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17-1

Conceptual Questions

17.1. No. There is not a transfer of energy due to a temperature difference. When the shuttle re-enters the

atmosphere there is friction between the molecules of the atmosphere and the shuttle. The mechanical energy of the

shuttle is transformed into thermal energy. The initial kinetic energy of the shuttle decreases.

17.2. (a) Temperature is a property of a system related to the thermal energy per molecule.

(b) Heat is the energy transferred between the system and the environment. Heat may cause the thermal energy to

change, but that doesn’t mean that heat and thermal energy are the same thing.

(c) Thermal energy is a state variable and property of the system. It continues to exist even if the system is isolated

and not interacting with the environment.

17.3. Greater than. A V A B P B 10 JQ n C T n C T and A B V A P B.n n C T C T Since P VC C

A B.T T But Ai Bi Af BfT .T T T

17.4. To use the least amount of heat energy you should heat the gas at constant volume. Then

V P V P.nC T nC T C C

17.5. In a constant-volume process all heat input is used to increase the temperature of the gas. But in a constant-

pressure process, since some work is done, some energy leaves the system as work. So more heat is needed to

produce the same P V: .T C C

17.6. (a) Lower.

(b) No. The temperature and thermal energy decrease because of the work done by the gas on the surroundings.

17.7. Even though the initial and final positions on the diagram are the same for both paths, the work done is not

equal. The work done on the gas is the negative of the area under the pV curve i f and V V . There is more area under

the A path than under the B path, so A B.W W

17.8. (a) There is one particular temperature corresponding to point i and another corresponding to point f, and the

T between the two will be the same no matter which path is taken.

(b) A A B B.W Q W Q The area under process A is larger than the area under B, so AW is more negative than B,W

so AQ has to be more positive than BQ to maintain the equality. Therefore, more heat energy is added for process A

than for process B.

WORK, HEAT, AND THE FIRST LAW OF THERMODYNAMICS

17

17-2 Chapter 17


17.9.

This is a case of gas compression and therefore W pdV is a positive quantity. The final point is on a lower

isotherm than the initial point, so f i.T T Heat energy is transferred out of the gas ( 0 J)Q and the thermal energy

of the gas decreases as the temperature falls. That is, th f th i.E E To bring about this process: (1) The locking pin is

removed so the piston can slide up and down. (2) The masses on the top of the piston are not changed. This keeps the

pressure constant. (3) The ice block is brought into contact with the bottom of the cylinder.

17.10.

For the isothermal process 0 K.T This means the first law of thermodynamics can only be satisfied if .W Q When

the gas compresses, 0 JW implying 0 J.Q Heat energy is transferred out of the gas, but the temperature of the gas

does not change. For the isochoric process, 0W J because the volume does not change. Since the final point is on a

lower isotherm, the final temperature is lower. That is, heat energy is transferred out of the gas ( 0)Q and hence the

thermal energy of the gas decreases. To bring about this process: (1) Place the cylinder on the ice. The gas will begin to

contract as heat energy is transferred to the ice. (2) Add masses on the top of the piston to increase the pressure and keep

pV constant as the volume decreases. (3) At the desired volume and pressure, insert a locking pin into the piston to fix the

volume. (4) Keep the ice in contact with the bottom of the cylinder until the original pressure is obtained in the gas.

17.11. (a) (i) 0T because the initial temperature is 0 C and the final temperature is 0 C. (ii) 0W because

the volume decreases. (iii) 0Q because energy is transferred due to contact with the cold ice.

(b)

Work, Heat, and the First Law of Thermodynamics 17-3


17.12. (a) (i) 0T because the gas expands adiabatically. (ii) 0W because the volume increases. (iii) 0Q

because the system is thermally insulated.

(b)

Exercises and Problems

Section 17.1 It’s All About Energy

Section 17.2 Work in Ideal-Gas Processes

17.1. Model: Assume the gas is ideal. The work done on a gas is the negative of the area under the pV curve.

Visualize: The gas is undergoing compression, so we expect the work done on it to be positive.

Solve: The work done on the gas is the negative of the area under the pV curve from iV to f ,V which is the area of the

rectangle and triangle. We have

f

i

6 3 3 6 3 312

(200 10 m )(200 10 Pa) (200 10 m )(200 10 Pa) 60 J

V

V

W pdV

where the inner negative signs come from the fact that f i.V V The work done on the gas by the environment is thus

60 J.W

Assess: The environment does positive work on the gas to compress it.


Visualize: The gas expands, so we expect the work done on the gas by the environment to be negative.

Solve: The work done on the gas is

f

i

6 3 6 3

(area under the curve)

[(400 kPa)(300 10 m 100 10 m )] 80 J

V

V

W pdV pV

Assess: The area under the curve is positive because the integration f i.V V Thus, the work is negative so the

environment does negative work on the gas to compress it, because the environmental pressure pushes against the

volume expansion.


Solve: The work done on gas in an isobaric process is

f i( )W p V p V V

Substituting into this equation, 3 4 3 3

1 1 i80 J (200 10 Pa)( 3 ) 2.0 10 m 200 cmV V V

Assess: The work done to compress a gas is positive.

17-4 Chapter 17


17.4. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes.

Solve: (a) Since the pressure i f( )p p p is constant the work done is

ion gas f i f i

i

3 3

3

( ) ( )

(1000 cm 2000 cm )(0.10 mol)(8.31 J/mol K)(573 K) 240 J

2000 cm

nRTW p V p V V V V

V

(b) For compression at a constant temperature,

on gas f i

6 3

6 3

ln( / )

1000 10 m(0.10 mol)(8.31J/mol K)(573 K)ln 330 J

2000 10 m

W nRT V V

Section 17.3 Heat

Section 17.4 The First Law of Thermodynamics

17.5. Visualize:

Solve: Because this is an isobaric process f i( )W pdV p V V . Since fV is greater than i ,V W is negative. That is,

the gas expands. Since the final point is on a higher isotherm than the initial point, f i.T T In other words, the thermal

energy increases. For this to happen, the heat energy transferred into the gas must be larger than the work done by the gas.

17.6. Visualize:

Solve: Because W pdV and this is an isochoric process, 0 J.W The final point is on a lower isotherm than

the initial point, so f i.T T Heat energy is thus transferred out of the gas ( 0)Q and the thermal energy of the gas

decreases th f th i( )E E as the temperature decreases.



17.7. Visualize:

Solve: This is an adiabatic process of gas compression so no heat energy is transferred between the gas and the

environment. That is, 0 J.Q According to the first law of thermodynamics, the work done on a gas in an adiabatic

process goes entirely to changing the thermal energy of the gas. The work W is positive because the gas is

compressed.

17.8. Visualize:

Solve: Because the process is isothermal, th th f th i 0 J.E E E According to the first law of thermodynamics,

th .E W Q This can only be satisfied if .W Q W is positive because the gas is compressing, hence Q is

negative. That is, heat energy is removed from the gas.

17.9. Solve: This is an isobaric process. 0W because the gas is compressed, which transfers energy into the

system. Also, 100 J of heat energy is transferred out of the gas. The first law of thermodynamics gives 5 6 3

th (4.0 10 Pa)(200 600) 10 m 100 J 60 JE W Q p V Q

Thermal energy increases by 60 J.

17.10. Solve: The first law of thermodynamics gives

th 200 J 500 J 700 JE W Q Q Q Q

The negative sign means a transfer of energy from the system to the environment.

Assess: Because 0W energy is transferred into the system, Q must be less than zero and larger in magnitude than

W, so th f th i.E E

Section 17.5 Thermal Properties of Matter

17.11. Model: The addition of heat to the aluminum increases its thermal energy and its temperature.

Solve: The heat needed to change an object’s temperature is .Q Mc T The mass of the aluminum cube is

3 3Al (2700 kg/m )(0.060 0.060 0.060)m 0.5832 kgM V

17-6 Chapter 17


The specific heat of ice from Table 17.2 is Al 900 J/kg K,c so

(0.5832 kg)(900 J/kg K)(100 K) 52 kJQ

Thus, the energy required to heat the Al block is 52 kJ.

Assess: The positive sign of Q means the system (i.e., Al block) gains energy.

17.12. Model: The spinning paddle wheel does work and changes the water’s thermal energy and its temperature.

Solve: (a) 0.Q No energy is transferred between the system and the environment because of a difference in

temperature.

