Skip to Content
OpenStax Logo
Buy book
  1. Preface
  2. Unit 1. Thermodynamics
    1. 1 Temperature and Heat
      1. Introduction
      2. 1.1 Temperature and Thermal Equilibrium
      3. 1.2 Thermometers and Temperature Scales
      4. 1.3 Thermal Expansion
      5. 1.4 Heat Transfer, Specific Heat, and Calorimetry
      6. 1.5 Phase Changes
      7. 1.6 Mechanisms of Heat Transfer
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 The Kinetic Theory of Gases
      1. Introduction
      2. 2.1 Molecular Model of an Ideal Gas
      3. 2.2 Pressure, Temperature, and RMS Speed
      4. 2.3 Heat Capacity and Equipartition of Energy
      5. 2.4 Distribution of Molecular Speeds
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 The First Law of Thermodynamics
      1. Introduction
      2. 3.1 Thermodynamic Systems
      3. 3.2 Work, Heat, and Internal Energy
      4. 3.3 First Law of Thermodynamics
      5. 3.4 Thermodynamic Processes
      6. 3.5 Heat Capacities of an Ideal Gas
      7. 3.6 Adiabatic Processes for an Ideal Gas
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 The Second Law of Thermodynamics
      1. Introduction
      2. 4.1 Reversible and Irreversible Processes
      3. 4.2 Heat Engines
      4. 4.3 Refrigerators and Heat Pumps
      5. 4.4 Statements of the Second Law of Thermodynamics
      6. 4.5 The Carnot Cycle
      7. 4.6 Entropy
      8. 4.7 Entropy on a Microscopic Scale
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Electricity and Magnetism
    1. 5 Electric Charges and Fields
      1. Introduction
      2. 5.1 Electric Charge
      3. 5.2 Conductors, Insulators, and Charging by Induction
      4. 5.3 Coulomb's Law
      5. 5.4 Electric Field
      6. 5.5 Calculating Electric Fields of Charge Distributions
      7. 5.6 Electric Field Lines
      8. 5.7 Electric Dipoles
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    2. 6 Gauss's Law
      1. Introduction
      2. 6.1 Electric Flux
      3. 6.2 Explaining Gauss’s Law
      4. 6.3 Applying Gauss’s Law
      5. 6.4 Conductors in Electrostatic Equilibrium
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 7 Electric Potential
      1. Introduction
      2. 7.1 Electric Potential Energy
      3. 7.2 Electric Potential and Potential Difference
      4. 7.3 Calculations of Electric Potential
      5. 7.4 Determining Field from Potential
      6. 7.5 Equipotential Surfaces and Conductors
      7. 7.6 Applications of Electrostatics
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 8 Capacitance
      1. Introduction
      2. 8.1 Capacitors and Capacitance
      3. 8.2 Capacitors in Series and in Parallel
      4. 8.3 Energy Stored in a Capacitor
      5. 8.4 Capacitor with a Dielectric
      6. 8.5 Molecular Model of a Dielectric
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 9 Current and Resistance
      1. Introduction
      2. 9.1 Electrical Current
      3. 9.2 Model of Conduction in Metals
      4. 9.3 Resistivity and Resistance
      5. 9.4 Ohm's Law
      6. 9.5 Electrical Energy and Power
      7. 9.6 Superconductors
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 10 Direct-Current Circuits
      1. Introduction
      2. 10.1 Electromotive Force
      3. 10.2 Resistors in Series and Parallel
      4. 10.3 Kirchhoff's Rules
      5. 10.4 Electrical Measuring Instruments
      6. 10.5 RC Circuits
      7. 10.6 Household Wiring and Electrical Safety
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 11 Magnetic Forces and Fields
      1. Introduction
      2. 11.1 Magnetism and Its Historical Discoveries
      3. 11.2 Magnetic Fields and Lines
      4. 11.3 Motion of a Charged Particle in a Magnetic Field
      5. 11.4 Magnetic Force on a Current-Carrying Conductor
      6. 11.5 Force and Torque on a Current Loop
      7. 11.6 The Hall Effect
      8. 11.7 Applications of Magnetic Forces and Fields
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 12 Sources of Magnetic Fields
      1. Introduction
      2. 12.1 The Biot-Savart Law
      3. 12.2 Magnetic Field Due to a Thin Straight Wire
      4. 12.3 Magnetic Force between Two Parallel Currents
      5. 12.4 Magnetic Field of a Current Loop
      6. 12.5 Ampère’s Law
      7. 12.6 Solenoids and Toroids
      8. 12.7 Magnetism in Matter
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    9. 13 Electromagnetic Induction
      1. Introduction
      2. 13.1 Faraday’s Law
      3. 13.2 Lenz's Law
      4. 13.3 Motional Emf
      5. 13.4 Induced Electric Fields
      6. 13.5 Eddy Currents
      7. 13.6 Electric Generators and Back Emf
      8. 13.7 Applications of Electromagnetic Induction
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 14 Inductance
      1. Introduction
      2. 14.1 Mutual Inductance
      3. 14.2 Self-Inductance and Inductors
      4. 14.3 Energy in a Magnetic Field
      5. 14.4 RL Circuits
      6. 14.5 Oscillations in an LC Circuit
      7. 14.6 RLC Series Circuits
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 15 Alternating-Current Circuits
      1. Introduction
      2. 15.1 AC Sources
      3. 15.2 Simple AC Circuits
      4. 15.3 RLC Series Circuits with AC
      5. 15.4 Power in an AC Circuit
      6. 15.5 Resonance in an AC Circuit
      7. 15.6 Transformers
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 16 Electromagnetic Waves
      1. Introduction
      2. 16.1 Maxwell’s Equations and Electromagnetic Waves
      3. 16.2 Plane Electromagnetic Waves
      4. 16.3 Energy Carried by Electromagnetic Waves
      5. 16.4 Momentum and Radiation Pressure
      6. 16.5 The Electromagnetic Spectrum
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
  12. Index

