauge indicate more gasoline in cold weather or in hot weather, or does the temperature not matter?

Thermal Stress
If you change the temperature of an object while preventing it from expanding or contracting, the object is subjected to stress that is compressive if the object would expand in the absence of constraint and tensile if it would contract. This stress resulting from temperature changes is known as thermal stress. It can be quite large and can cause damage.
To avoid this stress, engineers may design components so they can expand and contract freely. For instance, in highways, gaps are deliberately left between blocks to prevent thermal stress from developing. When no gaps can be left, engineers must consider thermal stress in their designs. Thus, the reinforcing rods in concrete are made of steel because steel’s coefficient of linear expansion is nearly equal to that of concrete.
To calculate the thermal stress in a rod whose ends are both fixed rigidly, we can think of the stress as developing in two steps. First, let the ends be free to expand (or contract) and find the expansion (or contraction). Second, find the stress necessary to compress (or extend) the rod to its original length by the methods you studied in Static Equilibrium and Elasticity on static equilibrium and elasticity. In other words, the of the thermal expansion equals the of the elastic distortion (except that the signs are opposite).

EXAMPLE 1.4

Calculating Thermal Stress

Concrete blocks are laid out next to each other on a highway without any space between them, so they cannot

expand. The construction crew did the work on a winter day when the temperature was . Find the stress in

the blocks on a hot summer day when the temperature is

. The compressive Young’s modulus of concrete

is

.

Strategy According to the chapter on static equilibrium and elasticity, the stress F/A is given by

where Y is the Young’s modulus of the material—concrete, in this case. In thermal expansion,

We combine these two equations by noting that the two

are equal, as stated above. Because we are not

given or A, we can obtain a numerical answer only if they both cancel out.

Solution We substitute the thermal-expansion equation into the elasticity equation to get

and as we hoped, has canceled and A appears only in F/A, the notation for the quantity we are calculating. Now we need only insert the numbers:

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1.4 • Heat Transfer, Specific Heat, and Calorimetry

17

Significance The ultimate compressive strength of concrete is the ultimate shear strength of concrete is only

so the blocks are unlikely to break. However, so some might chip off.

CHECK YOUR UNDERSTANDING 1.2
Two objects A and B have the same dimensions and are constrained identically. A is made of a material with a higher thermal expansion coefficient than B. If the objects are heated identically, will A feel a greater stress than B?

1.4 Heat Transfer, Specific Heat, and Calorimetry
Learning Objectives By the end of this section, you will be able to:
• Explain phenomena involving heat as a form of energy transfer • Solve problems involving heat transfer
We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of energy transfer that is caused by a temperature difference, and it can change the temperature of an object. As we learned earlier in this chapter, heat transfer is the movement of energy from one place or material to another as a result of a difference in temperature. Heat transfer is fundamental to such everyday activities as home heating and cooking, as well as many industrial processes. It also forms a basis for the topics in the remainder of this chapter.
We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We discuss another way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the study of the relationship of heat and work, which is the basis of engines and refrigerators and the central topic (and origin of the name) of thermodynamics.
Internal Energy and Heat
A thermal system has internal energy (also called thermal energy), which is the sum of the mechanical energies of its molecules. A system’s internal energy is proportional to its temperature. As we saw earlier in this chapter, if two objects at different temperatures are brought into contact with each other, energy is transferred from the hotter to the colder object until the bodies reach thermal equilibrium (that is, they are at the same temperature). No work is done by either object because no force acts through a distance (as we discussed in Work and Kinetic Energy). These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure 1.9 shows an example of heat transfer.

Figure 1.9 (a) Here, the soft drink has a higher temperature than the ice, so they are not in thermal equilibrium. (b) When the soft drink and ice are allowed to interact, heat is transferred from the drink to the ice due to the difference in temperatures until they reach the same

18

1 • Temperature and Heat

temperature, , achieving equilibrium. In fact, since the soft drink and ice are both in contact with the surrounding air and the bench, the ultimate equilibrium temperature will be the same as that of the surroundings.

The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say “the heat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy flow, whereas temperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature.

Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is

the calorie (cal), defined as the energy needed to change the temperature of 1.00 g of water by

—specifically, between

and

, since there is a slight temperature dependence. Also commonly

used is the kilocalorie (kcal), which is the energy needed to change the temperature of 1.00 kg of water by

. Since mass is most often specified in kilograms, the kilocalorie is convenient. Confusingly, food

calories (sometimes called “big calories,” abbreviated Cal) are actually kilocalories, a fact not easily

determined from package labeling.

Mechanical Equivalent of Heat

It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of a system. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889) performed many experiments to establish the mechanical equivalent of heat—the work needed to produce the same effects as heat transfer. In the units used for these two quantities, the value for this equivalence is

We consider this equation to represent the conversion between two units of energy. (Other numbers that you

may see refer to calories defined for temperature ranges other than

to

.)

Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can produce the same effects and measuring the mechanical equivalent of heat. It helped establish the principle of conservation of energy. Gravitational potential energy (U) was converted into kinetic energy (K), and then randomized by viscosity and turbulence into increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. Joule’s contributions to thermodynamics were so significant that the SI unit of energy was named after him.

Figure 1.10 Joule’s experiment established the equivalence of heat and work. As the masses descended, they caused the paddles to do

work,

, on the water. The result was a temperature increase, , measured by the thermometer. Joule found that was

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1.4 • Heat Transfer, Specific Heat, and Calorimetry

19

proportional to W and thus determined the mechanical equivalent of heat.

Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although a system has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A well-defined quantity that depends only on the current state of the system, rather than on the history of that system, is known as a state variable. Temperature and internal energy are state variables. To sum up this paragraph, heat and work are not state variables.

Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in the next section, the temperature does not change when a substance changes from one phase to another. An example is the melting of ice, which can be accomplished by adding heat or by doing frictional work, as when an ice cube is rubbed against a rough surface.
Temperature Change and Heat Capacity

We have noted that heat transfer often causes temperature change. Experiments show that with no phase change and no work done on or by the system, the transferred heat is typically directly proportional to the change in temperature and to the mass of the system, to a good approximation. (Below we show how to handle situations where the approximation is not valid.) The constant of proportionality depends on the substance and its phase, which may be gas, liquid, or solid. We omit discussion of the fourth phase, plasma, because although it is the most common phase in the universe, it is rare and short-lived on Earth.

We can understand the experimental facts by noting that the transferred heat is the change in the internal

energy, which is the total energy of the molecules. Under typical conditions, the total kinetic energy of the

molecules

is a constant fraction of the internal energy (for reasons and with exceptions that we’ll see in

the next chapter). The average kinetic energy of a molecule

is proportional to the absolute temperature.

Therefore, the change in internal energy of a system is typically proportional to the change in temperature and

to the number of molecules, N. Mathematically,

The dependence on the

substance results in large part from the different masses of atoms and molecules. We are considering its heat

capacity in terms of its mass, but as we will see in the next chapter, in some cases, heat capacities per molecule

are similar for different substances. The dependence on substance and phase also results from differences in

the potential energy associated with interactions between atoms and molecules.

Heat Transfer and Temperature Change

A practical approximation for the relationship between heat transfer and temperature change is:

1.5

where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and is the

change in temperature. The symbol c stands for the specific heat (also called “specific heat capacity”) and

depends on the material and phase. The specific heat is numerically equal to the amount of heat necessary

to change the temperature of kg of mass by

. The SI unit for specific heat is

or

. (Recall that the temperature change is the same in units of kelvin and degrees Celsius.)

Values of specific heat must generally be measured, because there is no simple way to calculate them precisely. Table 1.3 lists representative values of specific heat for various substances. We see from this table that the specific heat of water is five times that of glass and 10 times that of iron, which means that it takes five times as much heat to raise the temperature of water a given amount as for glass, and 10 times as much as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.
The specific heats of gases depend on what is maintained constant during the heating—typically either the volume or the pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the second (in parentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic theory of gases.

20

1 • Temperature and Heat

Substances

Specific Heat (c)

Solids

Aluminum

900

0.215

Asbestos

800

0.19

Concrete, granite (average)

840

0.20

Copper

387

0.0924

Glass

840

0.20

Gold

129

0.0308

Human body (average at

) 3500

0.83

Ice (average,

)

2090

0.50

Iron, steel

452

0.108

Lead

128

0.0305

Silver

235

0.0562

Wood

1700

0.40

Liquids

Benzene

1740

0.415

Ethanol

2450

0.586

Glycerin

2410

0.576

Mercury

139

0.0333

Water Gases[3]

4186

1.000

Air (dry)

721 (1015) 0.172 (0.242)

Ammonia

1670 (2190) 0.399 (0.523)

Carbon dioxide

638 (833)

0.152 (0.199)

Nitrogen

739 (1040) 0.177 (0.248)

Oxygen

651 (913)

0.156 (0.218)

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1.4 • Heat Transfer, Specific Heat, and Calorimetry

21

Substances

Specific Heat (c)

Steam

1520 (2020) 0.363 (0.482)

Table 1.3 Specific Heats of Various Substances[1] [1]The values for solids and liquids are at constant volume

and

, except as noted. [2]These values are identical in units of

[3]Specific heats at constant

volume and at

except as noted, and at 1.00 atm pressure. Values in parentheses are specific heats at a

constant pressure of 1.00 atm.

In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be

given in terms of an infinitesimal change in temperature. To do this, we note that

and replace

with d:

Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.

EXAMPLE 1.5

Calculating the Required Heat

A 0.500-kg aluminum pan on a stove and 0.250 L of water in it are heated from

to

. (a) How

much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the

water?

Strategy
We can assume that the pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and that of the pan are increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 1.3.

Solution

1. Calculate the temperature difference:

2. Calculate the mass of water. Because the density of water is

, 1 L of water has a mass of 1 kg,

and the mass of 0.250 L of water is

.

3. Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3:

4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3:

5. Find the total transferred heat:

Significance
In this example, the heat transferred to the water is more than the aluminum pan. Although the mass of the pan is twice that of the water, the specific heat of water is over four times that of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water as for the aluminum pan.

Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount

22

1 • Temperature and Heat

of energy had been added with a blowtorch instead of mechanically.)
EXAMPLE 1.6
Calculating the Temperature Increase from the Work Done on a Substance
Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material (Figure 1.11). This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. Since the mass of the truck is much greater than that of the brake material absorbing the energy, the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat.

Figure 1.11 The smoking brakes on a braking truck are visible evidence of the mechanical equivalent of heat.

Calculate the temperature increase of 10 kg of brake material with an average specific heat of

if

the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at

a constant speed.

Strategy
We calculate the gravitational potential energy (Mgh) that the entire truck loses in its descent, equate it to the increase in the brakes’ internal energy, and then find the temperature increase produced in the brake material alone.

Solution First we calculate the change in gravitational potential energy as the truck goes downhill:

Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energy is dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so take
. Then we calculate the temperature change from the heat transferred, using

where m is the mass of the brake material. Insert the given values to find

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1.4 • Heat Transfer, Specific Heat, and Calorimetry

23

Significance
If the truck had been traveling for some time, then just before the descent, the brake temperature would probably be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material very high, so this technique is not practical. Instead, the truck would use the technique of engine braking. A different idea underlies the recent technology of hybrid and electric cars, where mechanical energy (kinetic and gravitational potential energy) is converted by the brakes into electrical energy in the battery, a process called regenerative braking.

In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated from everything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out is called a calorimeter, and the use of a calorimeter to make measurements (typically of heat or specific heat capacity) is called calorimetry.
We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally isolated from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer between objects isolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due to conservation of energy:
1.6
We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered positive; the heat lost, negative.

EXAMPLE 1.7

Calculating the Final Temperature in Calorimetry

Suppose you pour 0.250 kg of

water (about a cup) into a 0.500-kg aluminum pan off the stove with a

temperature of

. Assume no heat transfer takes place to anything else: The pan is placed on an insulated

pad, and heat transfer to the air is neglected in the short time needed to reach equilibrium. Thus, this is a

calorimetry problem, even though no isolating container is specified. Also assume that a negligible amount of

water boils off. What is the temperature when the water and pan reach thermal equilibrium?

Strategy
Originally, the pan and water are not in thermal equilibrium: The pan is at a higher temperature than the water. Heat transfer restores thermal equilibrium once the water and pan are in contact; it stops once thermal equilibrium between the pan and the water is achieved. The heat lost by the pan is equal to the heat gained by the water—that is the basic principle of calorimetry.

Solution

1. Use the equation for heat transfer

to express the heat lost by the aluminum pan in terms of the

mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final

temperature:

2. Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water, and the final temperature:

3. Note that

and

and that as stated above, they must sum to zero:

4. Bring all terms involving on the left hand side and all other terms on the right hand side. Solving for

24

1 • Temperature and Heat

and insert the numerical values:

Significance

Why is the final temperature so much closer to

than to

? The reason is that water has a greater

specific heat than most common substances and thus undergoes a smaller temperature change for a given

heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature

appreciably. This explains why the temperature of a lake stays relatively constant during the day even when

the temperature change of the air is large. However, the water temperature does change over longer times (e.g.,

summer to winter).

CHECK YOUR UNDERSTANDING 1.3

If 25 kJ is necessary to raise the temperature of a rock from

the rock from

?

how much heat is necessary to heat

EXAMPLE 1.8

Temperature-Dependent Heat Capacity

At low temperatures, the specific heats of solids are typically proportional to . The first understanding of this behavior was due to the Dutch physicist Peter Debye, who in 1912, treated atomic oscillations with the quantum theory that Max Planck had recently used for radiation. For instance, a good approximation for the

specific heat of salt, NaCl, is

The constant 321 K is called the Debye

temperature of NaCl, and the formula works well when required to raise the temperature of 24.0 g of NaCl from 5 K to 15 K?

Using this formula, how much heat is

Solution Because the heat capacity depends on the temperature, we need to use the equation

We solve this equation for Q by integrating both sides: Then we substitute the given values in and evaluate the integral:

Significance If we had used the equation have gotten a very different value.

and the room-temperature specific heat of salt,

we would

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1.5 • Phase Changes

25

1.5 Phase Changes
Learning Objectives By the end of this section, you will be able to:
• Describe phase transitions and equilibrium between phases • Solve problems involving latent heat • Solve calorimetry problems involving phase changes

Phase transitions play an important theoretical and practical role in the study of heat flow. In melting (or “fusion”), a solid turns into a liquid; the opposite process is freezing. In evaporation, a liquid turns into a gas; the opposite process is condensation.

A substance melts or freezes at a temperature called its melting point, and boils (evaporates rapidly) or

condenses at its boiling point. These temperatures depend on pressure. High pressure favors the denser form,

so typically, high pressure raises the melting point and boiling point, and low pressure lowers them. For

example, the boiling point of water is

at 1.00 atm. At higher pressure, the boiling point is higher, and at

lower pressure, it is lower. The main exception is the melting and freezing of water, discussed in the next

section.

Phase Diagrams

The phase of a given substance depends on the pressure and temperature. Thus, plots of pressure versus temperature showing the phase in each region provide considerable insight into thermal properties of substances. Such a pT graph is called a phase diagram.

Figure 1.12 shows the phase diagram for water. Using the graph, if you know the pressure and temperature,

you can determine the phase of water. The solid curves—boundaries between phases—indicate phase

transitions, that is, temperatures and pressures at which the phases coexist. For example, the boiling point of

water is

at 1.00 atm. As the pressure increases, the boiling temperature rises gradually to

at a

pressure of 218 atm. A pressure cooker (or even a covered pot) cooks food faster than an open pot, because the

water can exist as a liquid at temperatures greater than

without all boiling away. (As we’ll see in the

next section, liquid water conducts heat better than steam or hot air.) The boiling point curve ends at a certain

point called the critical point—that is, a critical temperature, above which the liquid and gas phases cannot

be distinguished; the substance is called a supercritical fluid. At sufficiently high pressure above the critical

point, the gas has the density of a liquid but does not condense. Carbon dioxide, for example, is supercritical at

all temperatures above

. Critical pressure is the pressure of the critical point.

Figure 1.12 The phase diagram (pT graph) for water shows solid (s), liquid (l), and vapor (v) phases. At temperatures and pressure above

26

1 • Temperature and Heat

those of the critical point, there is no distinction between liquid and vapor. Note that the axes are nonlinear and the graph is not to scale. This graph is simplified—it omits several exotic phases of ice at higher pressures. The phase diagram of water is unusual because the melting-point curve has a negative slope, showing that you can melt ice by increasing the pressure.

Similarly, the curve between the solid and liquid regions in Figure 1.12 gives the melting temperature at

various pressures. For example, the melting point is

at 1.00 atm, as expected. Water has the unusual

property that ice is less dense than liquid water at the melting point, so at a fixed temperature, you can change

the phase from solid (ice) to liquid (water) by increasing the pressure. That is, the melting temperature of ice

falls with increased pressure, as the phase diagram shows. For example, when a car is driven over snow, the

increased pressure from the tires melts the snowflakes; afterwards, the water refreezes and forms an ice layer.