(b) The temperature change is f i 25 C 21 C 4 K.T T T The mass of the water is

6 3 3(200 10 m )(1000 kg/m ) 0.20 kgM

The work done is

th water (0.20 kg)(4190 J/kg K)(4 K) 3352 J 3.4 kJW E Mc T

17.13. Model: Heating the mercury changes its thermal energy and its temperature.

Solve: (a) The heat needed to change the mercury’s temperature is

HgHg

100 J35.7 K 36 C

(0.020 kg)(140 J/kg K)

QQ Mc T T

Mc

(b) The amount of heat required to raise the temperature of the same amount of water by the same amount is

water (0.020 kg)(4190 J/kg K)(35.7 K) 3000 JQ Mc T

Assess: Q is directly proportional to waterc and the specific heat for water is much higher than the specific heat for

mercury. This explains why water mercury.Q Q

17.14. Model: Heating the mercury at its boiling point changes its thermal energy without a change in temperature.

Solve: The mass of the mercury is 220 g 2.0 10 kg,M the specific heat mercury 140 J/kg K,c the boiling

point b 357 C,T and the heat of vaporization 5v 2.96 10 J/kg.L The heat required for the mercury to change to

the vapor phase is the sum of two steps. The first step is 2

1 mercury (2.0 10 kg)(140 J/kg K)(357 C 20 C) 940 JQ Mc T

The second step is 2 5

2 V (2.0 10 kg)(2.96 10 J/kg) 5920 JQ ML

The total heat needed is 940 J 5920 J 6860 J or, to two significant figures, 6.9 kJ.

17.15. Model: Changing liquid ethyl alcohol at 20 C to gaseous ethyl alcohol at its boiling point b 78 CT

requires two steps: raising the temperature to bT and then vaporizing the liquid at bT to a gas at b.T

Solve: The total heat is the sum of these two contributions, so

1 2 EA f i v

5

1000 J ( )

1000 J1000 J (2400 J/kg K)(78 K 20 K) (8.79 10 J/kg) 0.98 g

1,018,200 J/kg

Q Q Q Mc T T ML

M M M

The maximum mass of 20 C ethyl alcohol you can vaporize with 1000 J of heat is 0.98 g.

Section 17.6 Calorimetry

17.16. Model: We have a thermal interaction between the copper pellets and the water.

Solve: The conservation of energy equation Cu w 0Q W is

Cu Cu f w w f( 300 C) ( 20 C) 0 JM c T M c T



Solving this equation for the final temperature fT gives

Cu Cu w wf

Cu Cu w w

(300 C) (20 C)

(0.030 kg)(385 J/kg K )(300 C) (0.10 kg)(4190 J/kg K )(20 C)28 C

(0.030 kg)(385 J/kg K ) (0.10 kg)(4190 J/kg K )

M c M cT

M c M c

The final temperature of the water and the copper is 28C.

17.17. Model: We have a thermal interaction between the aluminum pan and the water.

Solve: The conservation of energy equation Al water 0 JQ Q gives

Al Al f i Al water water f i water[ ( ) ] [ ( ) ]M c T T M c T T

The pan and water reach a common final temperature f 24.0 CT

3 3 3i Al

i Al

(0.750 kg)(900 J/kg K)[24.0 C ( ) ] (10.0 10 m )(1000 kg/m )(4190 J/kg K)(24.0 C 20.0 C)

(675.0 J/K)[24.0 C ( ) ] 167,600 J 0 J

T

T

i Al( ) 272 C [(272)(9/5) 32] F 522 FT

17.18. Model: We have a thermal interaction between the thermometer and the water.

Solve: The conservation of energy equation thermo water 0 JQ Q gives

thermo thermo f i thermo water water f i water[ ( ) ] [ ( ) ] 0 JM c T T M c T T

The thermometer slightly cools the water until both have the same final temperature f 71.2 C.T Thus

6 3 3i water

i water

i water

0 J (0.0500 kg)(750 J/kg K)(71.2 C 20.0 C) (200 10 m )(1000 kg/m )(4190 J/kg K)[71.2 C ( ) ]

1920 J 838 (J/K)[71.2 C ( ) ]

( ) 73.5 C

T

T

T

Assess: The thermometer reads 71.2 C for a real temperature of 73.5 C. This is reasonable.

17.19. Model: We have a thermal interaction between the metal sphere and the mercury.

Solve: The conservation of energy equation metal Hg 0 JQ Q gives

metal metal f i metal Hg Hg f i Hg[ ( ) ] [ ( ) ] 0 JM c T T M c T T

The metal and mercury reach a common final temperature f 99.0 C.T Thus

6 3 3metal(0.500 kg) (99.0 C 300 C) (300 10 m )(13,600 kg/m )(140 J/kg K)(99.0 C 20.0 C) 0 Jc

We find that metal 449 J/kg K.c The metal is iron.

17.20. Model: We have a thermal interaction between the iron block and the water.

Solve: The mass of the iron block is

3 6 3 (7870 kg/m )(65 10 m ) = 0.512 kg.V The iron will heat all the water

to 100 C then cause a phase transformation of a mass sM of water into steam. The conservation of energy equation

Fe w 0 J givesQ Q

Fe Fe f i Fe w w f i w s v w[ ( ) ] [ ( ) ] ( ) 0 JM c T T M c T T M L

The final temperature of the remaining water and the iron block is 100C. Solving for sM gives

Fe Fe f i Fe w w f i ws

v w

5

2

[ ( ) ] [ ( ) ]

( )

(0.512 kg)(449 J/kg K)(100 C 800 C) (0.200 kg)(4190 J/kg K)(100 C 20 C)

22.6 10 J/kg

4.15 10 kg

M c T T M c T TM

L

Thus, the fraction f of the initial water mass that boils away is 2(4.15 10 kg)/(0.200 kg) 21%.f

17-8 Chapter 17


Section 17.7 The Specific Heat of Gases

17.21. Model: Use the models of isochoric and isobaric heating. Note that the change in temperature on the Kelvin

scale is the same as the change in temperature on the Celsius scale.

Solve: (a) The atomic mass number of argon is 40. That is, mol 40 g/mol.M The number of moles of argon gas in

the container is

mol

1.0 g0.025 mol

40 g/mol

Mn

M

The amount of heat is

V (0.025 mol)(12.5 J/mol K)(100 C) 31.25 J 31 JQ nC T

(b) For the isobaric process PQ nC T becomes

31.25 J (0.025 mol)(20.8 J/mol K) 60 CT T

17.22. Model: The heating processes are isobaric and isochoric. 2O is a diatomic ideal gas.

Solve: (a) The number of moles of oxygen is

mol

1.0 g0.03125 mol

32 g/mol

Mn

M

For the isobaric process,

p (0.03125 mol)(29.2 J/mol K)(100 C) 91.2 J 91 JQ nC T

(b) For the isochoric process,

V 91.2 J (0.03125 mol)(20.9 J/mol K) 140 CQ nC T T T

17.23. Model: 2N is a diatomic ideal gas. It condenses at 196 C (see Table 17.3).

Solve: The number of moles of nitrogen is

mol

7.0 g0.25 mol

28 g/mol

Mn

M

Less energy is required to cool the gas in an isochoric process because no work is done, so the minimum heat that

must be lost by the gas to cool it to the condensation point is

V (0.25 mol)(20.8 J/mol K)( 196 C 20 C) 1123 JQ nC T

To condense this to a liquid, the following additional heat of vaporization must be removed: 5

v (0.0070 kg)(1.99 10 J/kg) 1393 JQ ML

Thus, the total heat that must be removed from the gas is1123 J 1393 J 2.5 kJ, where we have taken positive sign

because this is the heat that we (the environment) must remove.

Assess: Because the gas loses this heat, the sign of its heat transfer is negative.