Check Your Understanding

1.1

The actual amount (mass) of gasoline left in the tank when the gauge hits “empty” is less in the summer than in the winter. The gasoline has the same volume as it does in the winter when the “add fuel” light goes on, but because the gasoline has expanded, there is less mass.

1.2

Not necessarily, as the thermal stress is also proportional to Young’s modulus.

1.3

To a good approximation, the heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case. (As we will see in the next section, the answer would have been different if the object had been made of some substance that changes phase anywhere between 30°C30°C and 50°C50°C.)

1.4

The ice and liquid water are in thermal equilibrium, so that the temperature stays at the freezing temperature as long as ice remains in the liquid. (Once all of the ice melts, the water temperature will start to rise.)

1.5

Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be transferred from the air, even if the air is above 0°C0°C.

1.6

Conduction: Heat transfers into your hands as you hold a hot cup of coffee. Convection: Heat transfers as the barista “steams” cold milk to make hot cocoa. Radiation: Heat transfers from the Sun to a jar of water with tea leaves in it to make “Sun tea.” A great many other answers are possible.

1.7

Because area is the product of two spatial dimensions, it increases by a factor of four when each dimension is doubled (Afinal=(2d)2=4d2=4Ainitial)(Afinal=(2d)2=4d2=4Ainitial). The distance, however, simply doubles. Because the temperature difference and the coefficient of thermal conductivity are independent of the spatial dimensions, the rate of heat transfer by conduction increases by a factor of four divided by two, or two:
Pfinal=kAfinal(ThTc)dfinal=k(4Afinal(ThTc))2dinitial=2kAfinal(ThTc)dinitial=2PinitialPfinal=kAfinal(ThTc)dfinal=k(4Afinal(ThTc))2dinitial=2kAfinal(ThTc)dinitial=2Pinitial.

1.8

Using a fan increases the flow of air: Warm air near your body is replaced by cooler air from elsewhere. Convection increases the rate of heat transfer so that moving air “feels” cooler than still air.

1.9

The radiated heat is proportional to the fourth power of the absolute temperature. Because T1=293KT1=293K and T2=313KT2=313K, the rate of heat transfer increases by about 30% of the original rate.

Conceptual Questions

1.

They are at the same temperature, and if they are placed in contact, no net heat flows between them.

3.

The reading will change.

5.

The cold water cools part of the inner surface, making it contract, while the rest remains expanded. The strain is too great for the strength of the material. Pyrex contracts less, so it experiences less strain.

7.

In principle, the lid expands more than the jar because metals have higher coefficients of expansion than glass. That should make unscrewing the lid easier. (In practice, getting the lid and jar wet may make gripping them more difficult.)

9.

After being heated, the length is (1+300α1+300α) (1m1m). After being cooled, the length is (1300α)(1+300α)(1m)(1300α)(1+300α)(1m). That answer is not 1 m, but it should be. The explanation is that even if αα is exactly constant, the relation ΔL=αLΔTΔL=αLΔT is strictly true only in the limit of small ΔTΔT. Since αα values are small, the discrepancy is unimportant in practice.