As you learned in the earlier section on thermometers and temperature scales, the triple point is the

combination of temperature and pressure at which ice, liquid water, and water vapor can coexist stably—that

is, all three phases exist in equilibrium. For water, the triple point occurs at

and 611.2 Pa;

that is a more accurate calibration temperature than the melting point of water at 1.00 atm, or

.

INTERACTIVE

View this video (https://openstax.org/l/21triplepoint) to see a substance at its triple point.

At pressures below that of the triple point, there is no liquid phase; the substance can exist as either gas or

solid. For water, there is no liquid phase at pressures below 0.00600 atm. The phase change from solid to gas is

called sublimation. You may have noticed that snow can disappear into thin air without a trace of liquid water,

or that ice cubes can disappear in a freezer. Both are examples of sublimation. The reverse also happens: Frost

can form on very cold windows without going through the liquid stage. Figure 1.13 shows the result, as well as

showing a familiar example of sublimation. Carbon dioxide has no liquid phase at atmospheric pressure. Solid

is known as dry ice because instead of melting, it sublimes. Its sublimation temperature at atmospheric

pressure is

. Certain air fresheners use the sublimation of a solid to spread a perfume around a room.

Some solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers to prevent

human exposure to their sublimation-produced vapors.

Figure 1.13 Direct transitions between solid and vapor are common, sometimes useful, and even beautiful. (a) Dry ice sublimes directly to carbon dioxide gas. The visible “smoke” consists of water droplets that condensed in the air cooled by the dry ice. (b) Frost forms patterns on a very cold window, an example of a solid formed directly from a vapor. (credit a: modification of work by Windell Oskay; credit b: modification of work by Liz West)
Equilibrium
At the melting temperature, the solid and liquid phases are in equilibrium. If heat is added, some of the solid will melt, and if heat is removed, some of the liquid will freeze. The situation is somewhat more complex for
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1.5 • Phase Changes

27

liquid-gas equilibrium. Generally, liquid and gas are in equilibrium at any temperature. We call the gas phase a

vapor when it exists at a temperature below the boiling temperature, as it does for water at

. Liquid in a

closed container at a fixed temperature evaporates until the pressure of the gas reaches a certain value, called

the vapor pressure, which depends on the gas and the temperature. At this equilibrium, if heat is added, some

of the liquid will evaporate, and if heat is removed, some of the gas will condense; molecules either join the

liquid or form suspended droplets. If there is not enough liquid for the gas to reach the vapor pressure in the

container, all the liquid eventually evaporates.

If the vapor pressure of the liquid is greater than the total ambient pressure, including that of any air (or other gas), the liquid evaporates rapidly; in other words, it boils. Thus, the boiling point of a liquid at a given pressure is the temperature at which its vapor pressure equals the ambient pressure. Liquid and gas phases are in equilibrium at the boiling temperature (Figure 1.14). If a substance is in a closed container at the boiling point, then the liquid is boiling and the gas is condensing at the same rate without net change in their amounts.

Figure 1.14 Equilibrium between liquid and gas at two different boiling points inside a closed container. (a) The rates of boiling and condensation are equal at this combination of temperature and pressure, so the liquid and gas phases are in equilibrium. (b) At a higher temperature, the boiling rate is faster, that is, the rate at which molecules leave the liquid and enter the gas is faster. This increases the number of molecules in the gas, which increases the gas pressure, which in turn increases the rate at which gas molecules condense and enter the liquid. The pressure stops increasing when it reaches the point where the boiling rate and the condensation rate are equal. The gas and liquid are in equilibrium again at this higher temperature and pressure.

For water,

is the boiling point at 1.00 atm, so water and steam should exist in equilibrium under these

conditions. Why does an open pot of water at

boil completely away? The gas surrounding an open pot is

not pure water: it is mixed with air. If pure water and steam are in a closed container at

and 1.00 atm,

they will coexist—but with air over the pot, there are fewer water molecules to condense, and water boils away.

Another way to see this is that at the boiling point, the vapor pressure equals the ambient pressure. However,

part of the ambient pressure is due to air, so the pressure of the steam is less than the vapor pressure at that

temperature, and evaporation continues. Incidentally, the equilibrium vapor pressure of solids is not zero, a

fact that accounts for sublimation.

CHECK YOUR UNDERSTANDING 1.4

Explain why a cup of water (or soda) with ice cubes stays at

even on a hot summer day.

Phase Change and Latent Heat
So far, we have discussed heat transfers that cause temperature change. However, in a phase transition, heat

28

1 • Temperature and Heat

transfer does not cause any temperature change.

For an example of phase changes, consider the addition of heat to a sample of ice at

(Figure 1.15) and

atmospheric pressure. The temperature of the ice rises linearly, absorbing heat at a constant rate of

until it reaches

Once at this temperature, the ice begins to melt and continues until it has

all melted, absorbing 333 kJ/kg of heat. The temperature remains constant at during this phase change.

Once all the ice has melted, the temperature of the liquid water rises, absorbing heat at a new constant rate of

At

the water begins to boil. The temperature again remains constant during this phase

change while the water absorbs 2256 kJ/kg of heat and turns into steam. When all the liquid has become

steam, the temperature rises again, absorbing heat at a rate of

. If we started with steam and

cooled it to make it condense into liquid water and freeze into ice, the process would exactly reverse, with the

temperature again constant during each phase transition.

Figure 1.15 Temperature versus heat. The system is constructed so that no vapor evaporates while ice warms to become liquid water, and so that, when vaporization occurs, the vapor remains in the system. The long stretches of constant temperatures at and reflect the large amounts of heat needed to cause melting and vaporization, respectively.
Where does the heat added during melting or boiling go, considering that the temperature does not change until the transition is complete? Energy is required to melt a solid, because the attractive forces between the molecules in the solid must be broken apart, so that in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Energy is needed to vaporize a liquid for similar reasons. Conversely, work is done by attractive forces when molecules are brought together during freezing and condensation. That energy must be transferred out of the system, usually in the form of heat, to allow the molecules to stay together (Figure 1.18). Thus, condensation occurs in association with cold objects—the glass in Figure 1.16, for example.
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1.5 • Phase Changes

29

Figure 1.16 Condensation forms on this glass of iced tea because the temperature of the nearby air is reduced. The air cannot hold as much water as it did at room temperature, so water condenses. Energy is released when the water condenses, speeding the melting of the ice in the glass. (credit: Jenny Downing)
The energy released when a liquid freezes is used by orange growers when the temperature approaches . Growers spray water on the trees so that the water freezes and heat is released to the growing oranges. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit (Figure 1.17).
Figure 1.17 The ice on these trees released large amounts of energy when it froze, helping to prevent the temperature of the trees from dropping below . Water is intentionally sprayed on orchards to help prevent hard frosts. (credit: Hermann Hammer)
The energy involved in a phase change depends on the number of bonds or force pairs and their strength. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The energy per unit mass required to change a substance from the solid phase to the liquid phase, or released when the substance changes from liquid to solid, is known as the heat of fusion. The energy per unit mass required to change a substance from the liquid phase to the vapor phase is known as the heat of vaporization. The strength of the forces depends on the type of molecules. The heat Q absorbed or released in a phase change in a sample of mass m is given by
1.7
1.8 where the latent heat of fusion and latent heat of vaporization are material constants that are determined experimentally. (Latent heats are also called latent heat coefficients and heats of transformation.)

30

1 • Temperature and Heat

These constants are “latent,” or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system, so in effect, the energy is hidden.

Figure 1.18 (a) Energy is required to partially overcome the attractive forces (modeled as springs) between molecules in a solid to form a liquid. That same energy must be removed from the liquid for freezing to take place. (b) Molecules become separated by large distances when going from liquid to vapor, requiring significant energy to completely overcome molecular attraction. The same energy must be removed from the vapor for condensation to take place.

Table 1.4 lists representative values of and in kJ/kg, together with melting and boiling points. Note that

in general,

. The table shows that the amounts of energy involved in phase changes can easily be

comparable to or greater than those involved in temperature changes, as Figure 1.15 and the accompanying

discussion also showed.

Substance Melting Point Helium[2]

Hydrogen

Nitrogen

Oxygen

Ethanol

–114

Ammonia –75

Mercury –38.9

Water

0.00

Sulfur

119

Lead

327

Antimony 631

kJ/kg kcal/kg Boiling Point

5.23 1.25

58.6 14.0

25.5 6.09

13.8 3.30

104 24.9

78.3

332 79.3

–33.4

11.8 2.82

357

334 79.8

100.0

38.1 9.10

444.6

24.5 5.85

1750

165 39.4

1440

kJ/kg kcal/kg

20.9

4.99

452

108

201

48.0

213

50.9

854

204

1370 327

272

65.0

2256[3] 539[4]

326

77.9

871

208

561

134

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1.5 • Phase Changes

31

Aluminum 660

380 90

2450

11400 2720

Silver

961

88.3 21.1

2193

2336 558

Gold

1063

64.5 15.4

2660

1578 377

Copper

1083

134 32.0

2595

5069 1211

Uranium 1133

84

20

3900

1900 454

Tungsten 3410

184 44

5900

4810 1150

Table 1.4 Heats of Fusion and Vaporization[1] [1]Values quoted at the normal melting and boiling temperatures

at standard atmospheric pressure (

). [2]Helium has no solid phase at atmospheric pressure. The melting

point given is at a pressure of 2.5 MPa. [3]At

(body temperature), the heat of vaporization for water is

2430 kJ/kg or 580 kcal/kg. [4]At

(body temperature), the heat of vaporization, for water is 2430 kJ/

kg or 580 kcal/kg.

Phase changes can have a strong stabilizing effect on temperatures that are not near the melting and boiling

points, since evaporation and condensation occur even at temperatures below the boiling point. For example,

air temperatures in humid climates rarely go above approximately

because most heat transfer goes

into evaporating water into the air. Similarly, temperatures in humid weather rarely fall below the dew

point—the temperature where condensation occurs given the concentration of water vapor in the air—because

so much heat is released when water vapor condenses.

More energy is required to evaporate water below the boiling point than at the boiling point, because the

kinetic energy of water molecules at temperatures below

is less than that at

, so less energy is

available from random thermal motions. For example, at body temperature, evaporation of sweat from the skin

requires a heat input of 2428 kJ/kg, which is about 10% higher than the latent heat of vaporization at

.

This heat comes from the skin, and this evaporative cooling effect of sweating helps reduce the body

temperature in hot weather. However, high humidity inhibits evaporation, so that body temperature might rise,

while unevaporated sweat might be left on your brow.

EXAMPLE 1.9

Calculating Final Temperature from Phase Change

Three ice cubes are used to chill a soda at

with mass

. The ice is at and each ice cube

has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored and that

the soda has the same specific heat as water. Find the final temperature when all ice has melted.

Strategy

The ice cubes are at the melting temperature of

Heat is transferred from the soda to the ice for melting.

Melting yields water at

so more heat is transferred from the soda to this water until the water plus soda

system reaches thermal equilibrium.

The heat transferred to the ice is

The heat given off by the soda is

32

1 • Temperature and Heat

Since no heat is lost,

as in Example 1.7, so that

Solve for the unknown quantity :

Solution First we identify the known quantities. The mass of ice is
Then we calculate the final temperature:

and the mass of soda is

Significance

This example illustrates the large energies involved during a phase change. The mass of ice is about 7% of the

mass of the soda but leads to a noticeable change in the temperature of the soda. Although we assumed that

the ice was at the freezing temperature, this is unrealistic for ice straight out of a freezer: The typical

temperature is

. However, this correction makes no significant change from the result we found. Can you

explain why?

Like solid-liquid and liquid-vapor transitions, direct solid-vapor transitions or sublimations involve heat. The

energy transferred is given by the equation

, where is the heat of sublimation, analogous to

and . The heat of sublimation at a given temperature is equal to the heat of fusion plus the heat of

vaporization at that temperature.

We can now calculate any number of effects related to temperature and phase change. In each case, it is necessary to identify which temperature and phase changes are taking place. Keep in mind that heat transfer and work can cause both temperature and phase changes.

PROBLEM-SOLVING STRATEGY

The Effects of Heat Transfer

1. Examine the situation to determine that there is a change in the temperature or phase. Is there heat

transfer into or out of the system? When it is not obvious whether a phase change occurs or not, you may

wish to first solve the problem as if there were no phase changes, and examine the temperature change

obtained. If it is sufficient to take you past a boiling or melting point, you should then go back and do the

problem in steps—temperature change, phase change, subsequent temperature change, and so on.

2. Identify and list all objects that change temperature or phase.

3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is

useful.

4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns). If

there is a temperature change, the transferred heat depends on the specific heat of the substance (Heat

Transfer, Specific Heat, and Calorimetry), and if there is a phase change, the transferred heat depends on

the latent heat of the substance (Table 1.4).

5. Solve the appropriate equation for the quantity to be determined (the unknown).

6. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions

complete with units. You may need to do this in steps if there is more than one state to the process, such as

a temperature change followed by a phase change. However, in a calorimetry problem, each step

corresponds to a term in the single equation

.

7. Check the answer to see if it is reasonable. Does it make sense? As an example, be certain that any

temperature change does not also cause a phase change that you have not taken into account.

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1.6 • Mechanisms of Heat Transfer

33

CHECK YOUR UNDERSTANDING 1.5
Why does snow often remain even when daytime temperatures are higher than the freezing temperature?
1.6 Mechanisms of Heat Transfer
Learning Objectives By the end of this section, you will be able to:
• Explain some phenomena that involve conductive, convective, and radiative heat transfer • Solve problems on the relationships between heat transfer, time, and rate of heat transfer • Solve problems using the formulas for conduction and radiation
Just as interesting as the effects of heat transfer on a system are the methods by which it occurs. Whenever there is a temperature difference, heat transfer occurs. It may occur rapidly, as through a cooking pan, or slowly, as through the walls of a picnic ice chest. So many processes involve heat transfer that it is hard to imagine a situation where no heat transfer occurs. Yet every heat transfer takes place by only three methods:
1. Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on a macroscopic scale—we know that thermal motion of the atoms and molecules occurs at any temperature above absolute zero.) Heat transferred from the burner of a stove through the bottom of a pan to food in the pan is transferred by conduction.
2. Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in a forced-air furnace and in weather systems, for example.
3. Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form of electromagnetic radiation is emitted or absorbed. An obvious example is the warming of Earth by the Sun. A less obvious example is thermal radiation from the human body.
In the illustration at the beginning of this chapter, the fire warms the snowshoers’ faces largely by radiation. Convection carries some heat to them, but most of the air flow from the fire is upward (creating the familiar shape of flames), carrying heat to the food being cooked and into the sky. The snowshoers wear clothes designed with low conductivity to prevent heat flow out of their bodies.
In this section, we examine these methods in some detail. Each method has unique and interesting characteristics, but all three have two things in common: They transfer heat solely because of a temperature difference, and the greater the temperature difference, the faster the heat transfer (Figure 1.19).

34

1 • Temperature and Heat

Figure 1.19 In a fireplace, heat transfer occurs by all three methods: conduction, convection, and radiation. Radiation is responsible for most of the heat transferred into the room. Heat transfer also occurs through conduction into the room, but much slower. Heat transfer by convection also occurs through cold air entering the room around windows and hot air leaving the room by rising up the chimney.
CHECK YOUR UNDERSTANDING 1.6
Name an example from daily life (different from the text) for each mechanism of heat transfer.
Conduction
As you walk barefoot across the living room carpet in a cold house and then step onto the kitchen tile floor, your feet feel colder on the tile. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation is explained by the different rates of heat transfer: The heat loss is faster for skin in contact with the tiles than with the carpet, so the sensation of cold is more intense. Some materials conduct thermal energy faster than others. Figure 1.20 shows a material that conducts heat slowly—it is a good thermal insulator, or poor heat conductor—used to reduce heat flow into and out of a house.
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1.6 • Mechanisms of Heat Transfer

35

Figure 1.20 Insulation is used to limit the conduction of heat from the inside to the outside (in winter) and from the outside to the inside (in summer). (credit: Giles Douglas)

A molecular picture of heat conduction will help justify the equation that describes it. Figure 1.21 shows

molecules in two bodies at different temperatures, and for “hot” and “cold.” The average kinetic energy

of a molecule in the hot body is higher than in the colder body. If two molecules collide, energy transfers from

the high-energy to the low-energy molecule. In a metal, the picture would also include free valence electrons

colliding with each other and with atoms, likewise transferring energy. The cumulative effect of all collisions is

a net flux of heat from the hotter body to the colder body. Thus, the rate of heat transfer increases with

increasing temperature difference

If the temperatures are the same, the net heat transfer rate

is zero. Because the number of collisions increases with increasing area, heat conduction is proportional to the

cross-sectional area—a second factor in the equation.

Figure 1.21 Molecules in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In this illustration, a molecule in the lowertemperature region (right side) has low energy before collision, but its energy increases after colliding with a high-energy molecule at the contact surface. In contrast, a molecule in the higher-temperature region (left side) has high energy before collision, but its energy

36

1 • Temperature and Heat

decreases after colliding with a low-energy molecule at the contact surface.
A third quantity that affects the conduction rate is the thickness of the material through which heat transfers. Figure 1.22 shows a slab of material with a higher temperature on the left than on the right. Heat transfers from the left to the right by a series of molecular collisions. The greater the distance between hot and cold, the more time the material takes to transfer the same amount of heat.