17.24. Model: The gas is an ideal gas that is subjected to an adiabatic process.

Solve: (a) For an adiabatic process,

f if if i

i f

ln(2.5)2.5 (2.0) 1.3

ln(2.0)

p Vp V pV

p V

(b) Equation 17.37 for an adiabatic process is 1

1 1 1.32 1 0.32f if if i

i f

(2.0) (2.0) 1.3T V

T V TVT V

17.25. Model: We assume the gas is an ideal gas and 1.40 for a diatomic gas.

Solve: Using the ideal-gas law,

3 3ii 5

i

(0.10 mol)(8.31J/mol K)(423 K)1.157 10 m

(3 1.013 10 Pa)

nRTV

p



(a) For an adiabatic process,

i fi fpV p V

1/ 1/1.40

3 3 3 3i if i

f i

(1.157 10 m ) 1.9 10 m0.5

p pV V

p p

(b) To find the final temperature, we use the ideal-gas law once again as follows:

3 3f f i

f i 3 3i i i

0.5 1.90 10 m(423 K) 346.9 K 74 C

1.157 10 m

p V pT T

p V p

17.26. Model: 2O gas has 1.40 and is an ideal gas.

Solve: (a) For an adiabatic process, pV remains a constant. That is,

1.40 1.40i i

i f f ii ff i

1(3.0 atm) (3.0 atm) 1.14 atm 1.1 atm

2 2

V VpV p V p p

V V

(b) Using the ideal-gas law, the final temperature of the gas is calculated as follows:

i i f f f f if i

i f i i i

1.14 atm 2(423 K) 321.5 K 48 C

3.0 atm

pV p V p V VT T

T T p V V

Section 17.8 Heat-Transfer Mechanisms

17.27. Model: We are asked for the heat-loss rate which is given by Equation 17.46:

Q Ak T

t L

We are given 210 m 14 m 140 m ,A 0 12 m,L . and 22 C 5 C 17 C 17 K.T D We look up the thermal

conductivity of concrete in Table 17.5: 0 8 W/m K.k .

Solve:

2140 m(0 8 W/m K) (17 K) 16 kW

0 12 m

Q Ak T .

t L .

Assess: The answer is in a reasonable range. The heat loss could be reduced with thicker concrete or adding a layer

of a different material.

17.28. Model: To determine the material we will solve for k in Equation 17.46:

Q Ak T

t L

We are given 0 20 mL . and 100K.T We compute 2 2 24 2(0 010m) 3 14 10 m .A r . .

We also convert the heat conduction to watts.

4 1 h4 5 10 J/h 12 5 W

3600 s. .

Solve: Solve the equation for .k

24 2

1 0 20 m 1(12 5 W) 80 W/m K

100 K3 14 10 m

Q L .k .

t A T .

Look this up in Table 17.5; the value corresponds to iron.

Assess: We are grateful that our answer was one of the entries in the table.

17.29. Model: Assume the lead sphere is an ideal radiator with 1.e Also assume that the highest temperature the

solid lead sphere can have is the melting temperature of lead.

Visualize: Use Equation 17.47. First look up the melting temperature of lead in Table 17.3: m 328 C 601 K.T

Then compute the surface area of the sphere: 2 2 24 4 (0 050 m) 0 0314 mA R . .

17-10 Chapter 17


Solve:

4 8 2 4 2 4(1)(5 67 10 W/m K )(0 0314 m )(601 K) 230 WQ

e AT . .t

Assess: If the sphere were larger it could radiate more power without melting.

17.30. Model: We will ignore the bottom of the head and model it with just the cylindrical sides and top, all

covered in skin. As instructed, assume an emissivity of 0 95.e .

Visualize: The area of the cylinder is

2 2 2side top 2 2 (0 10 m)(0 20 m) (0 10 m) 0 157 m .A A A rh r . . . .

Solve: Use Equation 17.49.

4 4 8 2 4 2 4 4net0( ) (0 95)(5 67 10 W/m K )(0 157 m )[(308 K) (278 K) ] 26 W

Qe A T T . . .

t

Assess: This is a significant amount of the heat lost by the body. Wearing a hat can help a lot.

17.31. Model: There are various steps to the problem. We must (1) raise the temperature of the ice to 0 C, (2)

melt the ice to liquid water at 0 C, (3) raise the water temperature to 100 C, (4) boil the water to produce steam at

100 C, and (5) raise the temperature of the steam to 200 C.

Solve: The heat needed for each step is

1 ice ice

52 f

3 water water

54 v

5 steam st

(0 0050 kg)(2090 J/kgK)(20 K) 209 J

(0 0050 kg)(3 33310 J/kg) 1665 J

(0 0050 kg)(4190 J/kgK)(100 K) 2095 J

(0 0050 kg)(22 6310 J/kg) 11,300 J

Q Mc T .

Q ML . .

Q Mc T .

Q ML . .

Q Mc T

eam (0 0050 kg)(2009 J/kgK)(100 K) 1005 J.

The total heat is

1 2 3 4 5 16 kJQ Q Q Q Q Q

Assess: It is interesting to note which step required the most heat: The boiling to change the liquid to the gas is by

far the largest contributor to the total.

17.32. Solve: The area of the garden pond is

2 2(2.5 m) 19.635 mA and its volume is

3(0.30 m) 5.891 m .V A

The mass of water in the pond is 3 3(1000 kg/m )(19.635 m ) 5891 kgM V

The water absorbs all the solar power, which is 2 2(400 w/m )(19.635 m ) 7854 W

This power is used to raise the temperature of the water. That is,

water(7854 W) (5891 kg)(4190 J/kg K)(10 K) 31,425 s 8.7 hQ t Mc T t

17.33. Model: The potential energy of the bowling ball is transferred into thermal energy of the mixture. We

assume the starting temperature of the bowling ball to be 0 C.

Solve: The potential energy of the bowling ball is 2 2

g ball (11 kg)(9.8 m/s ) (107.8 kgm/s )U M gh h h

This energy is transferred into the mixture of ice and water and melts 5 g of ice. That is, 5

2th w f 2

(0.005 kg)(3.33 10 J/kg)(107.8 kgm/s ) 15 m

(107.8 kgm/s )h E M L h

17.34. Model: Heating the water and the kettle raises the temperature of the water to the boiling point and raises

the temperature of the kettle to 100 C.

Solve: The amount of heat energy from the electric stove’s output in 3 minutes is 5(2000 W)(3.0 min 60 s/min) 3.6 10 JQ



This heat energy heats the kettle and brings the water to a boil. Thus,

water water kettle kettleQ M c T M c T

Substituting the given values into this equation,

5water

water

3.6 10 J (4190 J/kgK)(100 C 20 C) (0.750 kg)(449 J/kgK)(100 C 20 C)

0.994 kg

M

M

The volume of water in the kettle is

3 3 3 3water3

water

0.994 kg0.994 10 m 994 cm 990 cm

1000 kg/m

MV

Assess: 3 31L 10 cm , so 1L.V This is a reasonable volume of water.

17.35. Model: Model the alligator as a rectangular solid with an upper surface 2 22.9 0.60 m 1.74 m .

Solve: Apply Equation 17.18 to find the heat required to warm the alligator by 7 C:

(350 kg)(3400 J/kgK)(7K) 8.33 MJQ Mc T

To acquire this energy from sunlight would take approximately

4

2 2

8.33 MJ 1min 1h(9.57 10 s) 2.7 h

60 s 60 min(500 W/m )(1.74 m )

Qt

P

17.36. Model: Assume sweat is mostly water and that all the energy required to heat and vaporize the sweat comes

from the body (as opposed to the air surrounding the body).

Solve: To vaporize 30 g of 35°C water every minute requires a power

52 v

1

(0.030 kg)(24 10 J/kg)1200 J/s

(60 s) (60 s) 60 s

Q MLP

17.37. Model: Model evaporation in the same manner as for the previous problem.

Solve: At 12 breaths/min, we exhale every minute

6 6(12)(25 10 kg) 300 10 kgM of water. To vaporize 300 g

of water at 35°C every minute requires an approximate power of

6 52 v

2

(300 10 kg)(24 10 J/kg)12 J/s

(60 s) (60 s) 60 s

Q MLP

Assess: This is much less than the energy lost from the body by radiation.

17.38. Model: Each car’s kinetic energy is transformed into thermal energy.

Solve: For each car,

221

th car2car2

vK Mv E Mc T T

c

Assume car iron.c c The speed of the car is

280 1000 m (22.22 m/s)80 km/h = 22.22 m/s 0.55 C

3600 s 2(449 J/kgK)v T

Assess: Notice the answer is independent of the car’s mass.

