ferred to the environment. Show that the Clausius state-ment of the second law of thermodynamics is violated. (c) Find the energy input and work output of engine S as itputs out exhaust energy of 100 J. (d) Let engine S operateas in (c) and contribute 150 J of its work output to runningthe Carnot engine in reverse. Find the total energy the fire-box puts out as both engines operate together, the totalwork output, and the total energy transferred to the envi-ronment. Show that the Kelvin–Planck statement of the sec-ond law is violated. Thus our assumption about the effi-ciency of engine S must be false. (e) Let the enginesoperate together through one cycle as in part (d). Find thechange in entropy of the Universe. Show that the entropystatement of the second law is violated.

22. At point A in a Carnot cycle, 2.34 mol of a monatomicideal gas has a pressure of 1 400 kPa, a volume of 10.0 L,and a temperature of 720 K. It expands isothermally topoint B, and then expands adiabatically to point C whereits volume is 24.0 L. An isothermal compression brings itto point D, where its volume is 15.0 L. An adiabatic processreturns the gas to point A. (a) Determine all the unknownpressures, volumes and temperatures as you fill in thefollowing table:

26. A heat pump, shown in Figure P22.26, is essentially an airconditioner installed backward. It extracts energy fromcolder air outside and deposits it in a warmer room. Sup-pose that the ratio of the actual energy entering the roomto the work done by the device’s motor is 10.0% of the the-oretical maximum ratio. Determine the energy enteringthe room per joule of work done by the motor, given thatthe inside temperature is 20.0°C and the outside tempera-ture is " 5.00°C.

How much work does an ideal Carnot refrigeratorrequire to remove 1.00 J of energy from helium at 4.00 Kand reject this energy to a room-temperature (293-K)environment?

28. A refrigerator maintains a temperature of 0°C in the coldcompartment with a room temperature of 25.0°C. Itremoves energy from the cold compartment at the rate of8 000 kJ/h. (a) What minimum power is required tooperate the refrigerator? (b) The refrigerator exhaustsenergy into the room at what rate?

29. If a 35.0%-efficient Carnot heat engine (Fig. 22.2) is run inreverse so as to form a refrigerator (Fig. 22.5), what wouldbe this refrigerator’s coefficient of performance?

30. Two Carnot engines have the same efficiency. One engineruns in reverse as a heat pump, and the other runs in reverseas a refrigerator. The coefficient of performance of the heatpump is 1.50 times the coefficient of performance of therefrigerator. Find (a) the coefficient of performance of therefrigerator, (b) the coefficient of performance of the heatpump, and (c) the efficiency of each heat engine.

Section 22.5 Gasoline and Diesel EnginesIn a cylinder of an automobile engine, just after combus-tion, the gas is confined to a volume of 50.0 cm3 and hasan initial pressure of 3.00 ' 106 Pa. The piston moves out-ward to a final volume of 300 cm3, and the gas expandswithout energy loss by heat. (a) If * ! 1.40 for the gas,what is the final pressure? (b) How much work is done bythe gas in expanding?

32. A gasoline engine has a compression ratio of 6.00 anduses a gas for which * ! 1.40. (a) What is the efficiency

31.

27.

698 C H A P T E R 2 2 • Heat Engines, Entropy, and the Second Law of Thermodynamics

Q h

InsideTh

OutsideTc

Q cHeatpump

Figure P22.26

P V T

A 1 400 kPa 10.0 L 720 K

B

C 24.0 L

D 15.0 L

(b) Find the energy added by heat, the work done by theengine, and the change in internal energy for each of thesteps A : B, B : C, C : D, and D : A. (c) Calculate theefficiency Wnet/Q h . Show that it is equal to 1 " TC /TA ,the Carnot efficiency.

23. What is the coefficient of performance of a refrigeratorthat operates with Carnot efficiency between temperatures" 3.00°C and & 27.0°C?

24. What is the maximum possible coefficient of performanceof a heat pump that brings energy from outdoors at" 3.00°C into a 22.0°C house? Note that the work done torun the heat pump is also available to warm up the house.

An ideal refrigerator or ideal heat pump is equivalent to aCarnot engine running in reverse. That is, energy Q c istaken in from a cold reservoir and energy Q h is rejected toa hot reservoir. (a) Show that the work that must be sup-plied to run the refrigerator or heat pump is

(b) Show that the coefficient of performance of the idealrefrigerator is

COP !Tc

Th " Tc

W !Th " Tc

Tc Q c

25.