Figure 1.22 Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber.
All four of these quantities appear in a simple equation deduced from and confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in Figure 1.22, is given by

1.9

where P is the power or rate of heat transfer in watts or in kilocalories per second, A and d are its surface area

and thickness, as shown in Figure 1.22,

is the temperature difference across the slab, and k is the

thermal conductivity of the material. Table 1.5 gives representative values of thermal conductivity.

More generally, we can write

where x is the coordinate in the direction of heat flow. Since in Figure 1.22, the power and area are constant, dT/dx is constant, and the temperature decreases linearly from to

Substance

Thermal Conductivity k

Diamond

2000

Silver

420

Copper

390

Gold

318

Aluminum

220

Steel iron

80

Steel (stainless)

14

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1.6 • Mechanisms of Heat Transfer

37

Substance

Thermal Conductivity k

Ice

2.2

Glass (average)

0.84

Concrete brick

0.84

Water

0.6

Fatty tissue (without blood) 0.2

Asbestos

0.16

Plasterboard

0.16

Wood

0.08–0.16

Snow (dry)

0.10

Cork

0.042

Glass wool

0.042

Wool

0.04

Down feathers

0.025

Air

0.023

Polystyrene foam

0.010

Table 1.5 Thermal Conductivities of Common Substances Values are given for temperatures near .

EXAMPLE 1.10

Calculating Heat Transfer through Conduction

A polystyrene foam icebox has a total area of

and walls with an average thickness of 2.50 cm. The box

contains ice, water, and canned beverages at

The inside of the box is kept cold by melting ice. How much

ice melts in one day if the icebox is kept in the trunk of a car at

?

Strategy
This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time.

Solution First we identify the knowns.

for polystyrene foam;

;

;

38

1 • Temperature and Heat

Then we identify the unknowns. We need to solve for the mass of the ice, m. We also need to solve for the net heat transferred to melt the ice, Q. The rate of heat transfer by conduction is given by

The heat used to melt the ice is

.We insert the known values:

Multiplying the rate of heat transfer by the time we obtain We set this equal to the heat transferred to melt the ice,

and solve for the mass m:

Significance The result of 3.44 kg, or about 7.6 lb, seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.
Table 1.5 shows that polystyrene foam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goosedown feathers. Like polystyrene foam, these all contain many small pockets of air, taking advantage of air’s poor thermal conductivity.

In developing insulation, the smaller the conductivity k and the larger the thickness d, the better. Thus, the

ratio d/k, called the R factor, is large for a good insulator. The rate of conductive heat transfer is inversely

proportional to R. R factors are most commonly quoted for household insulation, refrigerators, and the like.

Unfortunately, in the United States, R is still in non-metric units of

, although the unit usually

goes unstated [1 British thermal unit (Btu) is the amount of energy needed to change the temperature of 1.0 lb

of water by

, which is 1055.1 J]. A couple of representative values are an R factor of 11 for 3.5-inch-thick

fiberglass batts (pieces) of insulation and an R factor of 19 for 6.5-inch-thick fiberglass batts (Figure 1.23). In

the US, walls are usually insulated with 3.5-inch batts, whereas ceilings are usually insulated with 6.5-inch

batts. In cold climates, thicker batts may be used.

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1.6 • Mechanisms of Heat Transfer

39

Figure 1.23 The fiberglass batt is used for insulation of walls and ceilings to prevent heat transfer between the inside of the building and the outside environment. (credit: Tracey Nicholls)
Note that in Table 1.5, most of the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, because they contain many free electrons that can transport thermal energy. (Diamond, an electrical insulator, conducts heat by atomic vibrations.) Cooking utensils are typically made from good conductors, but the handles of those used on the stove are made from good insulators (bad conductors).

EXAMPLE 1.11

Two Conductors End to End

A steel rod and an aluminum rod, each of diameter 1.00 cm and length 25.0 cm, are welded end to end. One

end of the steel rod is placed in a large tank of boiling water at

, while the far end of the aluminum rod is

placed in a large tank of water at

. The rods are insulated so that no heat escapes from their surfaces.

What is the temperature at the joint, and what is the rate of heat conduction through this composite rod?

Strategy
The heat that enters the steel rod from the boiling water has no place to go but through the steel rod, then through the aluminum rod, to the cold water. Therefore, we can equate the rate of conduction through the steel to the rate of conduction through the aluminum.

We repeat the calculation with a second method, in which we use the thermal resistance R of the rod, since it simply adds when two rods are joined end to end. (We will use a similar method in the chapter on directcurrent circuits.)

Solution

1. Identify the knowns and convert them to SI units.

The length of each rod is

the cross-sectional area of each rod is

the thermal conductivity of aluminum is

thermal conductivity of steel is

, the temperature at the hot end is

temperature at the cold end is

.

, the , and the

40

1 • Temperature and Heat

2. Calculate the heat-conduction rate through the steel rod and the heat-conduction rate through the aluminum rod in terms of the unknown temperature T at the joint:

3. Set the two rates equal and solve for the unknown temperature:

4. Calculate either rate:

5. If desired, check your answer by calculating the other rate.

Solution

1. Recall that

. Now

2. We know that

that rate of heat flow by P. Combine the equations:

. We also know that

and we denote

Thus, we can simply add R factors. Now,

.

3. Find the from the known quantities:

and

4. Substitute these values in to find

as before.

5. Determine for the aluminum rod (or for the steel rod) and use it to find T at the joint.

so

, as in Solution .

6. If desired, check by determining for the other rod.

Significance
In practice, adding R values is common, as in calculating the R value of an insulated wall. In the analogous situation in electronics, the resistance corresponds to AR in this problem and is additive even when the areas are unequal, as is common in electronics. Our equation for heat conduction can be used only when the areas are equal; otherwise, we would have a problem in three-dimensional heat flow, which is beyond our scope.

CHECK YOUR UNDERSTANDING 1.7
How does the rate of heat transfer by conduction change when all spatial dimensions are doubled?

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1.6 • Mechanisms of Heat Transfer

41

Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short times. For example, the temperature on Earth would be unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere were only through conduction. Also, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons. The next module discusses the important heat-transfer mechanism in such situations.
Convection
In convection, thermal energy is carried by the large-scale flow of matter. It can be divided into two types. In forced convection, the flow is driven by fans, pumps, and the like. A simple example is a fan that blows air past you in hot surroundings and cools you by replacing the air heated by your body with cooler air. A more complicated example is the cooling system of a typical car, in which a pump moves coolant through the radiator and engine to cool the engine and a fan blows air to cool the radiator.
In free or natural convection, the flow is driven by buoyant forces: hot fluid rises and cold fluid sinks because density decreases as temperature increases. The house in Figure 1.24 is kept warm by natural convection, as is the pot of water on the stove in Figure 1.25. Ocean currents and large-scale atmospheric circulation, which result from the buoyancy of warm air and water, transfer hot air from the tropics toward the poles and cold air from the poles toward the tropics. (Earth’s rotation interacts with those flows, causing the observed eastward flow of air in the temperate zones.)

Figure 1.24 Air heated by a so-called gravity furnace expands and rises, forming a convective loop that transfers energy to other parts of the room. As the air is cooled at the ceiling and outside walls, it contracts, eventually becoming denser than room air and sinking to the floor. A properly designed heating system using natural convection, like this one, can heat a home quite efficiently.
Figure 1.25 Natural convection plays an important role in heat transfer inside this pot of water. Once conducted to the inside, heat

42

1 • Temperature and Heat

transfer to other parts of the pot is mostly by convection. The hotter water expands, decreases in density, and rises to transfer heat to other regions of the water, while colder water sinks to the bottom. This process keeps repeating.
INTERACTIVE
Natural convection like that of Figure 1.24 and Figure 1.25, but acting on rock in Earth’s mantle, drives plate tectonics (https://openstax.org/l/21platetecton) that are the motions that have shaped Earth’s surface.
Convection is usually more complicated than conduction. Beyond noting that the convection rate is often approximately proportional to the temperature difference, we will not do any quantitative work comparable to the formula for conduction. However, we can describe convection qualitatively and relate convection rates to heat and time. Air is a poor conductor, so convection dominates heat transfer by air. Therefore, the amount of available space for airflow determines whether air transfers heat rapidly or slowly. There is little heat transfer in a space filled with air with a small amount of other material that prevents flow. The space between the inside and outside walls of a typical American house, for example, is about 9 cm (3.5 in.)—large enough for convection to work effectively. The addition of wall insulation prevents airflow, so heat loss (or gain) is decreased. On the other hand, the gap between the two panes of a double-paned window is about 1 cm, which largely prevents convection and takes advantage of air’s low conductivity reduce heat loss. Fur, cloth, and fiberglass also take advantage of the low conductivity of air by trapping it in spaces too small to support convection (Figure 1.26).

Figure 1.26 Fur is filled with air, breaking it up into many small pockets. Convection is very slow here, because the loops are so small. The low conductivity of air makes fur a very good lightweight insulator.
Some interesting phenomena happen when convection is accompanied by a phase change. The combination allows us to cool off by sweating even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is required for sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow caused by convection replaces the saturated air by dry air and evaporation continues.
EXAMPLE 1.12 Calculating the Flow of Mass during Convection
The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporate from the body to get rid of all this energy? (For simplicity, we assume this evaporation occurs when a person is sitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer by other methods.)
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Strategy Energy is needed for this phase change (

1.6 • Mechanisms of Heat Transfer

43

). Thus, the energy loss per unit time is

We divide both sides of the equation by to find that the mass evaporated per unit time is

Solution Insert the value of the latent heat from Table 1.4,

. This yields

Significance Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz.) per hour. If the air is very dry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place in the lungs and breathing passages.
Another important example of the combination of phase change and convection occurs when water evaporates from the oceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as clouds form, possibly far from the ocean, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the ocean to the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as much as 20.0 km into the stratosphere (Figure 1.27). Water vapor carried in by convection condenses, releasing tremendous amounts of energy. This energy causes the air to expand and rise to colder altitudes. More condensation occurs in these regions, which in turn drives the cloud even higher. This mechanism is an example of positive feedback, since the process reinforces and accelerates itself. It sometimes produces violent storms, with lightning and hail. The same mechanism drives hurricanes.
INTERACTIVE
This time-lapse video (https://openstax.org/l/21convthuncurr) shows convection currents in a thunderstorm, including “rolling” motion similar to that of boiling water.

Figure 1.27 Cumulus clouds are caused by water vapor that rises because of convection. The rise of clouds is driven by a positive feedback mechanism. (credit: “Amada44”/Wikimedia Commons)
CHECK YOUR UNDERSTANDING 1.8

44

1 • Temperature and Heat

Explain why using a fan in the summer feels refreshing.
Radiation
You can feel the heat transfer from the Sun. The space between Earth and the Sun is largely empty, so the Sun warms us without any possibility of heat transfer by convection or conduction. Similarly, you can sometimes tell that the oven is hot without touching its door or looking inside—it may just warm you as you walk by. In these examples, heat is transferred by radiation (Figure 1.28). That is, the hot body emits electromagnetic waves that are absorbed by the skin. No medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves of different wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Figure 1.28 Most of the heat transfer from this fire to the observers occurs through infrared radiation. The visible light, although dramatic,
transfers relatively little thermal energy. Convection transfers energy away from the observers as hot air rises, while conduction is negligibly
slow here. Skin is very sensitive to infrared radiation, so you can sense the presence of a fire without looking at it directly. (credit: Daniel
O’Neil)
The energy of electromagnetic radiation varies over a wide range, depending on the wavelength: A shorter wavelength (or higher frequency) corresponds to a higher energy. Because more heat is radiated at higher temperatures, higher temperatures produce more intensity at every wavelength but especially at shorter wavelengths. In visible light, wavelength determines color—red has the longest wavelength and violet the shortest—so a temperature change is accompanied by a color change. For example, an electric heating element on a stove glows from red to orange, while the higher-temperature steel in a blast furnace glows from yellow to white. Infrared radiation is the predominant form radiated by objects cooler than the electric element and the steel. The radiated energy as a function of wavelength depends on its intensity, which is represented in Figure 1.29 by the height of the distribution. (Electromagnetic Waves explains more about the electromagnetic spectrum, and Photons and Matter Waves discusses why the decrease in wavelength corresponds to an increase in energy.)

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1.6 • Mechanisms of Heat Transfer

45

Figure 1.29 (a) A graph of the spectrum of electromagnetic waves emitted from an ideal radiator at three different temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the spectrum shifts down in wavelength toward the visible and ultraviolet parts of the spectrum. The shaded portion denotes the visible part of the spectrum. It is apparent that the shift toward the ultraviolet with temperature makes the visible appearance shift from red to white to blue as temperature increases. (b) Note the variations in color corresponding to variations in flame temperature.

The rate of heat transfer by radiation also depends on the object’s color. Black is the most effective, and white

is the least effective. On a clear summer day, black asphalt in a parking lot is hotter than adjacent gray

sidewalk, because black absorbs better than gray (Figure 1.30). The reverse is also true—black radiates better

than gray. Thus, on a clear summer night, the asphalt is colder than the gray sidewalk, because black radiates

the energy more rapidly than gray. A perfectly black object would be an ideal radiator and an ideal absorber, as

it would capture all the radiation that falls on it. In contrast, a perfectly white object or a perfect mirror would

reflect all radiation, and a perfectly transparent object would transmit it all (Figure 1.31). Such objects would

not emit any radiation. Mathematically, the color is represented by the emissivity e. A “blackbody” radiator

would have an

, whereas a perfect reflector or transmitter would have

. For real examples, tungsten

light bulb filaments have an e of about 0.5, and carbon black (a material used in printer toner) has an

emissivity of about 0.95.

46

1 • Temperature and Heat

Figure 1.30 The darker pavement is hotter than the lighter pavement (much more of the ice on the right has melted), although both have been in the sunlight for the same time. The thermal conductivities of the pavements are the same.

Figure 1.31 A black object is a good absorber and a good radiator, whereas a white, clear, or silver object is a poor absorber and a poor radiator.
To see that, consider a silver object and a black object that can exchange heat by radiation and are in thermal equilibrium. We know from experience that they will stay in equilibrium (the result of a principle that will be discussed at length in Second Law of Thermodynamics). For the black object’s temperature to stay constant, it must emit as much radiation as it absorbs, so it must be as good at radiating as absorbing. Similar considerations show that the silver object must radiate as little as it absorbs. Thus, one property, emissivity, controls both radiation and absorption.
Finally, the radiated heat is proportional to the object’s surface area, since every part of the surface radiates. If you knock apart the coals of a fire, the radiation increases noticeably due to an increase in radiating surface area.
The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation:

where

is the Stefan-Boltzmann constant, a combination of fundamental

constants of nature; A is the surface area of the object; and T is its temperature in kelvins.

The proportionality to the fourth power of the absolute temperature is a remarkably strong temperature dependence. It allows the detection of even small temperature variations. Images called thermographs can be used medically to detect regions of abnormally high temperature in the body, perhaps indicative of disease. Similar techniques can be used to detect heat leaks in homes (Figure 1.32), optimize performance of blast furnaces, improve comfort levels in work environments, and even remotely map Earth’s temperature profile.

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1.6 • Mechanisms of Heat Transfer

47

Figure 1.32 A thermograph of part of a building shows temperature variations, indicating where heat transfer to the outside is most severe. Windows are a major region of heat transfer to the outside of homes. (credit: US Army)
The Stefan-Boltzmann equation needs only slight refinement to deal with a simple case of an object’s absorption of radiation from its surroundings. Assuming that an object with a temperature is surrounded by an environment with uniform temperature , the net rate of heat transfer by radiation is

1.10

where e is the emissivity of the object alone. In other words, it does not matter whether the surroundings are

white, gray, or black: The balance of radiation into and out of the object depends on how well it emits and

absorbs radiation. When

the quantity is positive, that is, the net heat transfer is from hot to cold.

Before doing an example, we have a complication to discuss: different emissivities at different wavelengths. If the fraction of incident radiation an object reflects is the same at all visible wavelengths, the object is gray; if the fraction depends on the wavelength, the object has some other color. For instance, a red or reddish object reflects red light more strongly than other visible wavelengths. Because it absorbs less red, it radiates less red when hot. Differential reflection and absorption of wavelengths outside the visible range have no effect on what we see, but they may have physically important effects. Skin is a very good absorber and emitter of infrared radiation, having an emissivity of 0.97 in the infrared spectrum. Thus, in spite of the obvious variations in skin color, we are all nearly black in the infrared. This high infrared emissivity is why we can so easily feel radiation on our skin. It is also the basis for the effectiveness of night-vision scopes used by law enforcement and the military to detect human beings.

EXAMPLE 1.13

Calculating the Net Heat Transfer of a Person

What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient

temperature is

? The person has a normal skin temperature of

and a surface area of

The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.