17.39. Model: There are three interacting systems: aluminum, copper, and ethyl alcohol.

Solve: The aluminum, copper, and alcohol form a closed system, so A1 cu eth 0 J.Q Q Q Q The mass of the

alcohol is

3 6 3eth (790 kg/m )(50 10 m ) 0.0395 kgM V

17-12 Chapter 17


Expressed in terms of specific heats and using the fact that f i ,T T T the 0 JQ condition is

Al Al Al Cu Cu Cu eth eth eth 0 JM c T M c T M c T

Substituting the appropriate values into this expression gives

(0.010 kg)(900 J/kgK)(298 K 473 K) (0.020 kg)(385 J/kgK)(298 K )

(0.0395 kg)(2400 J/kgK)(298 K 288 K) 1575 J (7.7 J/K)(298 ) 948 J 0 J

216.6 K 56.4 C 56 C

T

T

T

17.40. Model: There are two interacting systems: aluminum and ice. The system comes to thermal equilibrium in

four steps: (1) the ice temperature increases from 10 C to 0 C, (2) the ice becomes water at 0 C, (3) the water

temperature increases from 0 C to 20 C, and (4) the cup temperature decreases from 70 C to 20 C.

Solve: The aluminum and ice form a closed system, so 1 2 3 4 0 J.Q Q Q Q Q These quantities are

1 ice ice

52 ice f

3 ice water

4 Al Al Al Al

(0.100 kg)(2090 J/kgK)(10 K) 2090 J

(0.100 kg)(3.33 10 J/kg) 33,300 J

(0.100 kg)(4190 J/kgK)(20 K) 8380 J

(900 J/kgK)( 50 K) (45,000 J/kg)

Q M c T

Q M L

Q M c T

Q M c T M M

The 0 JQ equation now becomes

Al43,770 J (45,000 J/kg) = 0 JM

The solution to this is A1 0.97 kg.M

17.41. Model: There are three interacting systems: metal, aluminum, and water.

Solve: The metal, aluminum container, and water form a closed system, so m A1 w 0 J,Q Q Q where mQ is the

heat transferred to the metal sample. This equation can be written as

m m m Al Al Al w w 0 JM c T M c T M c T

Substituting in the given values gives

m

m

(0.512 kg) (351 K 288 K) (0.100 kg)(900 J/kgK)(351 K 371 K)

(0.325 kg)(4190 J/kgK)(351 K 371 K) 0 J

900 J/kgK

c

c

From Table 17.2, we see that this is the specific heat of aluminum.

17.42. Model: Heating the water raises its thermal energy and its temperature.

Solve: A 5.0 kW heater has power 5000 W.P That is, it supplies heat energy at the rate 5000 J/s. The heat

supplied in time is 5000 J.t Q t The temperature increase is C F(5/9) (5/9)(75 ) 41.67 C.T T Thus

w5000 J (150 kg)(4190 J/kg K)(41.67 C) 5283 s 87 minQ t Mc T t

Assess: A time of 1.5 hours to heat 40 gallons of water is reasonable.

17.43. Solve: From Table 17.4, we find the specific heat of 2O at constant volume is 20.9 J/mol K. The atomic

number of oxygen is 16, so 2O has 32 g/mol or 0.032 kg/mol. Converting the units gives the specific heat in J/kg K:

V

20.9 J/molK650 J/kgK

0.032 kg/molc

17.44. Model: Assume that your body heats the air in an isobaric process, and model air as consisting essentially of

nitrogen.

Solve: (a) Use the ideal gas to find the moles of air in a 3 33.0 L 3.0 10 miV volume at the given pressure:

i i

i

p Vn

RT



where i 0 C 273 KT and i 1 atm 101.3 kPa.p Use Equation 17.25 to find the energy required to change the

gas temperature: 3 3

i i PP

i

(101.3 kPa)(3.0 10 m )(29.1J/molK)(37 K)140 J

(8.31J/molK)(273 K)

pV C TQ nC T

RT

(b) Take the ratio of the ideal gas law at the initial and final temperature to find the volume increase:

i i 1 ff i

f f 2 i

310 K(3.0 L) 3.4 L

273 K

pV nRT TV V

p V nRT T

so the change in volume is 0.4 L.

17.45. Model: Heating the material increases its thermal energy.

Visualize: Heat raises the temperature of the substance from 40 C to 20 C, at which temperature a solid-to-

liquid phase change occurs. From 20 C, heat raises the liquid’s temperature up to 40 C. Boiling occurs at 40 C

where all of the liquid is converted to the vapor phase.

Solve: (a) In the solid phase,

1 20 kJ 12.0 kJ/kgK

20 K 0.50 kg

QQ Mc T c

T M

(b) In the liquid phase,

1 1 80 kJ2.7 kJ/kgK

0.50 kg 60 K

Qc

M T

(c) The melting point m 20 C T and the boiling point b 40 C.T

(d) The heat of fusion is

4f

20,000 J4.0 10 J/kg

0.50 kg

QL

M

The heat of vaporization is

5v

60,000 J1.2 10 J/kg

0.50 kg

QL

M

17.46. Model: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee temperature from

90 C to 60 C requires four steps: (1) raise the temperature of ice from 20 C to 0 C, (2) change ice at 0 C to

water at 0 C, (c) raise the water temperature from 0 C to 60 C, and (4) lower the coffee temperature from 90 C

to 60 C.

Solve: For the closed coffee-ice system,

ice coffee 1 2 3 4

1 ice ice ice ice

2 ice f ice

3 ice water ice ice

6 34 coffee coffee

( ) ( ) 0 J

(2090 J/kgK)(20 K) (41,800 J/kg)

(330,000 J/kg)

(4190 J/kgK)(60 K) (251,400 J/kg)

(300 10 m

Q Q Q Q Q Q Q

Q M c T M M

Q M L M

Q M c T M M

Q M c T

3)(1000 kg/m )(4190 J/kgK)( 30 K) 37,000 J

The 0 JQ equation thus becomes

ice ice(41,800 330,000 251,400) J/kg 37,710 J 0 J 0.061 kg 61 gM M

Assess: 61 g is the mass of approximately 1 ice cube.

17.47. Model: There are two interacting systems: the nuclear reactor and the water. The heat generated by the

nuclear reactor is used to raise the water temperature.

Solve: For the closed reactor-water system, energy conservation per second requires

reactor water 0 JQ Q Q

17-14 Chapter 17


The heat from the reactor in 1st is 9reactor 2000 MJ 2.0 10 JQ and the heat absorbed by the water is

water water water water (4190 J/kgK)(12 K)Q m c T m

9 4water water2.0 10 J (4190 J/kgK)(12 K) 0 J 3.98 10 kgm m

Each second,

43.98 10 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per minute is

34 6

3 3

kg 60 s 1m 1 L3.98 10 2.4 10 L/min

s min 1000 kg 10 m

Thus, 62.4 10 L of cooling water must flow through the reactor each minute.

17.48. Model: We have two interacting systems: the water and the gas. For the closed system comprised of water

and gas to come to equilibrium, heat is transferred from one interacting system to the other.

Solve: Energy conservation requires that

air water gas V f i gas water f i water0 J [ – ( ) ] [ – ( ) ] 0 JQ Q n C T T m c T T

Using the ideal-gas law, 5 6 3

gas gasi gas

gas

(10 1.013 10 Pa)(4000 10 m )( ) =1219 K

(0.40 mol)(8.31J/mol K)

p VT

n R

The energy conservation equation with i water( ) 293 KT becomes

3f(0.40 mol)(12.5 J/mol K)( 1219 K) (20 10 kg)(4190 J/kg K)( 293 K) 0 J 345 KT T T

We can now use the ideal-gas equation to find the final gas pressure.

i i f f ff i

i f i

345 K(10 atm) 2.8 atm

1219 K

pV p V Tp p

T T T

17.49. Model: These are isothermal and isobaric ideal-gas processes.

Solve: (a) The work done at constant temperature is

f f

i if i f i

13

(ln ln ) ln( / )

(2.0 mol)(8.31J/mol K)(303 K)ln 5.5 kJ

V V

V V

nRTW pdV dV nRT V V nRT V V

V

(b) The work done at constant pressure is

f

i

if i i i

2( )

3 3

2 2(2.0 mol)(8.31 J/mol K)(303 K) 3.4 kJ

3 3

V

V

VW pdV p V V p V pV

nRT

(c) For an isothermal process in which 1f i3

,V V the pressure changes to f i3 4.5p p atm.



17.50. Model: This is an isothermal process. The work done is positive for a compression.

Solve: For an isothermal process,

f iln( / )W nRT V V

For the first process,

12

500 J ln 721.35 JW nRT nRT

For the second process,

1 110 10

ln (721.35 J)ln 16.6 kJW nRT W

17.51. Model: The gas is an ideal gas, and its thermal energy is the total kinetic energy of the moving molecules.

Visualize: Please refer to Figure P17.51.