11.

Temperature differences cause heat transfer.

13.

No, it is stored as thermal energy. A thermodynamic system does not have a well-defined quantity of heat.

15.

It raises the boiling point, so the water, which the food gains heat from, is at a higher temperature.

17.

Yes, by raising the pressure above 56 atm.

19.

work

21.

0°C0°C (at or near atmospheric pressure)

23.

Condensation releases heat, so it speeds up the melting.

25.

Because of water’s high specific heat, it changes temperature less than land. Also, evaporation reduces temperature rises. The air tends to stay close to equilibrium with the water, so its temperature does not change much where there’s a lot of water around, as in San Francisco but not Sacramento.

27.

The liquid is oxygen, whose boiling point is above that of nitrogen but whose melting point is below the boiling point of liquid nitrogen. The crystals that sublime are carbon dioxide, which has no liquid phase at atmospheric pressure. The crystals that melt are water, whose melting point is above carbon dioxide’s sublimation point. The water came from the instructor’s breath.

29.

Increasing circulation to the surface will warm the person, as the temperature of the water is warmer than human body temperature. Sweating will cause no evaporative cooling under water or in the humid air immediately above the tub.

31.

It spread the heat over the area above the heating elements, evening the temperature there, but does not spread the heat much beyond the heating elements.

33.

Heat is conducted from the fire through the fire box to the circulating air and then convected by the air into the room (forced convection).

35.

The tent is heated by the Sun and transfers heat to you by all three processes, especially radiation.

37.

If shielded, it measures the air temperature. If not, it measures the combined effect of air temperature and net radiative heat gain from the Sun.

39.

Turn the thermostat down. To have the house at the normal temperature, the heating system must replace all the heat that was lost. For all three mechanisms of heat transfer, the greater the temperature difference between inside and outside, the more heat is lost and must be replaced. So the house should be at the lowest temperature that does not allow freezing damage.

41.

Air is a good insulator, so there is little conduction, and the heated air rises, so there is little convection downward.

Problems

43.

That must be Celsius. Your Fahrenheit temperature is 102°F.102°F. Yes, it is time to get treatment.

45.

a. ΔTC=22.2°CΔTC=22.2°C; b. We know that ΔTF=TF2TF1ΔTF=TF2TF1. We also know that TF2=95TC2+32TF2=95TC2+32 and TF1=95TC1+32.TF1=95TC1+32. So, substituting, we have ΔTF=(95TC2+32)(95TC1+32)ΔTF=(95TC2+32)(95TC1+32). Partially solving and rearranging the equation, we have ΔTF=95(TC2TC1)ΔTF=95(TC2TC1). Therefore, ΔTF=95ΔTCΔTF=95ΔTC.

47.

a. −40°−40°; b. 575 K

49.

Using Table 1.2 to find the coefficient of thermal expansion of marble:
L=L0+ΔL=L0(1+αΔT)=170m[1+(2.5×10−6/°C)(−45.0°C)]=169.98mL=L0+ΔL=L0(1+αΔT)=170m[1+(2.5×10−6/°C)(−45.0°C)]=169.98m.
(Answer rounded to five significant figures to show the slight difference in height.)

51.

Using Table 1.2 to find the coefficient of thermal expansion of mercury:
ΔL=αLΔT=(6.0×10−5/°C)(0.0300m)(3.00°C)=5.4×10−6mΔL=αLΔT=(6.0×10−5/°C)(0.0300m)(3.00°C)=5.4×10−6m.

53.

On the warmer day, our tape measure will expand linearly. Therefore, each measured dimension will be smaller than the actual dimension of the land. Calling these measured dimensions l'l' and w'w', we will find a new area, A. Let’s calculate these measured dimensions:
l'=l0Δl=(20m)(20°C)(20m)(1.2×10−5°C)=19.9952ml'=l0Δl=(20m)(20°C)(20m)(1.2×10−5°C)=19.9952m;
A'=l×w'=(29.9928m)(19.9952m)=599.71m2A'=l×w'=(29.9928m)(19.9952m)=599.71m2;
Cost change =(AA')($60,000m2)=((600599.71)m2)($60,000m2)=$17,000=(AA')($60,000m2)=((600599.71)m2)($60,000m2)=$17,000.
Because the area gets smaller, the price of the land decreases by about $17,000.