Strategy We can solve this by using the equation for the rate of radiative heat transfer.

Solution

Insert the temperature values

and

, so that

48

1 • Temperature and Heat

Significance
This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a person at rest may produce energy at the rate of 125 W and that conduction and convection are also transferring energy to the environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heat transfer to the environment by all mechanisms, because clothing slows down both conduction and convection, and has a lower emissivity (especially if it is light-colored) than skin.

The average temperature of Earth is the subject of much current discussion. Earth is in radiative contact with both the Sun and dark space, so we cannot use the equation for an environment at a uniform temperature. Earth receives almost all its energy from radiation of the Sun and reflects some of it back into outer space. Conversely, dark space is very cold, about 3 K, so that Earth radiates energy into the dark sky. The rate of heat transfer from soil and grasses can be so rapid that frost may occur on clear summer evenings, even in warm latitudes.

The average temperature of Earth is determined by its energy balance. To a first approximation, it is the temperature at which Earth radiates heat to space as fast as it receives energy from the Sun.

An important parameter in calculating the temperature of Earth is its emissivity (e). On average, it is about 0.65, but calculation of this value is complicated by the great day-to-day variation in the highly reflective cloud coverage. Because clouds have lower emissivity than either oceans or land masses, they emit some of the radiation back to the surface, greatly reducing heat transfer into dark space, just as they greatly reduce heat transfer into the atmosphere during the day. There is negative feedback (in which a change produces an effect that opposes that change) between clouds and heat transfer; higher temperatures evaporate more water to form more clouds, which reflect more radiation back into space, reducing the temperature.

The often-mentioned greenhouse effect is directly related to the variation of Earth’s emissivity with

wavelength (Figure 1.33). The greenhouse effect is a natural phenomenon responsible for providing

temperatures suitable for life on Earth and for making Venus unsuitable for human life. Most of the infrared

radiation emitted from Earth is absorbed by carbon dioxide ( ) and water ( ) in the atmosphere and

then re-radiated into outer space or back to Earth. Re-radiation back to Earth maintains its surface

temperature about

higher than it would be if there were no atmosphere. (The glass walls and roof of a

greenhouse increase the temperature inside by blocking convective heat losses, not radiative losses.)

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1.6 • Mechanisms of Heat Transfer

49

Figure 1.33 The greenhouse effect is the name given to the increase of Earth’s temperature due to absorption of radiation in the atmosphere. The atmosphere is transparent to incoming visible radiation and most of the Sun’s infrared. The Earth absorbs that energy and re-emits it. Since Earth’s temperature is much lower than the Sun’s, it re-emits the energy at much longer wavelengths, in the infrared. The atmosphere absorbs much of that infrared radiation and radiates about half of the energy back down, keeping Earth warmer than it would otherwise be. The amount of trapping depends on concentrations of trace gases such as carbon dioxide, and an increase in the concentration of these gases increases Earth’s surface temperature.
The greenhouse effect is central to the discussion of global warming due to emission of carbon dioxide and methane (and other greenhouse gases) into Earth’s atmosphere from industry, transportation, and farming. Changes in global climate could lead to more intense storms, precipitation changes (affecting agriculture), reduction in rain forest biodiversity, and rising sea levels.
INTERACTIVE
You can explore a simulation of the greenhouse effect (https://openstax.org/l/21simgreeneff) that takes the point of view that the atmosphere scatters (redirects) infrared radiation rather than absorbing it and reradiating it. You may want to run the simulation first with no greenhouse gases in the atmosphere and then look at how adding greenhouse gases affects the infrared radiation from the Earth and the Earth’s temperature.

PROBLEM-SOLVING STRATEGY

Effects of Heat Transfer
1. Examine the situation to determine what type of heat transfer is involved.

2. Identify the type(s) of heat transfer—conduction, convection, or radiation.

3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is

useful.

4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).

5. Solve the appropriate equation for the quantity to be determined (the unknown).

6. For conduction, use the equation

. Table 1.5 lists thermal conductivities. For convection,

determine the amount of matter moved and the equation

, along with

or

if

50

1 • Temperature and Heat

a substance changes phase. For radiation, the equation

gives the net heat transfer

rate.

7. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions

complete with units.

8. Check the answer to see if it is reasonable. Does it make sense?

CHECK YOUR UNDERSTANDING 1.9

How much greater is the rate of heat radiation when a body is at the temperature

temperature

?

than when it is at the

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1 • Chapter Review 51

CHAPTER REVIEW

Key Terms

absolute temperature scale scale, such as Kelvin,

with a zero point that is absolute zero

absolute zero temperature at which the average

kinetic energy of molecules is zero

calorie (cal) energy needed to change the

temperature of 1.00 g of water by

calorimeter container that prevents heat transfer

in or out

calorimetry study of heat transfer inside a

container impervious to heat

Celsius scale temperature scale in which the

freezing point of water is and the boiling

point of water is

coefficient of linear expansion ( ) material

property that gives the change in length, per unit

length, per

change in temperature; a

constant used in the calculation of linear

expansion; the coefficient of linear expansion

depends to some degree on the temperature of

the material

coefficient of volume expansion ( ) similar to

but gives the change in volume, per unit volume,

per

change in temperature

conduction heat transfer through stationary

matter by physical contact

convection heat transfer by the macroscopic

movement of fluid

critical point for a given substance, the

combination of temperature and pressure above

which the liquid and gas phases are

indistinguishable

critical pressure pressure at the critical point

critical temperature temperature at the critical

point

degree Celsius ( ) unit on the Celsius

temperature scale

degree Fahrenheit ( ) unit on the Fahrenheit

temperature scale

emissivity measure of how well an object radiates

Fahrenheit scale temperature scale in which the

freezing point of water is

and the boiling

point of water is

greenhouse effect warming of the earth that is due

to gases such as carbon dioxide and methane that

absorb infrared radiation from Earth’s surface

and reradiate it in all directions, thus sending

some of it back toward Earth

heat energy transferred solely due to a

temperature difference

heat of fusion energy per unit mass required to

change a substance from the solid phase to the liquid phase, or released when the substance changes from liquid to solid heat of sublimation energy per unit mass required to change a substance from the solid phase to the vapor phase heat of vaporization energy per unit mass required to change a substance from the liquid phase to the vapor phase heat transfer movement of energy from one place or material to another as a result of a difference in temperature Kelvin scale (K) temperature scale in which 0 K is the lowest possible temperature, representing absolute zero kilocalorie (kcal) energy needed to change the temperature of 1.00 kg of water between and latent heat coefficient general term for the heats of fusion, vaporization, and sublimation mechanical equivalent of heat work needed to produce the same effects as heat transfer net rate of heat transfer by radiation

phase diagram graph of pressure vs. temperature

of a particular substance, showing at which

pressures and temperatures the phases of the

substance occur

radiation energy transferred by electromagnetic

waves directly as a result of a temperature

difference

rate of conductive heat transfer rate of heat

transfer from one material to another

specific heat amount of heat necessary to change

the temperature of 1.00 kg of a substance by

; also called “specific heat capacity”

Stefan-Boltzmann law of radiation

where

is the

Stefan-Boltzmann constant, A is the surface area

of the object, T is the absolute temperature, and e

is the emissivity

sublimation phase change from solid to gas

temperature quantity measured by a

thermometer, which reflects the mechanical

energy of molecules in a system

thermal conductivity property of a material

describing its ability to conduct heat

thermal equilibrium condition in which heat no

longer flows between two objects that are in

contact; the two objects have the same

52 1 • Chapter Review

temperature thermal expansion change in size or volume of an
object with change in temperature thermal stress stress caused by thermal
expansion or contraction triple point pressure and temperature at which a
substance exists in equilibrium as a solid, liquid, and gas vapor gas at a temperature below the boiling
Key Equations

temperature vapor pressure pressure at which a gas coexists
with its solid or liquid phase zeroth law of thermodynamics law that states that
if two objects are in thermal equilibrium, and a third object is in thermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object

Linear thermal expansion

Thermal expansion in two dimensions

Thermal expansion in three dimensions

Heat transfer

Transfer of heat in a calorimeter

Heat due to phase change (melting and freezing)

Heat due to phase change (evaporation and condensation)

Rate of conductive heat transfer

Net rate of heat transfer by radiation

Summary
1.1 Temperature and Thermal Equilibrium
• Temperature is operationally defined as the quantity measured by a thermometer. It is proportional to the average kinetic energy of atoms and molecules in a system.
• Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy. Systems are in thermal equilibrium when they have the same temperature.
• The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with each other, and B is in thermal equilibrium with a third system C, then A is also in thermal equilibrium with C.
1.2 Thermometers and Temperature Scales
• Three types of thermometers are alcohol, liquid crystal, and infrared radiation (pyrometer).

• The three main temperature scales are Celsius, Fahrenheit, and Kelvin. Temperatures can be converted from one scale to another using temperature conversion equations.
• The three phases of water (ice, liquid water, and water vapor) can coexist at a single pressure and temperature known as the triple point.
1.3 Thermal Expansion
• Thermal expansion is the increase of the size (length, area, or volume) of a body due to a change in temperature, usually a rise. Thermal contraction is the decrease in size due to a change in temperature, usually a fall in temperature.
• Thermal stress is created when thermal expansion or contraction is constrained.

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1 • Chapter Review 53

1.4 Heat Transfer, Specific Heat, and Calorimetry
• Heat and work are the two distinct methods of energy transfer.
• Heat transfer to an object when its temperature changes is often approximated well by where m is the object’s mass and c is the specific heat of the substance.
1.5 Phase Changes
• Most substances have three distinct phases (under ordinary conditions on Earth), and they depend on temperature and pressure.
• Two phases coexist (i.e., they are in thermal equilibrium) at a set of pressures and temperatures.
• Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling, freezing (or melting), and sublimation points.
1.6 Mechanisms of Heat Transfer
• Heat is transferred by three different methods: conduction, convection, and radiation.
• Heat conduction is the transfer of heat between two objects in direct contact with each other.
• The rate of heat transfer P (energy per unit time) is proportional to the temperature difference and the contact area A and inversely
Conceptual Questions
1.1 Temperature and Thermal Equilibrium
1. What does it mean to say that two systems are in thermal equilibrium?
2. Give an example in which A has some kind of non-thermal equilibrium relationship with B, and B has the same relationship with C, but A does not have that relationship with C.
1.2 Thermometers and Temperature Scales
3. If a thermometer is allowed to come to equilibrium with the air, and a glass of water is not in equilibrium with the air, what will happen to the thermometer reading when it is placed in the water?
4. Give an example of a physical property that varies with temperature and describe how it is used to measure temperature.

proportional to the distance d between the

objects.

• Convection is heat transfer by the macroscopic

movement of mass. Convection can be natural

or forced, and generally transfers thermal

energy faster than conduction. Convection that

occurs along with a phase change can transfer

energy from cold regions to warm ones.

• Radiation is heat transfer through the emission

or absorption of electromagnetic waves.

• The rate of radiative heat transfer is

proportional to the emissivity e. For a perfect

blackbody,

, whereas a perfectly white,

clear, or reflective body has

, with real

objects having values of e between 1 and 0.

• The rate of heat transfer depends on the surface

area and the fourth power of the absolute

temperature:

where

is the

Stefan-Boltzmann constant and e is the

emissivity of the body. The net rate of heat

transfer from an object by radiation is

where is the temperature of the object surrounded by an environment with uniform temperature and e is the emissivity of the object.

1.3 Thermal Expansion
5. Pouring cold water into hot glass or ceramic cookware can easily break it. What causes the breaking? Explain why Pyrex®, a glass with a small coefficient of linear expansion, is less susceptible.
6. One method of getting a tight fit, say of a metal peg in a hole in a metal block, is to manufacture the peg slightly larger than the hole. The peg is then inserted when at a different temperature than the block. Should the block be hotter or colder than the peg during insertion? Explain your answer.
7. Does it really help to run hot water over a tight metal lid on a glass jar before trying to open it? Explain your answer.
8. When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why.
9. Calculate the length of a 1-meter rod of a material

54 1 • Chapter Review

with thermal expansion coefficient when the temperature is raised from 300 K to 600 K. Taking your answer as the new initial length, find the length after the rod is cooled back down to 300 K. Is your answer 1 meter? Should it be? How can you account for the result you got? 10. Noting the large stresses that can be caused by
thermal expansion, an amateur weapon inventor decides to use it to make a new kind of gun. He plans to jam a bullet against an aluminum rod inside a closed invar tube. When he heats the tube, the rod will expand more than the tube and a very strong force will build up. Then, by a method yet to be determined, he will open the tube in a split second and let the force of the rod launch the bullet at very high speed. What is he overlooking?

1.4 Heat Transfer, Specific Heat, and Calorimetry

17. Can carbon dioxide be liquefied at room

temperature (

)? If so, how? If not, why not?

11. How is heat transfer related to temperature?

(See the phase diagram in the preceding

12. Describe a situation in which heat transfer

problem.)

occurs.

18. What is the distinction between gas and vapor?

13. When heat transfers into a system, is the energy

19. Heat transfer can cause temperature and phase

stored as heat? Explain briefly.

changes. What else can cause these changes?

14. The brakes in a car increase in temperature by

20. How does the latent heat of fusion of water help

when bringing the car to rest from a speed

slow the decrease of air temperatures, perhaps

v. How much greater would be if the car

preventing temperatures from falling

initially had twice the speed? You may assume

significantly below

in the vicinity of large

the car stops fast enough that no heat transfers

bodies of water?

out of the brakes.

21. What is the temperature of ice right after it is

formed by freezing water?

1.5 Phase Changes

22. If you place ice into water in an

insulated container, what will the net result be?

15. A pressure cooker contains water and steam in

Will there be less ice and more liquid water, or

equilibrium at a pressure greater than

more ice and less liquid water, or will the

atmospheric pressure. How does this greater

amounts stay the same?

pressure increase cooking speed?

23. What effect does condensation on a glass of ice

16. As shown below, which is the phase diagram for

water have on the rate at which the ice melts?

carbon dioxide, what is the vapor pressure of solid

Will the condensation speed up the melting

carbon dioxide (dry ice) at

(Note that the

process or slow it down?

axes in the figure are nonlinear and the graph is not 24. In Miami, Florida, which has a very humid

to scale.)

climate and numerous bodies of water nearby, it

is unusual for temperatures to rise above about

(

). In the desert climate of Phoenix,

Arizona, however, temperatures rise above that

almost every day in July and August. Explain

how the evaporation of water helps limit high

temperatures in humid climates.

25. In winter, it is often warmer in San Francisco

than in Sacramento, 150 km inland. In summer,

it is nearly always hotter in Sacramento. Explain

how the bodies of water surrounding San

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1 • Chapter Review 55

Francisco moderate its extreme temperatures. 26. Freeze-dried foods have been dehydrated in a
vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result. 27. In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublime, with some crystals lingering for a while and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from Table 1.4.

1.6 Mechanisms of Heat Transfer

28. What are the main methods of heat transfer

from the hot core of Earth to its surface? From

Earth’s surface to outer space?

29. When our bodies get too warm, they respond by

sweating and increasing blood circulation to the

surface to transfer thermal energy away from

the core. What effect will those processes have

on a person in a

hot tub?

30. Shown below is a cut-away drawing of a

thermos bottle (also known as a Dewar flask),

which is a device designed specifically to slow

down all forms of heat transfer. Explain the

functions of the various parts, such as the

vacuum, the silvering of the walls, the thin-

walled long glass neck, the rubber support, the

air layer, and the stopper.

31. Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while the surface only a few centimeters away is safe to touch. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?
32. Loose-fitting white clothing covering most of the body, shown below, is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and night.

56 1 • Chapter Review

33. One way to make a fireplace more energyefficient is to have room air circulate around the outside of the fire box and back into the room. Detail the methods of heat transfer involved.
34. On cold, clear nights horses will sleep under the cover of large trees. How does this help them keep warm?
35. When watching a circus during the day in a large, dark-colored tent, you sense significant heat transfer from the tent. Explain why this occurs.
36. Satellites designed to observe the radiation from cold (3 K) dark space have sensors that are shaded from the Sun, Earth, and the Moon and are cooled to very low temperatures. Why must the sensors be at low temperature?
37. Why are thermometers that are used in weather stations shielded from the sunshine? What does a thermometer measure if it is shielded from the sunshine? What does it measure if it is not?
38. Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why.
39. Your house will be empty for a while in cold weather, and you want to save energy and

Problems

1.2 Thermometers and Temperature Scales

43. While traveling outside the United States, you

feel sick. A companion gets you a thermometer,

which says your temperature is 39. What scale

is that on? What is your Fahrenheit

temperature? Should you seek medical help?

44. What are the following temperatures on the

Kelvin scale?