Solve: (a) The piston is floating in static equilibrium, so the downward force of gravity on the piston’s mass must

exactly balance the upward force of the gas, gasF pA where 2A r is the area of the face of the piston. Since the

upper part of the cylinder is evacuated, there is no gas pressure force pushing downward. Thus,

piston Cu piston 3 2piston Cu (8920 kg/m )(9.80 m/s )(0.040 m) 3500 Pa

M g V gM g pA p gh

A A

(b) The gas volume is 2 2 4 31 (0.030) (0.20 m) 5.65 10 m .V r L The number of moles is

4 341 1

1

(3500 Pa)(5.65 10 m )8.12 10 mol

(8.31J/mol K)(293 K)

p Vn

RT

The number of molecules is 4 23 1 20

A (8.12 10 mol)(6.02 10 mol ) 4.9 10N nN

(c) The pressure in the gas is determined simply by the weight of the piston. That will not change as heat is added, so

the heating takes place at constant pressure with P .Q nC T The temperature increase is

4P

2.0 J85 K

(8.12 10 mol)(29.1 J/mol K)

QT

nC

This raises the gas temperature to 2 1 378 K 105 C.T T T To two significant figures, this is 110 C.

(d) Noting that the volume of a cylinder is 2V r L and that r doesn’t change, the ideal-gas relationship for an

isobaric process is

2 1 2 1 22 1

2 1 2 1 1

378 K20 cm 26 cm

293 K

V V L L TL L

T T T T T

(e) The work done by the gas is gas gas .W F y The force exerted on the piston by the gas is

2gas 9.90 NF pA p r

This force is applied through 5.8 cm m,y so the work done is

gas (9.90 N)(0.058 m) 0.57 JW

Thus, 0.57 J is the work done by the gas on the piston. The work done on the gas is −0.57 J.

17.52. Model: This is an isobaric process.

Visualize:

17-16 Chapter 17


Solve: (a) The initial conditions are 6 2 21 1 1 1 atm 1.013 10 Pa, 50 C 323 K, (0.050 m)p T V r L

3 3(0.20 m) 1.57 10 m . The gas is heated at a constant pressure, so heat and temperature change are related by

p .Q nC T From the ideal gas law, the number of moles of gas is

6 3 31 1

1

(1.013 10 Pa)(1.57 10 m )0.593 mol

(8.31J/mol K)(323 K)

p Vn

RT

The temperature change due to the addition of 2500 JQ of heat is thus

P

2500 J203 K

(0.593 mol)(20.8 J/mol K)

QT

nC

The final temperature is 2 1 526 K 253 C, or 250 CT T T to two significant figures.

(b) Noting that the volume of a cylinder is 2V r L and that r doesn’t change, the ideal gas relationship for an

isobaric process is

2 1 2 1 22 1

2 1 2 1 1

526 K(20 cm) 33 cm

323 K

V V L L TL L

T T T T T

17.53. Model: The process in part (a) is isochoric and the process in part (b) is isobaric.

Solve: (a) Initially 1 1m m L and 293 K.V T Helium has an atomic mass number 4,A

so 3 g of helium is mol/ 0.75n M M mole of helium. We can find the initial pressure from the ideal-gas law:

11 3

1

(0.75 mol)(8.31 J/mol K)(293 K)228 kPa 2.25 atm

0.0080 m

nRTp

V

Heating the gas will raise its temperature. A constant-volume process has V ,Q nC T so

V

1000 J107 K

(0.75 mol)(12.5 J/mol K)

QT

nC

This raises the final temperature to 2 1 400 K.T T T Because the process is isochoric,

2 1 22 1

2 1 1

400 K(2.25 atm) 3.1 atm

293 K

p p Tp p

T T T

(b) The initial conditions are the same as part a, but now P .Q nC T Thus,

P

1000 J64.1 K

(0.75 mol)(20.8 J/mol K)

QT

nC

Now the final temperature is 2 1 357 K.T T T Because the process is isobaric,

3 32 1 22 1

2 1 1

357 K(0.0080 m ) 0.0097 m 9.7 L

293 K

V V TV V

T T T

(c)



17.54. Model: Assume the air and the nitrogen are ideal gases. For the piston to be in equilibrium the forces on it

must sum to zero. Also assume that the temperature doesn’t change in the second part.

Solve: The area is 2 2 3 2(0 040 m) 5 0 10 m .A r . . 23 2 31 1 (5 0 10 m )(0 26 m) 0 0013 m .V Ah . . .

(a)

net in above

2 3 2above

in 3 2

0 N

(5 1 kg)(9 8 m/s ) (100 kPa)(5 0 10 m )110 kPa

5 0 10 m

F p A mg p A

mg p A . . .p

A .

(b) Since 1 2,T T then 1 1 2 2p V p V where 1p is inp from above, and 2p is the inside pressure after the new weight is

added. Compute 2p exactly as above replacing 5.1 kg with 5.1kg 3 5 kg 8 6 kg.. .

2 3 2above

2 3 2

(8 6 kg)(9 8 m/s ) (100 kPa)(5 0 10 m )117 kPa

5 0 10 m

mg p A . . .p

A .

33 31 1

22

(110 kPa)(0 0013 m )1 22 10 m

117 kPa

p V .V .

p

3 32

2 3 2

1 22 10 m0 2447 m 24 cm

5 0 10 m

V .h .

A .

Assess: The result seems to be a reasonable number; we expected the piston to be a little lower than before due to

the increased mass on it.

17.55. Model: This is an isothermal process.

Solve: (a) The final temperature is 2 1T T because the process is isothermal.

(b) The work done on the gas is

2 2

1 1

1 21 1

1

ln ln 2V V

V V

nRT VW pdV dV nRT nRT

V V

(c) From the first law of thermodynamics th 0 J becauseE W Q T Thus, the heat energy transferred to

the gas is 1 ln2.Q W nRT

17.56. Model: The gas is an ideal gas and it goes through an isobaric and an isochoric process.

Solve: (a) The initial conditions are 1 13.0 atm 304,000 Pa and 293Κ.p T Nitrogen has a molar mass

mol 28 g/mol,M so 5.0 g of nitrogen gas has mol/ 0.1786 mol.n M M From this, we can find the initial volume:

3 3 311

1

(0.1786 mol)(8.31J/mol K)(293 K)1.430 10 m 1400 cm

304,000 Pa

nRTV

p

The volume triples, so 32 13 4300 cm .V V The expansion is isobaric 2 1( 3.0 atm),p p so

2 1 22 1

2 1 1

(3)293 K 879 K 610 CV V V

T TT T V

(b) The process is isobaric, so

P (0.1786 mol)(29.1 J/mol K)(879 K 293 K) 3000 JQ nC T

(c) The pressure is decreased at constant volume 33 2( 4290 cm )V V until the original temperature is reached

3 1( 293 K).T T For an isochoric process,

3 2 33 2

3 2 2

293 K(3.0 atm) 1.0 atm

879 K

p p Tp p

T T T

(d) The process is isochoric, so

V (0.1786 mol)(20.8 J/mol K)(293 K 879 K) 2200 JQ nC T

So, 2200 J of heat was removed to decrease the pressure.

17-18 Chapter 17


(e)

17.57. Model: The gas is an ideal gas.

Visualize: In the figure, call the upper-left corner (on process A) 2 and the lower-right corner (on process B) 3.

Solve: The change in thermal energy is the same for any gas process that has the same ∆T. Processes A and B have

the same ∆T, since they start and end at the same points, so th A th B( ) ( ) .E E The first law then gives

th A A A th B B B A B B A( ) ( ) – –E Q W E Q W Q Q W W

In process B, work i i i i i(2 )W p V p V V pV is done during the isobaric process 3.i No work is done

during the isochoric process 3 .f Thus B i i.W pV Similarly, no work is done during the isochoric process

2i of process A, but i i i i i2 (2 ) 2W p V p V V pV is done during the isobaric process 2 .f Thus

A i i2 .W pV Combining these gives

A B B A i i i i i i– – – – ( – 2 ) Q Q W W pV pV pV

17.58. Model: The two processes are isochoric and isobaric.

Solve: Process A is isochoric which means 1

f i f i f i f i i i3/ / ( / ) (1 atm/3 atm)T T p p T T p p T T

From the ideal-gas equation, 5 6 3

i i 1i f i3

(3 1.013 10 Pa)(2000 10 m )731.4 K 243.8 K

(0.10 mol)(8.31 J/mol K)

pVT T T

nR

f i 487.6 KT T

Thus, the heat required for process A is

A V (0.10 mol)(20.8 J/mol K)( 487.6 K) 1000 JQ nC T

Process B is isobaric which means 3 3

f f i i f i f i i i/ / ( / ) (3000 cm /1000 cm ) 3T V T V T T V V T T

From the ideal-gas equation, 5 6 3

i ii

(2 1.013 10 Pa)(1000 10 m )243.8 K

(0.10 mol)(8.31 J/mol K)

pVT

nR

f i f i3 731.4 K 487.6 KT T T T

Thus, heat required for process B is

B P (0.10 mol)(29.1J/mol K)(487.6 K) 1400 JQ nC T

Assess: Heat is transferred out of the gas in process A, but transferred into the gas in process B.