Q (input) W (output) %E int

A : BB : CC : DD : AABCDA

of the engine if it operates in an idealized Otto cycle?(b) What If ? If the actual efficiency is 15.0%, whatfraction of the fuel is wasted as a result of friction andenergy losses by heat that could by avoided in a re-versible engine? (Assume complete combustion of theair–fuel mixture.)

33. A 1.60-L gasoline engine with a compression ratio of 6.20has a useful power output of 102 hp. Assuming the en-gine operates in an idealized Otto cycle, find the energytaken in and the energy exhausted each second. Assumethe fuel–air mixture behaves like an ideal gas with* ! 1.40.

34. The compression ratio of an Otto cycle, as shown in Figure22.13, is VA/VB ! 8.00. At the beginning A of the compres-sion process, 500 cm3 of gas is at 100 kPa and 20.0°C. Atthe beginning of the adiabatic expansion the temperatureis TC ! 750°C. Model the working fluid as an ideal gaswith E int ! nCVT ! 2.50nRT and * ! 1.40. (a) Fill in thetable below to follow the states of the gas:

38. In making raspberry jelly, 900 g of raspberry juice is com-bined with 930 g of sugar. The mixture starts at room tem-perature, 23.0°C, and is slowly heated on a stove until itreaches 220°F. It is then poured into heated jars and al-lowed to cool. Assume that the juice has the same specificheat as water. The specific heat of sucrose is 0.299 cal/g ( °C.Consider the heating process. (a) Which of the followingterms describe(s) this process: adiabatic, isobaric, isother-mal, isovolumetric, cyclic, reversible, isentropic? (b) Howmuch energy does the mixture absorb? (c) What is the mini-mum change in entropy of the jelly while it is heated?

39. What change in entropy occurs when a 27.9-g ice cube at" 12°C is transformed into steam at 115°C?

Section 22.7 Entropy Changes in IrreversibleProcesses

40. The temperature at the surface of the Sun is approxi-mately 5 700 K , and the temperature at the surface of theEarth is approximately 290 K. What entropy change occurswhen 1 000 J of energy is transferred by radiation from theSun to the Earth?

A 1 500-kg car is moving at 20.0 m/s. The driverbrakes to a stop. The brakes cool off to the temperature ofthe surrounding air, which is nearly constant at 20.0°C.What is the total entropy change?

42. A 1.00-kg iron horseshoe is taken from a forge at 900°Cand dropped into 4.00 kg of water at 10.0°C. Assumingthat no energy is lost by heat to the surroundings, deter-mine the total entropy change of the horseshoe-plus-watersystem.

43. How fast are you personally making the entropy of theUniverse increase right now? Compute an order-of-magni-tude estimate, stating what quantities you take as data andthe values you measure or estimate for them.

44. A rigid tank of small mass contains 40.0 g of argon, initiallyat 200°C and 100 kPa. The tank is placed into a reservoirat 0°C and allowed to cool to thermal equilibrium. (a) Cal-culate the volume of the tank. (b) Calculate the change ininternal energy of the argon. (c) Calculate the energytransferred by heat. (d) Calculate the change in entropy ofthe argon. (e) Calculate the change in entropy of the con-stant-temperature bath.

A 1.00-mol sample of H2 gas is contained in the left-handside of the container shown in Figure P22.45, which hasequal volumes left and right. The right-hand side is evacu-ated. When the valve is opened, the gas streams into theright-hand side. What is the final entropy change of thegas? Does the temperature of the gas change?

45.

41.

Problems 699

Valve

VacuumH2

Figure P22.45

T (K) P (kPa) V (cm3) E int

A 293 100 500BC 1 023DA

(c) Identify the energy input Q h , the energy exhaust Q c ,and the net output work Weng. (d) Calculate the thermalefficiency. (e) Find the number of crankshaft revolutionsper minute required for a one-cylinder engine to havean output power of 1.00 kW ! 1.34 hp. Note that thethermodynamic cycle involves four piston strokes.

Section 22.6 Entropy35. An ice tray contains 500 g of liquid water at 0°C. Calculate

the change in entropy of the water as it freezes slowly andcompletely at 0°C.

36. At a pressure of 1 atm, liquid helium boils at 4.20 K . Thelatent heat of vaporization is 20.5 kJ/kg. Determine theentropy change (per kilogram) of the helium resultingfrom vaporization.

Calculate the change in entropy of 250 g of water heatedslowly from 20.0°C to 80.0°C. (Suggestion: Note thatdQ ! mc dT.)

37.

(b) Fill in the table below to follow the processes:

644 Heat Engines, Entropy, and the Second Law of Thermodynamics

Section 22.5 Gasoline and Diesel Engines

P22.31 (a) PV P Vi i f fγ γ=

P PVVf i

i

f=FHGIKJ = ×

FHG

IKJ =

γ

3 00 1050 0300

24461.40

..