(a)

an indoor temperature sometimes

recommended for energy conservation in

winter

(b)

one of the highest atmospheric

temperatures ever recorded on Earth (Death

Valley, California, 1913)

(c)

the temperature of the surface of

the Sun

45. (a) Suppose a cold front blows into your locale

and drops the temperature by 40.0 Fahrenheit

degrees. How many degrees Celsius does the

temperature decrease when it decreases by

? (b) Show that any change in

temperature in Fahrenheit degrees is nine-

fifths the change in Celsius degrees

46. An Associated Press article on climate change

said, “Some of the ice shelf’s disappearance was

money. Should you turn the thermostat down to the lowest level that will protect the house from damage such as freezing pipes, or leave it at the normal temperature? (If you don’t like coming back to a cold house, imagine that a timer controls the heating system so the house will be warm when you get back.) Explain your answer. 40. You pour coffee into an unlidded cup, intending to drink it 5 minutes later. You can add cream when you pour the cup or right before you drink it. (The cream is at the same temperature either way. Assume that the cream and coffee come into thermal equilibrium with each other very quickly.) Which way will give you hotter coffee? What feature of this question is different from the previous one? 41. Broiling is a method of cooking by radiation, which produces somewhat different results from cooking by conduction or convection. A gas flame or electric heating element produces a very high temperature close to the food and above it. Why is radiation the dominant heattransfer method in this situation? 42. On a cold winter morning, why does the metal of a bike feel colder than the wood of a porch?

probably during times when the planet was 36

degrees Fahrenheit (2 degrees Celsius) to 37

degrees Fahrenheit (3 degrees Celsius) warmer

than it is today.” What mistake did the reporter

make?

47. (a) At what temperature do the Fahrenheit and

Celsius scales have the same numerical value?

(b) At what temperature do the Fahrenheit and

Kelvin scales have the same numerical value?

48. A person taking a reading of the temperature in

a freezer in Celsius makes two mistakes: first

omitting the negative sign and then thinking the

temperature is Fahrenheit. That is, the person

reads

as . Oddly enough, the result is

the correct Fahrenheit temperature. What is the

original Celsius reading? Round your answer to

three significant figures.

1.3 Thermal Expansion

49. The height of the Washington Monument is

measured to be 170.00 m on a day when the

temperature is

What will its height be

on a day when the temperature falls to

? Although the monument is made of

limestone, assume that its coefficient of thermal

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1 • Chapter Review 57

expansion is the same as that of marble. Give

your answer to five significant figures.

50. How much taller does the Eiffel Tower become

at the end of a day when the temperature has

increased by

Its original height is 321 m

and you can assume it is made of steel.

51. What is the change in length of a 3.00-cm-long

column of mercury if its temperature changes

from

to

, assuming the mercury

is constrained to a cylinder but unconstrained

in length? Your answer will show why

thermometers contain bulbs at the bottom

instead of simple columns of liquid.

52. How large an expansion gap should be left

between steel railroad rails if they may reach a

maximum temperature

greater than

when they were laid? Their original length is

10.0 m.

53. You are looking to buy a small piece of land in

Hong Kong. The price is “only” $60,000 per

square meter. The land title says the

dimensions are

. By how much

would the total price change if you measured

the parcel with a steel tape measure on a day

when the temperature was

above the

temperature that the tape measure was

designed for? The dimensions of the land do not

change.

54. Global warming will produce rising sea levels

partly due to melting ice caps and partly due to

the expansion of water as average ocean

temperatures rise. To get some idea of the size

of this effect, calculate the change in length of a

column of water 1.00 km high for a temperature

increase of

. Assume the column is not

free to expand sideways. As a model of the

ocean, that is a reasonable approximation, as

only parts of the ocean very close to the surface

can expand sideways onto land, and only to a

limited degree. As another approximation,

neglect the fact that ocean warming is not

uniform with depth.

55. (a) Suppose a meter stick made of steel and one

made of aluminum are the same length at .

What is their difference in length at

? (b)

Repeat the calculation for two 30.0-m-long

surveyor’s tapes.

56. (a) If a 500-mL glass beaker is filled to the brim

with ethyl alcohol at a temperature of

,

how much will overflow when the alcohol’s

temperature reaches the room temperature of

? (b) How much less water would

overflow under the same conditions?

57. Most cars have a coolant reservoir to catch

radiator fluid that may overflow when the

engine is hot. A radiator is made of copper and

is filled to its 16.0-L capacity when at

.

What volume of radiator fluid will overflow

when the radiator and fluid reach a temperature

of

given that the fluid’s volume

coefficient of expansion is

?

(Your answer will be a conservative estimate, as

most car radiators have operating temperatures

greater than

).

58. A physicist makes a cup of instant coffee and

notices that, as the coffee cools, its level drops

3.00 mm in the glass cup. Show that this

decrease cannot be due to thermal contraction

by calculating the decrease in level if the

of coffee is in a 7.00-cm-diameter cup

and decreases in temperature from

to

. (Most of the drop in level is actually due

to escaping bubbles of air.)

59. The density of water at is very nearly

(it is actually

),

whereas the density of ice at is

Calculate the pressure necessary to keep ice

from expanding when it freezes, neglecting the

effect such a large pressure would have on the

freezing temperature. (This problem gives you

only an indication of how large the forces

associated with freezing water might be.)

60. Show that

by calculating the

infinitesimal change in volume dV of a cube

with sides of length L when the temperature

changes by dT.

1.4 Heat Transfer, Specific Heat, and Calorimetry

61. On a hot day, the temperature of an 80,000-L

swimming pool increases by

. What is

the net heat transfer during this heating? Ignore

any complications, such as loss of water by

evaporation.

62. To sterilize a 50.0-g glass baby bottle, we must

raise its temperature from

to

.

How much heat transfer is required?

63. The same heat transfer into identical masses of

different substances produces different

temperature changes. Calculate the final

temperature when 1.00 kcal of heat transfers

into 1.00 kg of the following, originally at

: (a) water; (b) concrete; (c) steel; and (d)

mercury.

64. Rubbing your hands together warms them by

58 1 • Chapter Review

converting work into thermal energy. If a

woman rubs her hands back and forth for a total

of 20 rubs, at a distance of 7.50 cm per rub, and

with an average frictional force of 40.0 N, what

is the temperature increase? The mass of

tissues warmed is only 0.100 kg, mostly in the

palms and fingers.

65. A

block of a pure material is heated

from

to

by the addition of 4.35

kJ of energy. Calculate its specific heat and

identify the substance of which it is most likely

composed.

66. Suppose identical amounts of heat transfer into

different masses of copper and water, causing

identical changes in temperature. What is the

ratio of the mass of copper to water?

67. (a) The number of kilocalories in food is

determined by calorimetry techniques in which

the food is burned and the amount of heat

transfer is measured. How many kilocalories

per gram are there in a 5.00-g peanut if the

energy from burning it is transferred to 0.500

kg of water held in a 0.100-kg aluminum cup,

causing a

temperature increase?

Assume the process takes place in an ideal

calorimeter, in other words a perfectly insulated

container. (b) Compare your answer to the

following labeling information found on a

package of dry roasted peanuts: a serving of 33

g contains 200 calories. Comment on whether

the values are consistent.

68. Following vigorous exercise, the body

temperature of an 80.0 kg person is

. At

what rate in watts must the person transfer

thermal energy to reduce the body temperature

to

in 30.0 min, assuming the body

continues to produce energy at the rate of 150

W? 69. In a study of healthy young men1 , doing 20

push-ups in 1 minute burned an amount of

energy per kg that for a 70.0-kg man

corresponds to 8.06 calories (kcal). How much

would a 70.0-kg man’s temperature rise if he

did not lose any heat during that time?

70. A 1.28-kg sample of water at

is in a

calorimeter. You drop a piece of steel with a

mass of 0.385 kg at

into it. After the

sizzling subsides, what is the final equilibrium

temperature? (Make the reasonable

assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’t affect the outcome, as we’ll see in the next section.) 71. Repeat the preceding problem, assuming the water is in a glass beaker with a mass of 0.200 kg, which in turn is in a calorimeter. The beaker is initially at the same temperature as the water. Before doing the problem, should the answer be higher or lower than the preceding answer? Comparing the mass and specific heat of the beaker to those of the water, do you think the beaker will make much difference?

1.5 Phase Changes

72. How much heat transfer (in kilocalories) is

required to thaw a 0.450-kg package of frozen

vegetables originally at if their heat of

fusion is the same as that of water?

73. A bag containing ice is much more

effective in absorbing energy than one

containing the same amount of water. (a)

How much heat transfer is necessary to raise

the temperature of 0.800 kg of water from

to

? (b) How much heat transfer is

required to first melt 0.800 kg of ice and

then raise its temperature? (c) Explain how your

answer supports the contention that the ice is

more effective.

74. (a) How much heat transfer is required to raise

the temperature of a 0.750-kg aluminum pot

containing 2.50 kg of water from

to the

boiling point and then boil away 0.750 kg of

water? (b) How long does this take if the rate of

heat transfer is 500 W?

75. Condensation on a glass of ice water causes the

ice to melt faster than it would otherwise. If 8.00

g of vapor condense on a glass containing both

water and 200 g of ice, how many grams of the

ice will melt as a result? Assume no other heat

transfer occurs. Use for water at

as a

better approximation than for water at

.)

76. On a trip, you notice that a 3.50-kg bag of ice

lasts an average of one day in your cooler. What

is the average power in watts entering the ice if

it starts at and completely melts to

water in exactly one day?

77. On a certain dry sunny day, a swimming pool’s

1 JW Vezina, “An examination of the differences between two methods of estimating energy expenditure in resistance training activities,” Journal of Strength and Conditioning Research, April 28, 2014, http://www.ncbi.nlm.nih.gov/pubmed/24402448

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1 • Chapter Review 59

temperature would rise by

if not for

evaporation. What fraction of the water must

evaporate to carry away precisely enough

energy to keep the temperature constant?

78. (a) How much heat transfer is necessary to raise

the temperature of a 0.200-kg piece of ice from

to

, including the energy

needed for phase changes? (b) How much time

is required for each stage, assuming a constant

20.0 kJ/s rate of heat transfer? (c) Make a graph

of temperature versus time for this process.

79. In 1986, an enormous iceberg broke away from

the Ross Ice Shelf in Antarctica. It was an

approximately rectangular prism 160 km long,

40.0 km wide, and 250 m thick. (a) What is the

mass of this iceberg, given that the density of

ice is

? (b) How much heat transfer

(in joules) is needed to melt it? (c) How many

years would it take sunlight alone to melt ice

this thick, if the ice absorbs an average of

, 12.00 h per day?

80. How many grams of coffee must evaporate from

350 g of coffee in a 100-g glass cup to cool the

coffee and the cup from

to

?

Assume the coffee has the same thermal

properties as water and that the average heat of

vaporization is 2340 kJ/kg (560 kcal/g). Neglect

heat losses through processes other than

evaporation, as well as the change in mass of

the coffee as it cools. Do the latter two

assumptions cause your answer to be higher or

lower than the true answer?

81. (a) It is difficult to extinguish a fire on a crude

oil tanker, because each liter of crude oil

releases

of energy when burned.

To illustrate this difficulty, calculate the number

of liters of water that must be expended to

absorb the energy released by burning 1.00 L of

crude oil, if the water’s temperature rises from

to

, it boils, and the resulting

steam’s temperature rises to

at constant

pressure. (b) Discuss additional complications

caused by the fact that crude oil is less dense

than water.

82. The energy released from condensation in

thunderstorms can be very large. Calculate the

energy released into the atmosphere for a small

storm of radius 1 km, assuming that 1.0 cm of

rain is precipitated uniformly over this area.

83. To help prevent frost damage, 4.00 kg of water

at is sprayed onto a fruit tree. (a) How

much heat transfer occurs as the water freezes?

(b) How much would the temperature of the

200-kg tree decrease if this amount of heat

transferred from the tree? Take the specific heat

to be

, and assume that no phase

change occurs in the tree.

84. A 0.250-kg aluminum bowl holding

of

soup at

is placed in a freezer. What is

the final temperature if 388 kJ of energy is

transferred from the bowl and soup, assuming

the soup’s thermal properties are the same as

that of water?

85. A 0.0500-kg ice cube at

is placed in

0.400 kg of

water in a very well-

insulated container. What is the final

temperature?

86. If you pour 0.0100 kg of

water onto a

1.20-kg block of ice (which is initially at

), what is the final temperature? You

may assume that the water cools so rapidly that

effects of the surroundings are negligible.

87. Indigenous people sometimes cook in

watertight baskets by placing hot rocks into

water to bring it to a boil. What mass of

granite must be placed in 4.00 kg of

water to bring its temperature to

, if

0.0250 kg of water escapes as vapor from the

initial sizzle? You may neglect the effects of the

surroundings.

88. What would the final temperature of the pan

and water be in Example 1.7 if 0.260 kg of water

were placed in the pan and 0.0100 kg of the

water evaporated immediately, leaving the

remainder to come to a common temperature

with the pan?

1.6 Mechanisms of Heat Transfer

89. (a) Calculate the rate of heat conduction through

house walls that are 13.0 cm thick and have an

average thermal conductivity twice that of glass

wool. Assume there are no windows or doors.

The walls’ surface area is

and their

inside surface is at

, while their outside

surface is at

. (b) How many 1-kW room

heaters would be needed to balance the heat

transfer due to conduction?

90. The rate of heat conduction out of a window on

a winter day is rapid enough to chill the air next

to it. To see just how rapidly the windows

transfer heat by conduction, calculate the rate

of conduction in watts through a

window that is 0.634 cm thick (1/4 in.) if the

temperatures of the inner and outer surfaces

are

and

, respectively. (This

rapid rate will not be maintained—the inner

60 1 • Chapter Review

surface will cool, even to the point of frost

formation.)

91. Calculate the rate of heat conduction out of the

human body, assuming that the core internal

temperature is

, the skin temperature is

, the thickness of the fatty tissues

between the core and the skin averages 1.00

cm, and the surface area is

.

92. Suppose you stand with one foot on ceramic

flooring and one foot on a wool carpet, making

contact over an area of

with each foot.

Both the ceramic and the carpet are 2.00 cm

thick and are

on their bottom sides. At

what rate must heat transfer occur from each

foot to keep the top of the ceramic and carpet at

?

93. A man consumes 3000 kcal of food in one day,

converting most of it to thermal energy to

maintain body temperature. If he loses half this

energy by evaporating water (through breathing

and sweating), how many kilograms of water

evaporate?

94. A firewalker runs across a bed of hot coals

without sustaining burns. Calculate the heat

transferred by conduction into the sole of one

foot of a firewalker given that the bottom of the

foot is a 3.00-mm-thick callus with a

conductivity at the low end of the range for

wood and its density is

. The area of

contact is

the temperature of the

coals is

, and the time in contact is 1.00 s.

Ignore the evaporative cooling of sweat.

95. (a) What is the rate of heat conduction through

the 3.00-cm-thick fur of a large animal having a

surface area? Assume that the animal’s

skin temperature is

, that the air

temperature is

, and that fur has the

same thermal conductivity as air. (b) What food

intake will the animal need in one day to replace

this heat transfer?

96. A walrus transfers energy by conduction

through its blubber at the rate of 150 W when

immersed in

water. The walrus’s

internal core temperature is

, and it has

a surface area of

. What is the average

thickness of its blubber, which has the

conductivity of fatty tissues without blood?

97. Compare the rate of heat conduction through a

13.0-cm-thick wall that has an area of

and a thermal conductivity twice that of glass

wool with the rate of heat conduction through a

0.750-cm-thick window that has an area of

, assuming the same temperature

difference across each.

98. Suppose a person is covered head to foot by

wool clothing with average thickness of 2.00 cm

and is transferring energy by conduction

through the clothing at the rate of 50.0 W. What

is the temperature difference across the

clothing, given the surface area is

?

99. Some stove tops are smooth ceramic for easy

cleaning. If the ceramic is 0.600 cm thick and

heat conduction occurs through the same area

and at the same rate as computed in Example

1.11, what is the temperature difference across

it? Ceramic has the same thermal conductivity

as glass and brick.

100. One easy way to reduce heating (and cooling)

costs is to add extra insulation in the attic of a

house. Suppose a single-story cubical house

already had 15 cm of fiberglass insulation in

the attic and in all the exterior surfaces. If you

added an extra 8.0 cm of fiberglass to the attic,

by what percentage would the heating cost of

the house drop? Take the house to have

dimensions 10 m by 15 m by 3.0 m. Ignore air

infiltration and heat loss through windows and

doors, and assume that the interior is

uniformly at one temperature and the exterior

is uniformly at another.

101. Many decisions are made on the basis of the

payback period: the time it will take through

savings to equal the capital cost of an

investment. Acceptable payback times depend

upon the business or philosophy one has. (For

some industries, a payback period is as small

as 2 years.) Suppose you wish to install the

extra insulation in the preceding problem. If

energy cost

per million joules and the

insulation was $4.00 per square meter, then

calculate the simple payback time. Take the

average for the 120-day heating season to

be

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1 • Chapter Review 61

Additional Problems

102. In 1701, the Danish astronomer Ole Rømer

proposed a temperature scale with two fixed

points, freezing water at 7.5 degrees, and

boiling water at 60.0 degrees. What is the

boiling point of oxygen, 90.2 K, on the Rømer

scale?