17.59. Model: We have an adiabatic and an isothermal process.

Solve: For the adiabatic process, no heat is added or removed. That is 0 J.Q

Isothermal processes occur at a fixed temperature, so 0 K.T Thus th 0 J,E and the first law of thermodynamics

gives

f iln( / )Q W nRT V V



The temperature T can be obtained from the ideal-gas equation as follows:

5 6 3i i

i i

(1.013 10 Pa)(3000 10 m )366 K

(0.10 mol)(8.31 J/mol K)

pVpV nRT T

nR

Substituting into the equation for Q we get

6 3

6 3

1000 10 m(0.10 mol)(8.31J/mol K)(366 K)ln 330 J

3000 10 mQ

That is, 330 J of heat energy is removed from the gas.

17.60. Model: The monatomic gas is an ideal gas which is subject to isobaric and isochoric processes.

Solve: (a) For this isobaric process, 6 3 6 31 2 1 24 atm, 800 10 m , and 1600 10 m .p p V V The temperature

1T may be found from the ideal-gas law:

3 6 31 1

1

4(101.3 10 Pa)(800 10 m )390.6 K

(0.10 mol)(8.3 J/mol K)

p VT

nR

The temperature 2T may be obtained from the ideal-gas equation:

6 32

2 1 1 16 31

1600 10 m2.0 781.1K

800 10 m

VT T T T

V

Thus, the heat required for the process 1 2 is

P 2 1( ) (0.10 mol)(20.8 J/mol K)(781.1 K 390.6 K) 812 J 0.81kJQ nC T T

This is heat transferred to the gas.

(b) For the isochoric process, 6 32 3 2 3 31600 10 m , 4.0 atm 405.3 kPa, 2.0 atm 202.65 kPa.V V p p T can

be obtained by applying the ideal-gas equation to points 2 and 3:

2 2 3 3 33 2

2 3 2

2 atm(781.1 K) 390.6 K

4 atm

p V p V pT T

T T p

The heat required for the process 2 3 is

V 3 2( ) (0.10 mol)(12.5 J/mol K)(390.6 K 781.1 K) 488 J 0.49 kJQ nC T T

The negative sign means that 0.49 kJ of heat is removed from the gas.

(c) The change in the thermal energy of the gas is

th 1 2 2 3 1 2 2 3 1 2 1 1 2

6 3 6 3

( ) ( ) 812 J 488 J + 0 J 324.2 J

324.2 J – (405.3 kPa)(1600 10 m – 800 10 m ) 0.0 J

E Q Q W W W p V

Assess: This result was expected since 3 1.T T

17.61. Model: Model the gas as an ideal gas and the compression as an adiabatic process.

Solve: The gas temperature changes as the cylinder volume changes. Apply Equation 17.39:

11 i ii i

i

ln ( 1)ln ( 1)lnT V L

TV TVT V L

The cross section area of the cylinder doesn’t change, so we could replace the ratio of the volumes with the ratio of

the cylinder lengths. The initial values are i 20 cmL and i 21 C 294ΚT (room temperature). Note that the data

show how far the piston is pushed; the length is 20 cmL minus the push distance. If we plot iln( / )T T versus

iln( / ),L L the graph should be a straight line with slope 1. The plot is shown below. The experimental slope of

0.29 gives 1.29.

17-20 Chapter 17


17.62. Model: Assume that the gas is an ideal gas. A diatomic gas has 1.40.

Solve: (a) For container A,

4 3AiiA 5

Ai

(0.10 mol)(8.31 J/mol K)(300 K)8.20 10 m

3.0 1.013 10 Pa

nRTV

p

For an isothermal process Af Af Ai Ai.p V p V This means Af Ai 300 KT T and

4 3 3 3Af Ai Ai Af( / ) (8.20 10 m )(3.0 atm/1.0 atm) 2.5 10 mV V p p

The gas in container B starts with the same initial volume. For an adiabatic process, 1/ 1/1.40

4 3 3 3BiBf Bi Bf BiBf Bi

Bf

3.0 atm(8.20 10 m ) 1.8 10 m

1.0 atm

pp V p V V V

p

The final temperature BfT can now be obtained by using the ideal-gas equation:

3 3Bf Bf

Bf iB 4 3Bi Bi

1.0 atm 1.80 10 m(300 K) 220 K

3.0 atm 8.20 10 m

p VT T

p V

(b)

17.63. Model: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes.

Solve: (a) The initial conditions are 3 4 31 1 13.0 atm 304,000 Pa, 100 cm 1.0 10 m , and 100 Cp V T

The number of moles of gas is 4 3

31 1

1

(304,000 Pa)(1.0 10 m )9.81 10 mol

(8.31 J/mol K)(373 K)

p Vn

RT

At point 2 we have 32 1 2 13.0 atm and 300 cm .p p V V This is an isobaric process, so

2 1 22 1

2 1 1

3(373 K) 1119 KV V V

T TT T V



The gas is heated to raise the temperature from 1T to 2.T The amount of heat required is

3P (9.81 10 mol)(20.8 J/mol K)(1119 K 373 K) 0.15 kJQ nC T

This amount of heat is added during process 1 2.

(b) Point 3 returns to 3 100 C 373Κ.T This is an isochoric process, so

3V (9.81 10 mol)(12.5 J/mol K)(373 K 1119 K) 91 JQ nC T

This amount of heat is removed during process 2 3.

17.64. Assume the gas to be an ideal gas.

Solve: The work done on the gas is the negative of the area under the p-versus-V graph, that is

area under curve 50.7 J W

The change in thermal energy is

th V V f i( )E nC T nC T T

Using the ideal-gas law to calculate the initial and final temperatures, 5 6 3

i ii

(4.0 1.013 10 Pa)(100 10 m )325 K

(0.015 mol)(8.31J/mol K)

pVT

nR

5 6 3f f

f

(1.013 10 Pa)(300 10 m )244 K

(0.015 mol)(8.31J/mol K)

p VT

nR

th (0.015 mol)(12.5 J/mol K)(244 K 325 K) 15.2 JE

From the first law of thermodynamics,

th thW 15.2 J ( 50.7 J) 36 JE Q Q E W

That is, 36 J of heat energy is transferred to the gas.

17.65. Model: Assume that the gas is an ideal gas and that the work, heat, and thermal energy are connected by the

first law of thermodynamics.

Solve: (a) For point 1, 3 3 31 11000 cm 1.0 10 m , 133 C 406Κ,V T and the number of moles is

3

mol

120 10 g0.030 mol

4 g/mol

Mn

M

Thus, the pressure 1p is

511

1

1.012 10 Pa 1.0 atmnRT

pV

The process 1 2 is isochoric 2 1( )V V and 2 15 5.0 atm.p p Thus,

2 1 2 1( / ) (406 K)(5) 2030 K 1757 CT T p p

The process 2 3 is isothermal 2 3( ),T T so

33 2 2 3 2 2 1 2( / ) ( / ) 5 5000 cmV V p p V p p V

p (atm) ( C)T 3(cm )V

Point 1 1.0 133 1000

Point 2 5.0 1757 1000

Point 3 1.0 1757 5000

(b) The work 1 2 0W J because it is an isochoric process. The work in process 2 3 can be found using

Equation 17.14 as follows:

2 3 2 3 2ln( / ) (0.030 mol)(8.31 J/mol K)(2030 K)ln(5) 815 JW nRT V V

The work in the isobaric process 3 1 is 5 3 3 3 3

3 1 f i( ) (1.012 10 Pa)(1.0 10 m 5.0 10 m ) 405 JW p V V

17-22 Chapter 17


(c) The heat transferred in process 1 2 is

1 2 V (0.030 mol)(12.5 J/mol K)(2030 K 406 K) 609 JQ nC T

The heat transferred in the isothermal process 2 3 is 2 3 2 3 815 J.Q W The heat transferred in the isobaric

process 3 1 is

3 1 P (0.030 mol)(20.8 J/mol K)(406 K 2030 K) 1.01 kJQ nC T

17.66. Model: Air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.

Solve: (a) Equation 17.38 for an adiabatic process is 1

11 1 f if if i

i f

V TT V TV

V T

For the temperature to increase from i f20 C 293 K to 1000 C 273 K,T T the compression ratio will be