55.

a. Use Table 1.2 to find the coefficients of thermal expansion of steel and aluminum. Then ΔLAlΔLsteel=(αAlαsteel)L0ΔT=(2.5×10−5°C1.2×10−5°C)(1.00m)(22°C)=2.9×10−4mΔLAlΔLsteel=(αAlαsteel)L0ΔT=(2.5×10−5°C1.2×10−5°C)(1.00m)(22°C)=2.9×10−4m.
b. By the same method with L0=30.0mL0=30.0m, we have ΔL=8.6×10−3mΔL=8.6×10−3m.

57.

ΔV=0.475LΔV=0.475L

59.

If we start with the freezing of water, then it would expand to (1m3)(1000kg/m3917kg/m3)=1.09m3=1.98×108N/m2(1m3)(1000kg/m3917kg/m3)=1.09m3=1.98×108N/m2 of ice.

61.

m=5.20×108Jm=5.20×108J

63.

Q=mcΔTΔT=QmcQ=mcΔTΔT=Qmc; a. 21.0°C21.0°C; b. 25.0°C25.0°C; c. 29.3°C29.3°C; d. 50.0°C50.0°C

65.

Q=mcΔTc=QmΔT=1.04kcal(0.250kg)(45.0°C)=0.0924kcal/kg·°CQ=mcΔTc=QmΔT=1.04kcal(0.250kg)(45.0°C)=0.0924kcal/kg·°C. It is copper.

67.

a. Q=mwcwΔT+mA1cA1ΔT=(mwcw+mA1cA1)ΔTQ=mwcwΔT+mA1cA1ΔT=(mwcw+mA1cA1)ΔT;
Q=[(0.500kg)(1.00kcal/kg·°C)+(0.100kg)(0.215kcal/kg·°C)](54.9°C)=28.63kcalQ=[(0.500kg)(1.00kcal/kg·°C)+(0.100kg)(0.215kcal/kg·°C)](54.9°C)=28.63kcal;
Qmp=28.63kcal5.00g=5.73kcal/gQmp=28.63kcal5.00g=5.73kcal/g; b. Qmp=200kcal33g=6kcal/g,Qmp=200kcal33g=6kcal/g, which is consistent with our results to part (a), to one significant figure.

69.

0.139°C0.139°C

71.

It should be lower. The beaker will not make much difference: 16.3°C16.3°C

73.

a. 1.00×105J1.00×105J; b. 3.68×105J3.68×105J; c. The ice is much more effective in absorbing heat because it first must be melted, which requires a lot of energy, and then it gains the same amount of heat as the bag that started with water. The first 2.67×105J2.67×105J of heat is used to melt the ice, then it absorbs the 1.00×105J1.00×105J of heat as water.

75.

58.1 g

77.

Let M be the mass of pool water and m be the mass of pool water that evaporates.
McΔT=mLV(37°C)mM=cΔTLV(37°C)=(1.00kcal/kg·°C)(1.50°C)580kcal/kg=2.59×10−3McΔT=mLV(37°C)mM=cΔTLV(37°C)=(1.00kcal/kg·°C)(1.50°C)580kcal/kg=2.59×10−3;
(Note that LVLV for water at 37°C37°C is used here as a better approximation than LVLV for 100°C100°C water.)

79.

a. 1.47×1015kg1.47×1015kg; b. 4.90×1020J4.90×1020J; c. 48.5 y

81.

a. 9.67 L; b. Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less able to absorb the heat generated by the oil.

83.

a. 319 kcal; b. 2.00°C2.00°C

85.

First bring the ice up to 0°C0°C and melt it with heat Q1:Q1: 4.74 kcal. This lowers the temperature of water by ΔT2:ΔT2: 23.15°C23.15°C. Now, the heat lost by the hot water equals that gained by the cold water (TfTf is the final temperature): 20.6°C20.6°C

87.

Let the subscripts r, e, v, and w represent rock, equilibrium, vapor, and water, respectively.
mrcr(T1Te)=mVLV+mWcW(TeT2)mrcr(T1Te)=mVLV+mWcW(TeT2);
mr=mVLV+mWcW(TeT2)cr(T1Te)=(0.0250kg)(2256×103J/kg)+(3.975kg)(4186×103J/kg·°C)(100°C15°C)(840J/kg·°C)(500°C100°C)=4.38kgmr=mVLV+mWcW(TeT2)cr(T1Te)=(0.0250kg)(2256×103J/kg)+(3.975kg)(4186×103J/kg·°C)(100°C15°C)(840J/kg·°C)(500°C100°C)=4.38kg

89.

a. 1.01×103W1.01×103W; b. One 1-kilowatt room heater is needed.