103. What is the percent error of thinking the

melting point of tungsten is

instead of

the correct value of 3695 K?

104. An engineer wants to design a structure in

which the difference in length between a steel

beam and an aluminum beam remains at

0.500 m regardless of temperature, for

ordinary temperatures. What must the lengths

of the beams be?

105. How much stress is created in a steel beam if

its temperature changes from

to

but it cannot expand? For steel, the Young’s

modulus

from Stress,

Strain, and Elastic Modulus. (Ignore the

change in area resulting from the expansion.)

106. A brass rod

with a

diameter of 0.800 cm and a length of 1.20 m

when the temperature is

, is fixed at both

ends. At what temperature is the force in it at

36,000 N?

107. A mercury thermometer still in use for

meteorology has a bulb with a volume of

and a tube for the mercury to

expand into of inside diameter 0.130 mm. (a)

Neglecting the thermal expansion of the glass,

what is the spacing between marks apart?

(b) If the thermometer is made of ordinary

glass (not a good idea), what is the spacing?

108. Even when shut down after a period of normal

use, a large commercial nuclear reactor

transfers thermal energy at the rate of 150 MW

by the radioactive decay of fission products.

This heat transfer causes a rapid increase in

temperature if the cooling system fails

or

and

(a) Calculate the rate of

temperature increase in degrees Celsius per

second

if the mass of the reactor core is

and it has an average specific

heat of

. (b) How long would it

take to obtain a temperature increase of

, which could cause some metals

holding the radioactive materials to melt? (The

initial rate of temperature increase would be

greater than that calculated here because the

heat transfer is concentrated in a smaller

mass. Later, however, the temperature

increase would slow down because the

500,000-kg steel containment vessel would

also begin to heat up.)

109. You leave a pastry in the refrigerator on a plate

and ask your roommate to take it out before

you get home so you can eat it at room

temperature, the way you like it. Instead, your

roommate plays video games for hours. When

you return, you notice that the pastry is still

cold, but the game console has become hot.

Annoyed, and knowing that the pastry will not

be good if it is microwaved, you warm up the

pastry by unplugging the console and putting

it in a clean trash bag (which acts as a perfect

calorimeter) with the pastry on the plate. After

a while, you find that the equilibrium

temperature is a nice, warm

. You know

that the game console has a mass of 2.1 kg.

Approximate it as having a uniform initial

temperature of

. The pastry has a mass of

0.16 kg and a specific heat of

and is at a uniform initial temperature of

. The plate is at the same temperature

and has a mass of 0.24 kg and a specific heat of

. What is the specific heat of

the console?

110. Two solid spheres, A and B, made of the same

material, are at temperatures of and

, respectively. The spheres are placed in

thermal contact in an ideal calorimeter, and

they reach an equilibrium temperature of

. Which is the bigger sphere? What is the

ratio of their diameters?

62 1 • Chapter Review

111. In some countries, liquid nitrogen is used on

dairy trucks instead of mechanical

refrigerators. A 3.00-hour delivery trip

requires 200 L of liquid nitrogen, which has a

density of

(a) Calculate the heat

transfer necessary to evaporate this amount of

liquid nitrogen and raise its temperature to

. (Use and assume it is constant over

the temperature range.) This value is the

amount of cooling the liquid nitrogen supplies.

(b) What is this heat transfer rate in kilowatt-

hours? (c) Compare the amount of cooling

obtained from melting an identical mass of

ice with that from evaporating the liquid

nitrogen.

112. Some gun fanciers make their own bullets,

which involves melting lead and casting it into

lead slugs. How much heat transfer is needed

to raise the temperature and melt 0.500 kg of

lead, starting from

?

113. A 0.800-kg iron cylinder at a temperature of

is dropped into an insulated

chest of 1.00 kg of ice at its melting point.

What is the final temperature, and how much

ice has melted?

114. Repeat the preceding problem with 2.00 kg of

ice instead of 1.00 kg.

115. Repeat the preceding problem with 0.500 kg of

ice, assuming that the ice is initially in a

copper container of mass 1.50 kg in

equilibrium with the ice.

116. A 30.0-g ice cube at its melting point is

dropped into an aluminum calorimeter of

mass 100.0 g in equilibrium at

with

300.0 g of an unknown liquid. The final

temperature is

. What is the heat

capacity of the liquid?

117. (a) Calculate the rate of heat conduction

through a double-paned window that has a

area and is made of two panes of

0.800-cm-thick glass separated by a 1.00-cm

air gap. The inside surface temperature is

while that on the outside is

(Hint: There are identical temperature drops

across the two glass panes. First find these and

then the temperature drop across the air gap.

This problem ignores the increased heat

transfer in the air gap due to convection.) (b)

Calculate the rate of heat conduction through a

1.60-cm-thick window of the same area and

with the same temperatures. Compare your

answer with that for part (a).

118. (a) An exterior wall of a house is 3 m tall and

10 m wide. It consists of a layer of drywall with

an R factor of 0.56, a layer 3.5 inches thick

filled with fiberglass batts, and a layer of

insulated siding with an R factor of 2.6. The

wall is built so well that there are no leaks of

air through it. When the inside of the wall is at

and the outside is at

, what is the

rate of heat flow through the wall? (b) More

realistically, the 3.5-inch space also contains

2-by-4 studs—wooden boards 1.5 inches by 3.5

inches oriented so that 3.5-inch dimension

extends from the drywall to the siding. They

are “on 16-inch centers,” that is, the centers of

the studs are 16 inches apart. What is the heat

current in this situation? Don’t worry about

one stud more or less.

119. For the human body, what is the rate of heat

transfer by conduction through the body’s

tissue with the following conditions: the tissue

thickness is 3.00 cm, the difference in

temperature is

, and the skin area is

. How does this compare with the

average heat transfer rate to the body resulting

from an energy intake of about 2400 kcal per

day? (No exercise is included.)

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1 • Chapter Review 63

120. You have a Dewar flask (a laboratory vacuum

flask) that has an open top and straight sides,

as shown below. You fill it with water and put it

into the freezer. It is effectively a perfect

insulator, blocking all heat transfer, except on

the top. After a time, ice forms on the surface

of the water. The liquid water and the bottom

surface of the ice, in contact with the liquid

water, are at . The top surface of the ice is

at the same temperature as the air in the

freezer,

Set the rate of heat flow

through the ice equal to the rate of loss of heat

of fusion as the water freezes. When the ice

layer is 0.700 cm thick, find the rate in m/s at

which the ice is thickening.

121. An infrared heater for a sauna has a surface

area of

and an emissivity of 0.84.

What temperature must it run at if the

required power is 360 W? Neglect the

temperature of the environment.

122. (a) Determine the power of radiation from the

Sun by noting that the intensity of the

radiation at the distance of Earth is

. Hint: That intensity will be found

everywhere on a spherical surface with radius

equal to that of Earth’s orbit. (b) Assuming that

the Sun’s temperature is 5780 K and that its

emissivity is 1, find its radius.

Challenge Problems

123. A pendulum is made of a rod of length L and

negligible mass, but capable of thermal

expansion, and a weight of negligible size. (a)

Show that when the temperature increases by

dT, the period of the pendulum increases by a

fraction

. (b) A clock controlled by a

brass pendulum keeps time correctly at

.

If the room temperature is

, does the

clock run faster or slower? What is its error in

seconds per day?

124. At temperatures of a few hundred kelvins the

specific heat capacity of copper approximately

follows the empirical formula

where

and

How much heat is

needed to raise the temperature of a 2.00-kg

piece of copper from

to

?

125. In a calorimeter of negligible heat capacity,

200 g of steam at

and 100 g of ice at

are mixed. The pressure is maintained

at 1 atm. What is the final temperature, and

how much steam, ice, and water are present?

64 1 • Chapter Review

126. An astronaut performing an extra-vehicular

activity (space walk) shaded from the Sun is

wearing a spacesuit that can be approximated

as perfectly white

except for a

patch in the form of the

astronaut’s national flag. The patch has

emissivity 0.300. The spacesuit under the

patch is 0.500 cm thick, with a thermal

conductivity

, and its inner

surface is at a temperature of

. What is

the temperature of the patch, and what is the

rate of heat loss through it? Assume the patch

is so thin that its outer surface is at the same

temperature as the outer surface of the

spacesuit under it. Also assume the

temperature of outer space is 0 K. You will get

an equation that is very hard to solve in closed

form, so you can solve it numerically with a

graphing calculator, with software, or even by

trial and error with a calculator.

127. Find the growth of an ice layer as a function of

time in a Dewar flask as seen in Exercise

1.120. Call the thickness of the ice layer L. (a)

Derive an equation for dL/dt in terms of L , the

temperature T above the ice, and the

properties of ice (which you can leave in

symbolic form instead of substituting the

numbers). (b) Solve this differential equation

assuming that at

, you have

If you

have studied differential equations, you will

know a technique for solving equations of this

type: manipulate the equation to get dL/dt

multiplied by a (very simple) function of L on

one side, and integrate both sides with respect

to time. Alternatively, you may be able to use

your knowledge of the derivatives of various

functions to guess the solution, which has a

simple dependence on t. (c) Will the water

eventually freeze to the bottom of the flask?

128. As the very first rudiment of climatology,

estimate the temperature of Earth. Assume it

is a perfect sphere and its temperature is

uniform. Ignore the greenhouse effect.

Thermal radiation from the Sun has an

intensity (the “solar constant” S) of about

at the radius of Earth’s orbit. (a)

Assuming the Sun’s rays are parallel, what

area must S be multiplied by to get the total

radiation intercepted by Earth? It will be

easiest to answer in terms of Earth’s radius, R.

(b) Assume that Earth reflects about 30% of

the solar energy it intercepts. In other words,

Earth has an albedo with a value of

. In

terms of S, A, and R, what is the rate at which

Earth absorbs energy from the Sun? (c) Find

the temperature at which Earth radiates

energy at the same rate. Assume that at the

infrared wavelengths where it radiates, the

emissivity e is 1. Does your result show that

the greenhouse effect is important? (d) How

does your answer depend on the the area of

Earth?

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129. Let’s stop ignoring the greenhouse effect and

incorporate it into the previous problem in a

very rough way. Assume the atmosphere is a

single layer, a spherical shell around Earth,

with an emissivity

(chosen simply to

give the right answer) at infrared wavelengths

emitted by Earth and by the atmosphere.

However, the atmosphere is transparent to the

Sun’s radiation (that is, assume the radiation is

at visible wavelengths with no infrared), so the

Sun’s radiation reaches the surface. The

greenhouse effect comes from the difference

between the atmosphere’s transmission of

visible light and its rather strong absorption of

infrared. Note that the atmosphere’s radius is

not significantly different from Earth’s, but

since the atmosphere is a layer above Earth, it

emits radiation both upward and downward,

so it has twice Earth’s area. There are three

radiative energy transfers in this problem:

solar radiation absorbed by Earth’s surface;

infrared radiation from the surface, which is

absorbed by the atmosphere according to its

emissivity; and infrared radiation from the

atmosphere, half of which is absorbed by Earth

and half of which goes out into space. Apply

the method of the previous problem to get an

equation for Earth’s surface and one for the

atmosphere, and solve them for the two

unknown temperatures, surface and

atmosphere.

a. In terms of Earth’s radius, the constant ,

and the unknown temperature of the

surface, what is the power of the infrared

radiation from the surface?

b. What is the power of Earth’s radiation

absorbed by the atmosphere?

c. In terms of the unknown temperature

of the atmosphere, what is the power

radiated from the atmosphere?

d. Write an equation that says the power of

the radiation the atmosphere absorbs

from Earth equals the power of the

radiation it emits.

e. Half of the power radiated by the

atmosphere hits Earth. Write an equation

that says that the power Earth absorbs

from the atmosphere and the Sun equals

the power that it emits.

f. Solve your two equations for the unknown

temperature of Earth.

For steps that make this model less crude,

see for example the lectures

1 • Chapter Review 65
(https://openstax.org/l/21paulgormlec) by Paul O’Gorman.

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CHAPTER 2
The Kinetic Theory of Gases
Figure 2.1 A volcanic eruption releases tons of gas and dust into the atmosphere. Most of the gas is water vapor, but several other gases are common, including greenhouse gases such as carbon dioxide and acidic pollutants such as sulfur dioxide. However, the emission of volcanic gas is not all bad: Many geologists believe that in the earliest stages of Earth’s formation, volcanic emissions formed the early atmosphere. (credit: modification of work by “Boaworm”/Wikimedia Commons)
Chapter Outline
2.1 Molecular Model of an Ideal Gas 2.2 Pressure, Temperature, and RMS Speed 2.3 Heat Capacity and Equipartition of Energy 2.4 Distribution of Molecular Speeds INTRODUCTION Gases are literally all around us—the air that we breathe is a mixture of gases. Other gases include those that make breads and cakes soft, those that make drinks fizzy, and those that burn to heat many homes. Engines and refrigerators depend on the behaviors of gases, as we will see in later chapters. As we discussed in the preceding chapter, the study of heat and temperature is part of an area of physics known as thermodynamics, in which we require a system to be macroscopic, that is, to consist of a huge number (such as ) of molecules. We begin by considering some macroscopic properties of gases: volume, pressure, and temperature. The simple model of a hypothetical “ideal gas” describes these properties of a gas very accurately under many conditions. We move from the ideal gas model to a more widely applicable approximation, called the Van der Waals model.

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2 • The Kinetic Theory of Gases

To understand gases even better, we must also look at them on the microscopic scale of molecules. In gases, the molecules interact weakly, so the microscopic behavior of gases is relatively simple, and they serve as a good introduction to systems of many molecules. The molecular model of gases is called the kinetic theory of gases and is one of the classic examples of a molecular model that explains everyday behavior.
2.1 Molecular Model of an Ideal Gas
Learning Objectives By the end of this section, you will be able to:
• Apply the ideal gas law to situations involving the pressure, volume, temperature, and the number of molecules of a gas
• Use the unit of moles in relation to numbers of molecules, and molecular and macroscopic masses • Explain the ideal gas law in terms of moles rather than numbers of molecules • Apply the van der Waals gas law to situations where the ideal gas law is inadequate
In this section, we explore the thermal behavior of gases. Our word “gas” comes from the Flemish word meaning “chaos,” first used for vapors by the seventeenth-century chemist J. B. van Helmont. The term was more appropriate than he knew, because gases consist of molecules moving and colliding with each other at random. This randomness makes the connection between the microscopic and macroscopic domains simpler for gases than for liquids or solids.
How do gases differ from solids and liquids? Under ordinary conditions, such as those of the air around us, the difference is that the molecules of gases are much farther apart than those of solids and liquids. Because the typical distances between molecules are large compared to the size of a molecule, as illustrated in Figure 2.2, the forces between them are considered negligible, except when they come into contact with each other during collisions. Also, at temperatures well above the boiling temperature, the motion of molecules is fast, and the gases expand rapidly to occupy all of the accessible volume. In contrast, in liquids and solids, molecules are closer together, and the behavior of molecules in liquids and solids is highly constrained by the molecules’ interactions with one another. The macroscopic properties of such substances depend strongly on the forces between the molecules, and since many molecules are interacting, the resulting “many-body problems” can be extremely complicated (see Condensed Matter Physics).

Figure 2.2 Atoms and molecules in a gas are typically widely separated. Because the forces between them are quite weak at these distances, the properties of a gas depend more on the number of atoms per unit volume and on temperature than on the type of atom.
The Gas Laws
In the previous chapter, we saw one consequence of the large intermolecular spacing in gases: Gases are easily compressed. Table 1.2 shows that gases have larger coefficients of volume expansion than either solids or liquids. These large coefficients mean that gases expand and contract very rapidly with temperature changes. We also saw (in the section on thermal expansion) that most gases expand at the same rate or have the same coefficient of volume expansion, . This raises a question: Why do all gases act in nearly the same way, when all the various liquids and solids have widely varying expansion rates?
To study how the pressure, temperature, and volume of a gas relate to one another, consider what happens when you pump air into a deflated car tire. The tire’s volume first increases in direct proportion to the amount of air injected, without much increase in the tire pressure. Once the tire has expanded to nearly its full size, the tire’s walls limit its volume expansion. If we continue to pump air into the tire, the pressure increases. When
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2.1 • Molecular Model of an Ideal Gas

69

the car is driven and the tires flex, their temperature increases, and therefore the pressure increases even further (Figure 2.3).

Figure 2.3 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) When the tire is filled to a certain point, the tire walls resist further expansion, and the pressure increases with more air. (c) Once the tire is inflated, its pressure increases with temperature.
Figure 2.4 shows data from the experiments of Robert Boyle (1627–1691), illustrating what is now called Boyle’s law: At constant temperature and number of molecules, the absolute pressure of a gas and its volume are inversely proportional. (Recall from Fluid Mechanics that the absolute pressure is the true pressure and the gauge pressure is the absolute pressure minus the ambient pressure, typically atmospheric pressure.) The graph in Figure 2.4 displays this relationship as an inverse proportionality of volume to pressure.