1

1.4 1f max

i min

293 K 10.02542 39.3 39

1273 K 0.02542

V V

V V

(b) From the Equation 17.37,

1.4f if if i

i f

(39.3) 170p V

p V pVp V

17.67. Model: Model the gas as an ideal gas.

Solve: Because the two gases have the same initial conditions, the ideal-gas law tells us that they have the same

number of moles. Apply Equation 17.25 for constant pressure to find the heat energy required by the first process:

P 1.Q nC T Insert this into Equation 17.25 for a constant volume process to find the temperature increase:

P 1 PV 2 2 1 1

V V V

Q nC T CQ nC T T T T

nC nC C

For a diatomic gas, 1.40 (see Equation 17.35), so the temperature change for the second container of gas is

2 1.40(20 C) 28 CT

17.68. Model: The nitrogen gas is assumed to be an ideal gas that is subjected to an adiabatic process.

Solve: (a) The number of moles in 14.0 g of N2 gas is

mol

14.0 g0.50 mol

28 g/mol

Mn

M

At 5i i273 K and 1.0 atm 10 Pa,T p the gas has a volume

3ii

i

0.0112 m 11.2 LnRT

Vp

For an adiabatic process that compresses to a pressure f 20 atm,p we can use Equation 17.37 and Equation 17.38

as follows: 1

1

1 1 f i f if i f if i f i

i f i f

T V p V

T V TV p V pVT V p V

Combining the above two equations yields 1 1.4 1.0

1.4f i f i f i i( / ) ( / ) (20) (2.3535) 643 KT T p p T T T

(b) The work done on the gas is

th V f i( ) (0.50 mol)(20.8 J/mol K)(643 K 273 K) 3.8 kJW E nC T T

(c) The heat input to the gas is 0 J.Q



(d) From the above equation, 1

1

i f max1.4

f i min

(20) 8.5V p V

V p V

(e)

17.69. Model: The gas is assumed to be an ideal gas that is subjected to an isochoric process.

Solve: (a) The number of moles in 14.0 g of 2N gas is

mol

14.0 g0.50 mol

28 g/mol

Mn

M

At 5i i273 K and atm 1.013 10 Pa,T p the gas has a volume

3ii

i

0.0112 m 11.2 LnRT

Vp

For an isochoric process i f( ),V V

f ff

i i

20 atm20 20(273 K) 5.5 kK

1 atm

T pT

T p

(b) The work done on the gas is 0 J.W p V

(c) The heat input to the gas is 4

V f i( ) (0.50 mol)(20.8 J/mol K)(5460 K 273 K) 5.4 10 JQ nC T T

(d) The pressure ratio is

max f

min i

20 atm20

1 atm

p p

p p

(e)

17-24 Chapter 17


17.70. Model: The air is assumed to be an ideal gas. Because the air is compressed without time to exchange heat

with its surroundings, the compression is an adiabatic process.

Solve: The initial pressure of air in the mountains behind Los Angeles is 3i i60 10 Pa at 273 K.p T The pressure

of this air when it is carried down to the elevation near sea level is 3f 100 10 Pa.p The adiabatic compression of a

gas leads to an increase in temperature according to Equations 17.38 and 17.39, which are 1

1

1 1f i f if i f if i f i

i f i f

p V T V

p V pV T V TVp V T V

Combining these two equations, 1.4 1

1 0.2863 1.4

f i f i f i 3

100 10 Pa 5( / ) ( / ) (273 K) 316 K 43 C 109 F

360 10 PaT T p p T T

17.71. Model: Assume the collector has emissivity 1.e We will use Equation 17.48 and divide both sides by A

to get intensity in 2W/m . At the equilibrium temperature the collector will absorb 2800 W/m on the sun side and

must radiate the same amount from the same side. 0 20 C 293 K.T

Solve:

4 40

/( )

Q te T T

A

Solve this for T.

4 40

/Q tT T

Ae

24 44 4

0 8 2 4

/ 800 W/m(293 K) 383 K 110 C

(1 0)(5 67 10 W/m K )

Q tT T

Ae . .

2

Assess: 110 C is hot enough to boil water, so this seems feasible as a hot water heater, at least during the 5 hours a

day the collector gets 2800 W/m . You can store the hot water in an insulated tank for use at other times of the day.

17.72. Model: This is a problem about conduction of the heat from inside the box to outside. 100 W of heat is

generated by the light bulb inside the box, so in equilibrium that is how much must be conducted away through the

sides of the box.

Visualize: We are given 0 012 m,L . / 100 WQ t and the thermal conductivity of concrete 0 8 W/mK.k . We

compute 26(0 20 m 0 20 m) 0 24 m .A . . . C 20 C 293 K.T

Solve: Use Equation 17.46 for the rate of heat transfer by conduction.

H C( )Q A

k T Tt L

Solve for H.T

H C( )Q L

T Tt kA

H C 2

0 012 m(100 W) 293 K 299 K 26 C

(0 8 W/mK)(0 24 m )

Q L .T T

t kA . .

Assess: We expected HT to be a few degrees hotter than C ,T which it is.

17.73. Model: Refer to Example 17.11. Assume the surface of the earth is an ideal radiator with 1.e

Visualize: Only half the earth faces the sun at a time, so the earth intercepts 21370 W/m over a cross section of 2 6 2 14 2(6 37 10 m) 1 275 10 m .eR . . If the earth’s surface absorbs 70% of the incident power then the total

power absorbed is 2 14 2 17(0 70)(1370 W/m )(1 275 10 m ) 1 222 10 W.. . . This much power must also be radiated

away from the earth in equilibrium.



Solve: The power is radiated from the whole spherical surface of the earth: 2e4 .A R Use Equation 17.47 to find the

temperature of the earth. 1/4 1/4

17

2 8 2 6 2e

/ 1 222 10 W255 K 18 C

(4 ) (1)(5 67 10 W/m K)4 (6 37 10 m)

Q t .T

e R . .

3

Assess: Without a moderate greenhouse effect the earth would be too cold for mammal life. On the other hand, when

the greenhouse effect gets out of hand, such as on Venus, it can be too hot for life.

17.74. Solve: (a) 50 J of work are done on a gas to compress it to one-third of its original volume at a constant

temperature of 77 C. How many moles of the gas are in the sample?

(b) The number of moles is

13

50 J0.0156 mol

(8.31 J/mol K)(350 K) lnn

17.75. Solve: (a) A heated 500 g iron slug is dropped into a 200 cm3 pool of mercury at 15 C. If the mercury

temperature rises to 90 C, what was the initial temperature of the iron slug?

(b) The initial temperature was 217 C.

17.76. Solve: (a) A diatomic gas is adiabatically compressed from 1 atm pressure to 10 atm pressure. What is the

compression ratio max min/ ?V V

(b) The ratio is 1/1.4max min/ 10 5.18.V V

17.77. Model: Assume the helium gas to be an ideal gas. The gas is subjected to isothermal, isochoric, and

adiabatic processes.

Visualize: Please refer to Figure CP17.77. The gas at point 1 has volume 3 3 31 1000 cm 1.0 10 mV and pressure

1 3.0 atm.p At point 2,

3 3 32 23000 cm 3.0 10 m and 1.0 atm.V p These values mean that 2 1,T T so process

1 2 is an isothermal process. The process 2 3 occurs at constant volume and is thus an isochoric process.

Finally, because temperature 3T is lower than 1T or 2 ,T the process 3 1 is an adiabatic process.