91.

84.0 W

93.

2.59 kg

95.

a. 39.7 W; b. 820 kcal

97.

Qt=kA(T2T1)dQt=kA(T2T1)d, so that
(Q/t)wall(Q/t)window=kwallAwalldwindowkwindowAwindowdwall=(2×0.042J/s·m·°C)(10.0m2)(0.750×10−2m)(0.84J/s·m·°C)(2.00m2)(13.0×10−2m)(Q/t)wall(Q/t)window=kwallAwalldwindowkwindowAwindowdwall=(2×0.042J/s·m·°C)(10.0m2)(0.750×10−2m)(0.84J/s·m·°C)(2.00m2)(13.0×10−2m)
This gives 0.0288 wall: window, or 35:1 window: wall

99.

Qt=kA(T2T1)d=kAΔTdQt=kA(T2T1)d=kAΔTd
ΔT=d(Q/t)kA=(6.00×10−3m)(2256W)(0.84J/s·m·°C)(1.54×10−2m2)=1046°C=1.05×103KΔT=d(Q/t)kA=(6.00×10−3m)(2256W)(0.84J/s·m·°C)(1.54×10−2m2)=1046°C=1.05×103K

101.

We found in the preceding problem that P=126ΔTW·°CP=126ΔTW·°C as baseline energy use. So the total heat loss during this period is Q=(126J/s·°C)(15.0°C)(120days)(86.4×103s/day)=1960×106JQ=(126J/s·°C)(15.0°C)(120days)(86.4×103s/day)=1960×106J. At the cost of $1/MJ, the cost is $1960. From an earlier problem, the savings is 12% or $235/y. We need 150m2150m2 of insulation in the attic. At $4/m2$4/m2, this is a $500 cost. So the payback period is $600/($235/y)=2.6years$600/($235/y)=2.6years (excluding labor costs).

Additional Problems

103.

7.39%7.39%

105.

FA=(210×109Pa)(12×10−6/°C)(40°C(−15°C))=1.4×108N/m2FA=(210×109Pa)(12×10−6/°C)(40°C(−15°C))=1.4×108N/m2.

107.

a. 1.061.06 cm; b. 1.111.11 cm

109.

1.7kJ/(kg·ºC)1.7kJ/(kg·ºC)

111.

a. 1.57×104kcal1.57×104kcal; b. 18.3kW·h18.3kW·h; c. 1.29×104kcal1.29×104kcal

113.

6.3°C6.3°C. All of the ice melted.

115.

63.9°C63.9°C, all the ice melted

117.

a. 83 W; b. 1.97×103W1.97×103W; The single-pane window has a rate of heat conduction equal to 1969/83, or 24 times that of a double-pane window.

119.

The rate of heat transfer by conduction is 20.0 W. On a daily basis, this is 1,728 kJ/day. Daily food intake is 2400kcal/d×4186J/kcal=10,050kJ/day2400kcal/d×4186J/kcal=10,050kJ/day. So only 17.2% of energy intake goes as heat transfer by conduction to the environment at this ΔTΔT.

121.

620 K

Challenge Problems

123.

Denoting the period by P, we know P=2πL/g.P=2πL/g. When the temperature increases by dT, the length increases by αLdTαLdT. Then the new length is a. P=2πL+αLdTg=2πLg(1+αdT)=2πLg(1+12αdT)=P(1+12αdT)P=2πL+αLdTg=2πLg(1+αdT)=2πLg(1+12αdT)=P(1+12αdT)
by the binomial expansion. b. The clock runs slower, as its new period is 1.00019 s. It loses 16.4 s per day.

125.

The amount of heat to melt the ice and raise it to 100°C100°C is not enough to condense the steam, but it is more than enough to lower the steam’s temperature by 50°C50°C, so the final state will consist of steam and liquid water in equilibrium, and the final temperature is 100°C100°C; 9.5 g of steam condenses, so the final state contains 49.5 g of steam and 40.5 g of liquid water.

127.

a. dL/dT=kT/ρLdL/dT=kT/ρL; b. L=2kTt/ρLfL=2kTt/ρLf; c. yes

129.

a. 4(πR2)Ts44(πR2)Ts4; b. 4eσπR2Ts44eσπR2Ts4; c. 8eσπR2Te48eσπR2Te4; d. Ts4=2Te4Ts4=2Te4; e. eσTs4+14(1A)S=σTs4eσTs4+14(1A)S=σTs4; f. 288K288K

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction
Citation information

© Oct 6, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.