Figure 2.4 Robert Boyle and his assistant found that volume and pressure are inversely proportional. Here their data are plotted as V versus 1/p; the linearity of the graph shows the inverse proportionality. The number shown as the volume is actually the height in inches of air in a cylindrical glass tube. The actual volume was that height multiplied by the cross-sectional area of the tube, which Boyle did not publish. The data are from Boyle’s book A Defence of the Doctrine Touching the Spring and Weight of the Air…, p. 60.1
Figure 2.5 shows experimental data illustrating what is called Charles’s law, after Jacques Charles (1746–1823). Charles’s law states that at constant pressure and number of molecules, the volume of a gas is proportional to its absolute temperature.
1 http://bvpb.mcu.es/en/consulta/registro.cmd?id=406806

70

2 • The Kinetic Theory of Gases

Figure 2.5 Experimental data showing that at constant pressure, volume is approximately proportional to temperature. The best-fit line passes approximately through the origin.2
Similar is Amonton’s or Gay-Lussac’s law, which states that at constant volume and number of molecules, the pressure is proportional to the temperature. That law is the basis of the constant-volume gas thermometer, discussed in the previous chapter. (The histories of these laws and the appropriate credit for them are more complicated than can be discussed here.) It is known experimentally that for gases at low density (such that their molecules occupy a negligible fraction of the total volume) and at temperatures well above the boiling point, these proportionalities hold to a good approximation. Not surprisingly, with the other quantities held constant, either pressure or volume is proportional to the number of molecules. More surprisingly, when the proportionalities are combined into a single equation, the constant of proportionality is independent of the composition of the gas. The resulting equation for all gases applies in the limit of low density and high temperature; it’s the same for oxygen as for helium or uranium hexafluoride. A gas at that limit is called an ideal gas; it obeys the ideal gas law, which is also called the equation of state of an ideal gas.
Ideal Gas Law
The ideal gas law states that 2.1
where p is the absolute pressure of a gas, V is the volume it occupies, N is the number of molecules in the gas, and T is its absolute temperature.
The constant is called the Boltzmann constant in honor of the Austrian physicist Ludwig Boltzmann (1844–1906) and has the value
The ideal gas law describes the behavior of any real gas when its density is low enough or its temperature high enough that it is far from liquefaction. This encompasses many practical situations. In the next section, we’ll see why it’s independent of the type of gas.
2 http://chemed.chem.purdue.edu/genchem/history/charles.html
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2.1 • Molecular Model of an Ideal Gas

71

In many situations, the ideal gas law is applied to a sample of gas with a constant number of molecules; for instance, the gas may be in a sealed container. If N is constant, then solving for N shows that pV /T is constant. We can write that fact in a convenient form:

2.2

where the subscripts 1 and 2 refer to any two states of the gas at different times. Again, the temperature must be expressed in kelvin and the pressure must be absolute pressure, which is the sum of gauge pressure and atmospheric pressure.

EXAMPLE 2.1

Calculating Pressure Changes Due to Temperature Changes

Suppose your bicycle tire is fully inflated, with an absolute pressure of

(a gauge pressure of just

under

) at a temperature of

What is the pressure after its temperature has risen to

on a hot day? Assume there are no appreciable leaks or changes in volume.

Strategy

The pressure in the tire is changing only because of changes in temperature. We know the initial pressure

the initial temperature

and the final temperature

We must

find the final pressure Since the number of molecules is constant, we can use the equation

Since the volume is constant, and are the same and they divide out. Therefore, We can then rearrange this to solve for

where the temperature must be in kelvin. Solution 1. Convert temperatures from degrees Celsius to kelvin
2. Substitute the known values into the equation,
Significance The final temperature is about greater than the original temperature, so the final pressure is about greater as well. Note that absolute pressure (see Fluid Mechanics) and absolute temperature (see Temperature and Heat) must be used in the ideal gas law.

72

2 • The Kinetic Theory of Gases

EXAMPLE 2.2

Calculating the Number of Molecules in a Cubic Meter of Gas
How many molecules are in a typical object, such as gas in a tire or water in a glass? This calculation can give us an idea of how large N typically is. Let’s calculate the number of molecules in the air that a typical healthy young adult inhales in one breath, with a volume of 500 mL, at standard temperature and pressure (STP), which is defined as and atmospheric pressure. (Our young adult is apparently outside in winter.)

Strategy Because pressure, volume, and temperature are all specified, we can use the ideal gas law, N.

to find

Solution

1. Identify the knowns.

2. Substitute the known values into the equation and solve for N.

Significance

N is huge, even in small volumes. For example,

of a gas at STP contains

again, note that our result for N is the same for all types of gases, including mixtures.

molecules. Once

As we observed in the chapter on fluid mechanics, pascals are

, so

Thus, our result

for N is dimensionless, a pure number that could be obtained by counting (in principle) rather than measuring.

As it is the number of molecules, we put “molecules” after the number, keeping in mind that it is an aid to

communication rather than a unit.

Moles and Avogadro’s Number

It is often convenient to measure the amount of substance with a unit on a more human scale than molecules.

The SI unit for this purpose was developed by the Italian scientist Amedeo Avogadro (1776–1856). (He worked

from the hypothesis that equal volumes of gas at equal pressure and temperature contain equal numbers of

molecules, independent of the type of gas. As mentioned above, this hypothesis has been confirmed when the

ideal gas approximation applies.) A mole (abbreviated mol) is defined as the amount of any substance that

contains as many molecules as there are atoms. (Technically, we should say “formula units,” not “molecules,”

but this distinction is irrelevant for our purposes.) The number of molecules in one mole is called Avogadro’s

number

and the value of Avogadro’s number is now known to be

We can now write

, where n represents the number of moles of a substance.

Avogadro’s number relates the mass of an amount of substance in grams to the number of protons and

neutrons in an atom or molecule (12 for a carbon-12 atom), which roughly determine its mass. It’s natural to

define a unit of mass such that the mass of an atom is approximately equal to its number of neutrons and

protons. The unit of that kind accepted for use with the SI is the unified atomic mass unit (u), also called the

dalton. Specifically, a carbon-12 atom has a mass of exactly 12 u, so that its molar mass M in grams per mole is

numerically equal to the mass of one carbon-12 atom in u. That equality holds for any substance. In other

words, is not only the conversion from numbers of molecules to moles, but it is also the conversion from u

to grams:

See Figure 2.6.

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2.1 • Molecular Model of an Ideal Gas

73

Figure 2.6 How big is a mole? On a macroscopic level, Avogadro’s number of table tennis balls would cover Earth to a depth of about 40 km.

Now letting stand for the mass of a sample of a substance, we have of a molecule, we have

Letting m stand for the mass

CHECK YOUR UNDERSTANDING 2.1

The recommended daily amount of vitamin or niacin,

for women who are not pregnant or

nursing, is 14 mg. Find the number of molecules of niacin in that amount.

CHECK YOUR UNDERSTANDING 2.2

The density of air in a classroom

and

is

if the temperature is kept constant?

. At what pressure is the density

The Ideal Gas Law Restated using Moles
A very common expression of the ideal gas law uses the number of moles in a sample, n, rather than the number of molecules, N. We start from the ideal gas law,

and multiply and divide the right-hand side of the equation by Avogadro’s number This gives us

Note that

is the number of moles. We define the universal gas constant as

the ideal gas law in terms of moles.

Ideal Gas Law (in terms of moles)
In terms of number of moles n, the ideal gas law is written as

In SI units,

and obtain 2.3

In other units, You can use whichever value of R is most convenient for a particular problem.

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2 • The Kinetic Theory of Gases

EXAMPLE 2.3

Density of Air at STP and in a Hot Air Balloon

Calculate the density of dry air (a) under standard conditions and (b) in a hot air balloon at a temperature of

. Dry air is approximately

and

.

Strategy and Solution

a. We are asked to find the density, or mass per cubic meter. We can begin by finding the molar mass. If we have a hundred molecules, of which 78 are nitrogen, 21 are oxygen, and 1 is argon, the average molecular

mass is

, or the mass of each constituent multiplied by its percentage. The same

applies to the molar mass, which therefore is

Now we can find the number of moles per cubic meter. We use the ideal gas law in terms of moles,

with

,

,

, and

. The most convenient choice

for R in this case is

because the known quantities are in SI units:

Then, the mass of that air is Finally the density of air at STP is

b. The air pressure inside the balloon is still 1 atm because the bottom of the balloon is open to the

atmosphere. The calculation is the same except that we use a temperature of

, which is 393 K. We

can repeat the calculation in (a), or simply observe that the density is proportional to the number of moles,

which is inversely proportional to the temperature. Then using the subscripts 1 for air at STP and 2 for the

hot air, we have

Significance

Using the methods of Archimedes’ Principle and Buoyancy, we can find that the net force on

of air at

is

or enough to lift about 867 kg. The mass

density and molar density of air at STP, found above, are often useful numbers. From the molar density, we can

easily determine another useful number, the volume of a mole of any ideal gas at STP, which is 22.4 L.

CHECK YOUR UNDERSTANDING 2.3
Liquids and solids have densities on the order of 1000 times greater than gases. Explain how this implies that the distances between molecules in gases are on the order of 10 times greater than the size of their molecules.

The ideal gas law is closely related to energy: The units on both sides of the equation are joules. The right-hand

side of the ideal gas law equation is

This term is roughly the total translational kinetic energy (which,

when discussing gases, refers to the energy of translation of a molecule, not that of vibration of its atoms or

rotation) of N molecules at an absolute temperature T, as we will see formally in the next section. The left-hand

side of the ideal gas law equation is pV. As mentioned in the example on the number of molecules in an ideal

gas, pressure multiplied by volume has units of energy. The energy of a gas can be changed when the gas does

work as it increases in volume, something we explored in the preceding chapter, and the amount of work is

related to the pressure. This is the process that occurs in gasoline or steam engines and turbines, as we’ll see

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2.1 • Molecular Model of an Ideal Gas

75

in the next chapter.

PROBLEM-SOLVING STRATEGY

The Ideal Gas Law
Step 1. Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal unless they are close to the boiling point or at pressures far above atmospheric pressure.

Step 2. Make a list of what quantities are given or can be inferred from the problem as stated (identify the known quantities).

Step 3. Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written list is useful.

Step 4. Determine whether the number of molecules or the number of moles is known or asked for to decide

whether to use the ideal gas law as

where N is the number of molecules, or

where n

is the number of moles.

Step 5. Convert known values into proper SI units (K for temperature, Pa for pressure, for volume, molecules for N, and moles for n). If the units of the knowns are consistent with one of the non-SI values of R, you can leave them in those units. Be sure to use absolute temperature and absolute pressure.

Step 6. Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratio of final states to initial states to eliminate the unknown quantities that are kept fixed.

Step 7. Substitute the known quantities, along with their units, into the appropriate equation and obtain numerical solutions complete with units.

Step 8. Check the answer to see if it is reasonable: Does it make sense?

The Van der Waals Equation of State
We have repeatedly noted that the ideal gas law is an approximation. How can it be improved upon? The van der Waals equation of state (named after the Dutch physicist Johannes van der Waals, 1837−1923) improves it by taking into account two factors. First, the attractive forces between molecules, which are stronger at higher density and reduce the pressure, are taken into account by adding to the pressure a term equal to the square of the molar density multiplied by a positive coefficient a. Second, the volume of the molecules is represented by a positive constant b, which can be thought of as the volume of a mole of molecules. This is subtracted from the total volume to give the remaining volume that the molecules can move in. The constants a and b are determined experimentally for each gas. The resulting equation is

2.4

In the limit of low density (small n), the a and b terms are negligible, and we have the ideal gas law, as we

should for low density. On the other hand, if

is small, meaning that the molecules are very close

together, the pressure must be higher to give the same nRT, as we would expect in the situation of a highly

compressed gas. However, the increase in pressure is less than that argument would suggest, because at high

density the

term is significant. Since it’s positive, it causes a lower pressure to give the same nRT.

The van der Waals equation of state works well for most gases under a wide variety of conditions. As we’ll see in the next module, it even predicts the gas-liquid transition.

pV Diagrams

We can examine aspects of the behavior of a substance by plotting a pV diagram, which is a graph of pressure

versus volume. When the substance behaves like an ideal gas, the ideal gas law

describes the

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2 • The Kinetic Theory of Gases

relationship between its pressure and volume. On a pV diagram, it’s common to plot an isotherm, which is a

curve showing p as a function of V with the number of molecules and the temperature fixed. Then, for an ideal

gas,

For example, the volume of the gas decreases as the pressure increases. The resulting

graph is a hyperbola.

However, if we assume the van der Waals equation of state, the isotherms become more interesting, as shown in Figure 2.7. At high temperatures, the curves are approximately hyperbolas, representing approximately ideal behavior at various fixed temperatures. At lower temperatures, the curves look less and less like hyperbolas—that is, the gas is not behaving ideally. There is a critical temperature at which the curve has a point with zero slope. Below that temperature, the curves do not decrease monotonically; instead, they each have a “hump,” meaning that for a certain range of volume, increasing the volume increases the pressure.

Figure 2.7 pV diagram for a Van der Waals gas at various temperatures. The red curves are calculated at temperatures above the critical temperature and the blue curves at temperatures below it. The blue curves have an oscillation in which volume (V) increases with increasing pressure (P), an impossible situation, so they must be corrected as in Figure 2.8. (credit: “Eman”/Wikimedia Commons)
Such behavior would be completely unphysical. Instead, the curves are understood as describing a liquid-gas phase transition. The oscillating part of the curve is replaced by a horizontal line, showing that as the volume increases at constant temperature, the pressure stays constant. That behavior corresponds to boiling and condensation; when a substance is at its boiling temperature for a particular pressure, it can increase in volume as some of the liquid turns to gas, or decrease as some of the gas turns to liquid, without any change in temperature or pressure. Figure 2.8 shows similar isotherms that are more realistic than those based on the van der Waals equation. The steep parts of the curves to the left of the transition region show the liquid phase, which is almost incompressible—a slight decrease in volume requires a large increase in pressure. The flat parts show the liquid-gas transition; the blue regions that they define represent combinations of pressure and volume where liquid and gas can coexist.
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2.1 • Molecular Model of an Ideal Gas

77

Figure 2.8 pV diagrams. (a) Each curve (isotherm) represents the relationship between p and V at a fixed temperature; the upper curves are at higher temperatures. The lower curves are not hyperbolas because the gas is no longer an ideal gas. (b) An expanded portion of the pV diagram for low temperatures, where the phase can change from a gas to a liquid. The term “vapor” refers to the gas phase when it exists at a temperature below the boiling temperature.

The isotherms above do not go through the liquid-gas transition. Therefore, liquid cannot exist above that

temperature, which is the critical temperature (described in the chapter on temperature and heat). At

sufficiently low pressure above that temperature, the gas has the density of a liquid but will not condense; the

gas is said to be supercritical. At higher pressure, it is solid. Carbon dioxide, for example, has no liquid phase

at a temperature above

. The critical pressure is the maximum pressure at which the liquid can exist.

The point on the pV diagram at the critical pressure and temperature is the critical point (which you learned

about in the chapter on temperature and heat). Table 2.1 lists representative critical temperatures and

pressures.

Substance

Critical temperature

Critical pressure

K

Pa

atm

Water

647.4 374.3

219.0

Sulfur dioxide 430.7 157.6

78.0

Ammonia

405.5 132.4

111.7

Carbon dioxide 304.2 31.1

73.2

Oxygen

154.8 –118.4

50.3

Nitrogen

126.2 –146.9

33.6

Hydrogen

33.3

–239.9

12.9

Helium

5.3

–267.9

2.27

Table 2.1 Critical Temperatures and Pressures for Various Substances

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2 • The Kinetic Theory of Gases

2.2 Pressure, Temperature, and RMS Speed
Learning Objectives By the end of this section, you will be able to:
• Explain the relations between microscopic and macroscopic quantities in a gas • Solve problems involving mixtures of gases • Solve problems involving the distance and time between a gas molecule’s collisions
We have examined pressure and temperature based on their macroscopic definitions. Pressure is the force divided by the area on which the force is exerted, and temperature is measured with a thermometer. We can gain a better understanding of pressure and temperature from the kinetic theory of gases, the theory that relates the macroscopic properties of gases to the motion of the molecules they consist of. First, we make two assumptions about molecules in an ideal gas.
1. There is a very large number N of molecules, all identical and each having mass m. 2. The molecules obey Newton’s laws and are in continuous motion, which is random and isotropic, that is,
the same in all directions.
To derive the ideal gas law and the connection between microscopic quantities such as the energy of a typical molecule and macroscopic quantities such as temperature, we analyze a sample of an ideal gas in a rigid container, about which we make two further assumptions:
3. The molecules are much smaller than the average distance between them, so their total volume is much less than that of their container (which has volume V). In other words, we take the Van der Waals constant b, the volume of a mole of gas molecules, to be negligible compared to the volume of a mole of gas in the container.
4. The molecules make perfectly elastic collisions with the walls of the container and with each other. Other forces on them, including gravity and the attractions represented by the Van der Waals constant a, are negligible (as is necessary for the assumption of isotropy).
The collisions between molecules do not appear in the derivation of the ideal gas law. They do not disturb the derivation either, since collisions between molecules moving with random velocities give new random velocities. Furthermore, if the velocities of gas molecules in a container are initially not random and isotropic, molecular collisions are what make them random and isotropic.
We make still further assumptions that simplify the calculations but do not affect the result. First, we let the container be a rectangular box. Second, we begin by considering monatomic gases, those whose molecules consist of single atoms, such as helium. Then, we can assume that the atoms have no energy except their translational kinetic energy; for instance, they have neither rotational nor vibrational energy. (Later, we discuss the validity of this assumption for real monatomic gases and dispense with it to consider diatomic and polyatomic gases.)
Figure 2.9 shows a collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (by Newton’s third law). These collisions are the source of pressure in a gas. As the number of molecules increases, the number of collisions, and thus the pressure, increases. Similarly, if the average velocity of the molecules is higher, the gas pressure is higher.