Solve: (a) The number of moles of gas is

0.120 g0.030 mols

4 g/moln

The temperature 1T can be calculated to be

5 3 31 1

1

(3.0 1.013 10 Pa)(1.0 10 m )1219 K 946 C

(0.030 mol)(8.31J/mol K)

p VT

nR

For the isothermal process 21 2, 1219 K.T For the adiabatic process 3 1,

1.67

3 0.671 13 1 1 3 3 1 1 33 3

1000 cm( / ) (3.0 atm) 0.48 atm ( / ) (1219 K) 583 K 310 C

3000 cmp p V V T T V V

Point p (atm) ( C)T 3(cm )V

1 3.0 946 1000

2 1.0 946 3000

3 0.48 310 3000

Note that the values obtained above are consistent with the isochoric process 2 3, for which

2 32 2 3 3

2 3

( / ) (1219 K/583 K)(0.48 atm) 1 atmp p

p T T pT T

17-26 Chapter 17


(b) From Equation 17.13,

21 2 1

1

ln (0.030 mol)(8.31J/mol K)(1219 K)ln(3) 334 JV

W nRTV

The work done in the isochoric process is 2 3 0 J.W The work done in the adiabatic process is

3 1 V 1 3( ) (0.030 mol)(12.5 J/mol K)(1219 K 583 K) 239 JW nC T T

(c) For the process th1 2, 0 K 0 J 334 JT E Q W

For the process 2 3, 0 J,W

th V 3 2( ) 239 JE Q nC T T

For the process 3 1, 0 J.Q

17.78. Model: The air is assumed to be an ideal diatomic gas that is subjected to an adiabatic process.

Solve: The air admitted into the cylinder at 0 030 C 303 K and 1atm PaT p has a volume

6 30 600 10 mV and contains

0 0

0

0.024 molp V

nRT

Using Equation 17.34 and the fact that 0 JQ for an adiabatic process,

th V VE Q W nC T W nC T

f f400 J (0.024 mol)(20.8 J/mol K)( – 303 K) 1100 KT T

For an adiabatic process Equation 17.37 is 1 1

1 1.4 11 1 4 3 5 3 30f 0 f 00f

f

303 K(6.0 10 m ) 2.39 10 m 24 cm

1100 K

TT V T V V V

T

Assess: Note that W is positive because the environment does work on the gas.

17.79. Model: There is a thermal interaction between the iron, assumed to be initially at room temperature (20 C),

and the liquid nitrogen. The boiling point of liquid nitrogen is 196 C 77 K.

Solve: The piece of iron has mass iron 197 g kgM and volume iron iron iron/ ( kg)/V M

3 3 3(7870 kg/m ) m 25 cm 25 mL. The heat lost by the iron is

4iron iron iron (0.197 kg)(449 J/kgK)(77 K – 293 K) 1.911 10 JQ M c T

This heat causes mass M of the liquid nitrogen to boil. Energy conservation requires 4

ironiron N2 iron f 5

f

1.911 10 J0 0.0960 kg

1.99 10 J/kg

QQ Q Q ML M

L

The volume of liquid nitrogen boiled away is thus

4 3boil 3

N2

0.0960 kg1.19 10 m 119 mL

810 kg/m

MV

Now the volume of nitrogen gas (at 77 K) is 1500 mL before the iron is dropped in. The volume of the piece of iron

excludes 25 mL of gas, so the initial gas volume, when the lid is sealed and the liquid starts to boil, is 1 1475 mL.V

The pressure is 1 1.0 atmp and the temperature is 1 77 K.T Thus the number of moles of nitrogen gas is

3 31 1

11

(101,300 Pa)(1.475 10 m )0.234 mol

(8.31 J/mol K)(77 K)

p Vn

RT

119 mL of liquid boils away, so the gas volume increases to 2 1 boil 1475 mL mL mL.V V V The

temperature is still 2 77 K,T but the number of moles of gas has been increased by the liquid that boiled. The

number of moles that boiled away is

boil

96.0 g3.429 mol

28 mol/gn



Thus the number of moles of nitrogen gas increases to 2 1 boil 0.234 mol 3.429 mol 3.663 mol.n n n

Consequently, the gas pressure increases to

62 22 3 3

2

(3.663 mol)(8.31 J/mol K)(77 K)1.470 10 Pa 15 atm

1.594 10 m

n RTp

V

Assess: Don’t try this! The large pressure increase could cause a flask of liquid nitrogen to explode, leading to serious injuries.

17.80. Model: Ignore any radiation and assume all of the heat is transferred by conduction through the compound

rod from the hot end to the cold end.

Visualize: Look up the thermal conductivities of the two metals in Table 17.5: Cu 400 W/m Kk and

Fe 80 W/mK.k H 373 KT and C 273 K.T Equation 17.46 will be applied to each rod and all of the heat must go

through both rods at the same rate.

Cu Fe

Q Q

t t

Solve: Since A and L are the same for the two rods, then

Cu Cu Fe Fek T k T

Label the temperature at the joining point in the middle with a subscript M.

Cu H M Fe M C( ) ( )k T T k T T

Solve for M.T

Cu H Cu M Fe M Fe C

Cu H Fe C Fe M Cu M

Fe Cu M Cu H Fe C

Cu H Fe CM

Cu Fe

( )

k T k T k T k T

k T k T k T k T

k k T k T k T

k T k TT

k k

M

(400 W/m K)(373 K) (80 W/m K)(273 K)356 K 83 C

400 W/m K 80 W/m KT

Assess: Had we not arrived at an answer between 100 C and 0 C we would have been very worried. Furthermore,

because the conductivity of copper is greater than the conductivity of iron we expected the answer to be above 50 C.

17.81. Model: Assume the gas is ideal. The process is not adiabatic, despite Equation 17.36, because the exponent

given in the problem is not equal to . The gas is identified as diatomic which means that 1.40, but the exponent

is explicitly given as 2, so while we are given 2 constant,pV it is not true that constant.pV We will use the

fact that the gas is diatomic to deduce 5V 2

20 8J/mol k,C R . although we won’t specifically need 1.40.

Solve: (a) We are given 0 020 mol,n . i 293 K,T 3i 1500 cm ,V and 3

f 500 cm .V The strategy to find Q will

be to use the first law th .Q E W We will use th VE nC T and d .W p V

th V (0 020 mol)(20 8 J/mol K)(879 K 293 K) 243 8 JE nC T . . .

To do the W pdV integral we need to know what the constant is in

2 constant.pV Use the ideal gas law to

compute i.p

ii 3 3

i

(0 020 mol)(8 31 J/mol K)(293 K)32,460 Pa

1 5 10 m

nRT . .p

V .

2 3 3 2 6i iconstant (32,460 Pa)(1 5 10 m ) 0 0730 Pa mpV . .

3 3f

33i

0 00050m 0.00050m6 6constant

2 20.0015m0 0015m

d 12 d (0 0730 Pa m ) (0 0730 Pa m ) 97 4 J

V .

VV .

vW pdV V . . .

VV

th 243 8 J 97 4 J 146 4 J 150 JQ E W . . .

Assess: Both Q and W are positive because heat was added to the system and work was done on the system.

17-28 Chapter 17


17.82. Model: Assume the gas is ideal.

Solve: Apply the first law of thermodynamics, th V .E W Q nC T Consider adding a small amount of heat

energy, so that the gas expands by an infinitesimal amount. This process can be considered isobaric, so we can use

Equation 17.11, from which we see that the infinitesimal work done on the gas is

f

i

V

V

dW pdV W pdV

The gas pressure is simply the force of the spring times the piston area, or / .p kx A Inserting this into the expression

for work and integrating gives

f f

i i

2 2i f

2

V x

V x

kx kW dV kxdx x x

A

where we have used .dV Adx Now, use the ideal gas law to find the temperature change. The initial temperature is

1 300 K.T The final temperature is

f f f ff i

i i

p V p VT T

nR pV

Thus, the temperature change is

f ff i i

i i

1p V

T T T TpV

The last piece needed is the number of moles n, which may be found from the ideal gas law:

i i

i

p Vn

RT

Inserting these components into the first-law expression gives

2 2 2 2i i f f V i i f fV V i i f i f

i i i i i

1 12 2

pV p V k C pV p V kW Q nC T Q C T x x x x

RT pV R pV

Because the gas in monatomic, v 12.5 J/mol KC (see Table 17.4). From the problem statement, we know that

spring springi fi f i f, , , ( 6.0 cm)

F Fkx kxp p V LA V L A

A A A A

with 2i f20 cm, 8.0 cm, and 8.0 cm .x x A Inserting these values into the expression for the heat transferred gives

2 2V i fi f

i

2 2

( 6.0 cm)1

2

(12.5 J/mol K)(2000 N/m)(0.020 m)(0.100 m) (0.080 m)(0.160 m) (2000 N/m)1 (0.020 m) (0.080 m)

(8.3 J/mol K) (0.020 m)(0.100 m) 2

39 J

C kx L x L kQ x x

R x L

instructor solution manual

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