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2.2 • Pressure, Temperature, and RMS Speed

79

Figure 2.9 When a molecule collides with a rigid wall, the component of its momentum perpendicular to the wall is reversed. A force is
thus exerted on the wall, creating pressure.
In a sample of gas in a container, the randomness of the molecular motion causes the number of collisions of molecules with any part of the wall in a given time to fluctuate. However, because a huge number of molecules collide with the wall in a short time, the number of collisions on the scales of time and space we measure fluctuates by only a tiny, usually unobservable fraction from the average. We can compare this situation to that of a casino, where the outcomes of the bets are random and the casino’s takings fluctuate by the minute and the hour. However, over long times such as a year, the casino’s takings are very close to the averages expected from the odds. A tank of gas has enormously more molecules than a casino has bettors in a year, and the molecules make enormously more collisions in a second than a casino has bets.
A calculation of the average force exerted by molecules on the walls of the box leads us to the ideal gas law and to the connection between temperature and molecular kinetic energy. (In fact, we will take two averages: one over time to get the average force exerted by one molecule with a given velocity, and then another average over molecules with different velocities.) This approach was developed by Daniel Bernoulli (1700–1782), who is best known in physics for his work on fluid flow (hydrodynamics). Remarkably, Bernoulli did this work before Dalton established the view of matter as consisting of atoms.
Figure 2.10 shows a container full of gas and an expanded view of an elastic collision of a gas molecule with a wall of the container, broken down into components. We have assumed that a molecule is small compared with the separation of molecules in the gas, and that its interaction with other molecules can be ignored. Under these conditions, the ideal gas law is experimentally valid. Because we have also assumed the wall is rigid and the particles are points, the collision is elastic (by conservation of energy—there’s nowhere for a particle’s kinetic energy to go). Therefore, the molecule’s kinetic energy remains constant, and hence, its speed and the magnitude of its momentum remain constant as well. This assumption is not always valid, but the results in the rest of this module are also obtained in models that let the molecules exchange energy and momentum with the wall.

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2 • The Kinetic Theory of Gases

Figure 2.10 Gas in a box exerts an outward pressure on its walls. A molecule colliding with a rigid wall has its velocity and momentum in the x-direction reversed. This direction is perpendicular to the wall. The components of its velocity momentum in the y- and z-directions are not changed, which means there is no force parallel to the wall.

If the molecule’s velocity changes in the x-direction, its momentum changes from

to

Thus, its

change in momentum is

According to the impulse-momentum theorem

given in the chapter on linear momentum and collisions, the force exerted on the ith molecule, where i labels

the molecules from 1 to N, is given by

(In this equation alone, p represents momentum, not pressure.) There is no force between the wall and the

molecule except while the molecule is touching the wall. During the short time of the collision, the force

between the molecule and wall is relatively large, but that is not the force we are looking for. We are looking for

the average force, so we take to be the average time between collisions of the given molecule with this wall,

which is the time in which we expect to find one collision. Let l represent the length of the box in the

x-direction. Then is the time the molecule would take to go across the box and back, a distance 2l, at a

speed of Thus

and the expression for the force becomes

This force is due to one molecule. To find the total force on the wall, F, we need to add the contributions of all N molecules:

We now use the definition of the average, which we denote with a bar, to find the force:

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2.2 • Pressure, Temperature, and RMS Speed

81

We want the force in terms of the speed v, rather than the x-component of the velocity. Note that the total velocity squared is the sum of the squares of its components, so that

With the assumption of isotropy, the three averages on the right side are equal, so

Substituting this into the expression for F gives

The pressure is F/A, so we obtain

where we used

for the volume. This gives the important result

2.5

Combining this equation with

gives

We can get the average kinetic energy of a molecule, out N and multiplying by 3/2.

, from the left-hand side of the equation by dividing

Average Kinetic Energy per Molecule
The average kinetic energy of a molecule is directly proportional to its absolute temperature: 2.6

The equation

is the average kinetic energy per molecule. Note in particular that nothing in this

equation depends on the molecular mass (or any other property) of the gas, the pressure, or anything but the temperature. If samples of helium and xenon gas, with very different molecular masses, are at the same temperature, the molecules have the same average kinetic energy.

The internal energy of a thermodynamic system is the sum of the mechanical energies of all of the molecules

in it. We can now give an equation for the internal energy of a monatomic ideal gas. In such a gas, the

molecules’ only energy is their translational kinetic energy. Therefore, denoting the internal energy by

we

simply have

or

2.7

Often we would like to use this equation in terms of moles:

We can solve

for a typical speed of a molecule in an ideal gas in terms of temperature to

determine what is known as the root-mean-square (rms) speed of a molecule.

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2 • The Kinetic Theory of Gases

RMS Speed of a Molecule
The root-mean-square (rms) speed of a molecule, or the square root of the average of the square of the speed , is
2.8

The rms speed is not the average or the most likely speed of molecules, as we will see in Distribution of Molecular Speeds, but it provides an easily calculated estimate of the molecules’ speed that is related to their kinetic energy. Again we can write this equation in terms of the gas constant R and the molar mass M in kg/ mol:

2.9

We digress for a moment to answer a question that may have occurred to you: When we apply the model to

atoms instead of theoretical point particles, does rotational kinetic energy change our results? To answer this

question, we have to appeal to quantum mechanics. In quantum mechanics, rotational kinetic energy cannot

take on just any value; it’s limited to a discrete set of values, and the smallest value is inversely proportional to

the rotational inertia. The rotational inertia of an atom is tiny because almost all of its mass is in the nucleus,

which typically has a radius less than

. Thus the minimum rotational energy of an atom is much more

than

for any attainable temperature, and the energy available is not enough to make an atom rotate. We

will return to this point when discussing diatomic and polyatomic gases in the next section.

EXAMPLE 2.4

Calculating Kinetic Energy and Speed of a Gas Molecule

(a) What is the average kinetic energy of a gas molecule at

(room temperature)? (b) Find the rms speed

of a nitrogen molecule

at this temperature.

Strategy (a) The known in the equation for the average kinetic energy is the temperature:

Before substituting values into this equation, we must convert the given temperature into kelvin: We can find the rms speed of a nitrogen molecule by using the equation

but we must first find the mass of a nitrogen molecule. Obtaining the molar mass of nitrogen periodic table, we find

from the

Solution
a. The temperature alone is sufficient for us to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives

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2.2 • Pressure, Temperature, and RMS Speed

83

b. Substituting this mass and the value for into the equation for yields

Significance
Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. On the other hand, it is much greater than the typical difference in gravitational potential energy when a molecule moves from, say, the top to the bottom of a room, so our neglect of gravitation is justified in typical real-world situations. The rms speed of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule moves on average between collisions, discussed a bit later in this section) of molecules in air is very small, so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, which is about 340 m/s at room temperature. The higher the rms speed of air molecules, the faster sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium (see Figure 2.11).

Figure 2.11 (a) In an ordinary gas, so many molecules move so fast that they collide billions of times every second. (b) Individual molecules do not move very far in a small amount of time, but disturbances like sound waves are transmitted at speeds related to the molecular speeds.
EXAMPLE 2.5
Calculating Temperature: Escape Velocity of Helium Atoms
To escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel away from Earth at 11.1 km/s. This speed is called the escape velocity. At what temperature would helium atoms have an rms speed equal to the escape velocity? Strategy Identify the knowns and unknowns and determine which equations to use to solve the problem. Solution 1. Identify the knowns: v is the escape velocity, 11.1 km/s.

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2 • The Kinetic Theory of Gases

2. Identify the unknowns: We need to solve for temperature, T. We also need to solve for the mass m of the helium atom.
3. Determine which equations are needed. ◦ To get the mass m of the helium atom, we can use information from the periodic table:

◦ To solve for temperature T, we can rearrange

to yield

4. Substitute the known values into the equations and solve for the unknowns,

and

Significance
This temperature is much higher than atmospheric temperature, which is approximately 250 K at high elevation. Very few helium atoms are left in the atmosphere, but many were
present when the atmosphere was formed, and more are always being created by radioactive decay (see the chapter on nuclear physics). The reason for the loss of helium atoms is that a small number of helium atoms have speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one collision to the next, so that at any instant, there is a small but nonzero chance that the atom’s speed is greater than the escape velocity. The chance is high enough that over the lifetime of Earth, almost all the helium atoms that have been in the atmosphere have reached escape velocity at high altitudes and escaped from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water, have smaller rms speeds, and so it is much less likely that any of them will have speeds greater than the escape velocity. In fact, the likelihood is so small that billions of years are required to lose significant amounts of heavier molecules from the atmosphere. Figure 2.12 shows the effect of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is much weaker, it has lost almost its entire atmosphere. The atmospheres of Earth and other bodies are compared in this chapter’s exercises.

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2.2 • Pressure, Temperature, and RMS Speed

85

Figure 2.12 This photograph of Apollo 17 Commander Eugene Cernan driving the lunar rover on the Moon in 1972 looks as though it was taken at night with a large spotlight. In fact, the light is coming from the Sun. Because the acceleration due to gravity on the Moon is so low (about 1/6 that of Earth), the Moon’s escape velocity is much smaller. As a result, gas molecules escape very easily from the Moon, leaving it with virtually no atmosphere. Even during the daytime, the sky is black because there is no gas to scatter sunlight. (credit: Harrison H. Schmitt/NASA)

CHECK YOUR UNDERSTANDING 2.4
If you consider a very small object, such as a grain of pollen, in a gas, then the number of molecules striking its surface would also be relatively small. Would you expect the grain of pollen to experience any fluctuations in pressure due to statistical fluctuations in the number of gas molecules striking it in a given amount of time?

Vapor Pressure, Partial Pressure, and Dalton’s Law

The pressure a gas would create if it occupied the total volume available is called the gas’s partial pressure. If two or more gases are mixed, they will come to thermal equilibrium as a result of collisions between molecules; the process is analogous to heat conduction as described in the chapter on temperature and heat. As we have seen from kinetic theory, when the gases have the same temperature, their molecules have the same average kinetic energy. Thus, each gas obeys the ideal gas law separately and exerts the same pressure on the walls of a container that it would if it were alone. Therefore, in a mixture of gases, the total pressure is the sum of partial pressures of the component gases, assuming ideal gas behavior and no chemical reactions between the components. This law is known as Dalton’s law of partial pressures, after the English scientist John Dalton (1766–1844) who proposed it. Dalton’s law is consistent with the fact that pressures add according to Pascal’s principle.

In a mixture of ideal gases in thermal equilibrium, the number of molecules of each gas is proportional to its

partial pressure. This result follows from applying the ideal gas law to each in the form

Because

the right-hand side is the same for any gas at a given temperature in a container of a given volume, the left-

hand side is the same as well.

• Partial pressure is the pressure a gas would create if it existed alone. • Dalton’s law states that the total pressure is the sum of the partial pressures of all of the gases present. • For any two gases (labeled 1 and 2) in equilibrium in a container,

An important application of partial pressure is that, in chemistry, it functions as the concentration of a gas in

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2 • The Kinetic Theory of Gases

determining the rate of a reaction. Here, we mention only that the partial pressure of oxygen in a person’s lungs is crucial to life and health. Breathing air that has a partial pressure of oxygen below 0.16 atm can impair coordination and judgment, particularly in people not acclimated to a high elevation. Lower partial pressures of have more serious effects; partial pressures below 0.06 atm can be quickly fatal, and permanent damage is likely even if the person is rescued. However, the sensation of needing to breathe, as when holding one’s breath, is caused much more by high concentrations of carbon dioxide in the blood than by low concentrations of oxygen. Thus, if a small room or closet is filled with air having a low concentration of oxygen, perhaps because a leaking cylinder of some compressed gas is stored there, a person will not feel any “choking” sensation and may go into convulsions or lose consciousness without noticing anything wrong. Safety engineers give considerable attention to this danger.
Another important application of partial pressure is vapor pressure, which is the partial pressure of a vapor at which it is in equilibrium with the liquid (or solid, in the case of sublimation) phase of the same substance. At any temperature, the partial pressure of the water in the air cannot exceed the vapor pressure of the water at that temperature, because whenever the partial pressure reaches the vapor pressure, water condenses out of the air. Dew is an example of this condensation. The temperature at which condensation occurs for a sample of air is called the dew point. It is easily measured by slowly cooling a metal ball; the dew point is the temperature at which condensation first appears on the ball.
The vapor pressures of water at some temperatures of interest for meteorology are given in Table 2.2.

T

Vapor Pressure (Pa)

0

610.5

3

757.9

5

872.3

8

1073

10

1228

13

1497

15

1705

18

2063

20

2338

23

2809

25

3167

30

4243

35

5623

40

7376

Table 2.2 Vapor Pressure of Water at Various Temperatures

The relative humidity (R.H.) at a temperature T is defined by

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2.2 • Pressure, Temperature, and RMS Speed

87

A relative humidity of

means that the partial pressure of water is equal to the vapor pressure; in other

words, the air is saturated with water.

EXAMPLE 2.6

Calculating Relative Humidity
What is the relative humidity when the air temperature is

and the dew point is

?

Strategy We simply look up the vapor pressure at the given temperature and that at the dew point and find the ratio.

Solution

Significance R.H. is important to our comfort. The value of comfort indoors.

is within the range of

recommended for

As noted in the chapter on temperature and heat, the temperature seldom falls below the dew point, because when it reaches the dew point or frost point, water condenses and releases a relatively large amount of latent heat of vaporization.

Mean Free Path and Mean Free Time

We now consider collisions explicitly. The usual first step (which is all we’ll take) is to calculate the mean free

path, the average distance a molecule travels between collisions with other molecules, and the mean free

time , the average time between the collisions of a molecule. If we assume all the molecules are spheres with a

radius r, then a molecule will collide with another if their centers are within a distance 2r of each other. For a

given particle, we say that the area of a circle with that radius, , is the “cross-section” for collisions. As the

particle moves, it traces a cylinder with that cross-sectional area. The mean free path is the length such that

the expected number of other molecules in a cylinder of length and cross-section

is 1. If we temporarily

ignore the motion of the molecules other than the one we’re looking at, the expected number is the number

density of molecules, N/V, times the volume, and the volume is

, so we have

or

Taking the motion of all the molecules into account makes the calculation much harder, but the only change is a factor of The result is

2.10

In an ideal gas, we can substitute

to obtain

2.11

The mean free time is simply the mean free path divided by a typical speed, and the usual choice is the rms speed. Then

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2 • The Kinetic Theory of Gases

2.12

EXAMPLE 2.7

Calculating Mean Free Time
Find the mean free time for argon atoms atm. Take the radius of an argon atom to be

at a temperature of and a pressure of 1.00

Solution

1. Identify the knowns and convert into SI units. We know the molar mass is 0.0399 kg/mol, the temperature

is 273 K, the pressure is

and the radius is

2. Find the rms speed:

.

3. Substitute into the equation for the mean free time:

Significance We can hardly compare this result with our intuition about gas molecules, but it gives us a picture of molecules colliding with extremely high frequency.

CHECK YOUR UNDERSTANDING 2.5
Which has a longer mean free path, liquid water or water vapor in the air?

2.3 Heat Capacity and Equipartition of Energy
Learning Objectives By the end of this section, you will be able to:
• Solve problems involving heat transfer to and from ideal monatomic gases whose volumes are held constant • Solve similar problems for non-monatomic ideal gases based on the number of degrees of freedom of a
molecule • Estimate the heat capacities of metals using a model based on degrees of freedom

In the chapter on temperature and heat, we defined the specific heat capacity with the equation

or

. However, the properties of an ideal gas depend directly on the number of moles in a sample,

so here we define specific heat capacity in terms of the number of moles, not the mass. Furthermore, when

talking about solids and liquids, we ignored any changes in volume and pressure with changes in

temperature—a good approximation for solids and liquids, but for gases, we have to make some condition on

volume or pressure changes. Here, we focus on the heat capacity with the volume held constant. We can

calculate it for an ideal gas.

Heat Capacity of an Ideal Monatomic Gas at Constant Volume

We define the molar heat capacity at constant volume as

This is often expressed in the form

2.13

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