Pa cm cm

kPa3

3e j

(b) W PdVV

V

i

i

= z P PVVi

i= FHGIKJγ

Integrating,

W PVVVi i

i

f=

−FHGIKJ −

FHGIKJ

L

NMM

O

QPP = × × −

FHG

IKJ

LNMM

OQPP

=

−

−11

1 2 50 3 00 10 5 00 10 150 0300

192

1

6 50 400

γ

γ

. . ..

.

a fe je j Pa m cm cm

J

33

3

P22.32 Compression ratio = 6 00. , γ = 1 40.

(a) Efficiency of an Otto-engine eVV

= −FHGIKJ

−

1 2

1

1γ

e = − FHGIKJ =1

16 00

51 2%0 400

..

.

.

(b) If actual efficiency ′ =e 15 0%. losses in system are e e− ′ = 36 2%. .

P22.33 eV V

Otto = − = − = −− −1

11

1

6 201

1

6 201 21 7 5 1 0 400b g a f a fb gγ

. . .

eOtto = 0 518.

We have assumed the fuel-air mixture to behave like a diatomic gas.

Now eW

Q

W t

Q th h= =

eng eng

Qt

W t

eQt

Q W Q

Qt

Qt

W

tQt

h

h

h c

c h

c

= =

=

= +

= −

= × −FHG

IKJ =

eng

eng

eng

hp746 W 1 hp

kW

W hp746 W1 hp

kW

1020 518

146

146 10 102 70 83

.

.

Chapter 22 645

P22.34 (a), (b) The quantity of gas is

nP VRT

E nRT P V

A A

A

A A A A

= =× ×

⋅=

= = = × × =

−

−

100 10 500 10

8 314 2930 020 5

52

52

52

100 10 500 10 125

3 6

3 6

Pa m

J mol K K mol

Pa m J

3

int, 3

e je jb ga f

e je j.

.

In process AB, P PVVB A

A

B=FHGIKJ = × = ×γ

100 10 8 00 1 84 103 1.40 6 Pa Pae ja f. .

TP VnR

E nRT

BB B

B B

= =× ×

⋅=

= = ⋅ =

−1 84 10 500 10 8 00

0 020 5 8 314673

52

52

0 020 5 8 314 673 287

6 6. .

. .

. .

Pa m

mol J mol K K

mol J mol K K J

3

int,

e je jb gb gb gb ga f

so ∆E Q W WABint, out out J J J= − = = − = −287 125 162 0 WAB = −162 J

Process BC takes us to:

PnRT

V

E nRT

E Q W Q

Q

CC

C

C

BC

BC

= =⋅

×= ×

= = ⋅ =

= − = = − = −

=

−

0 020 5 8 314 1 023

62 5 102 79 10

52

52

0 020 5 8 314 1 023 436

436 287 149 0

149

66. .

..

. .

mol J mol K K

m Pa

mol J mol K K J

J J J

J

3

int, C

int, out

b gb gb g

b gb gb g

In process CD:

P PVV

TP V

nR

E nRT

E Q W W

W

D CC

D

DD D

D D

CD

CD

=FHGIKJ = × F

HGIKJ = ×

= =× ×

⋅=

= = ⋅ =

= − = − = − = −

=

−

γ

2 79 101

8 001 52 10

1 52 10 500 10

0 020 5 8 314445

52

52

0 020 5 8 314 445 190

190 436 246 0

246

61.40

5

5 6

..

.

.

. .

. .

Pa Pa

Pa m

mol J mol K K

mol J mol K K J

J J J

J

3

int,

int, out out

e j

e je jb gb gb gb ga f

∆

and ∆E E E Q W QDA A Dint, int, int, out J J J= − = − = − = − = −125 190 65 0 0.

QDA = −65 0. J

continued on next page

646 Heat Engines, Entropy, and the Second Law of Thermodynamics

For the entire cycle, ∆Eint, net J= + − − =162 149 246 65 0 0. . The net work is

W

Q

eng

net

J J J

J J J

= − + + + =

= + + − =

162 0 246 0 84 3

0 149 0 65 0 84 3

.

. .

The tables look like:

State T(K) P(kPa) V(cm3 ) Eint (J)A 293 100 500 125B 673 1 840 62.5 287C 1 023 2 790 62.5 436D 445 152 500 190A 293 100 500 125

Process Q(J) output W(J) ∆Eint (J)AB 0 –162 162BC 149 0 149CD 0 246 –246DA –65.0 0 –65.0

ABCDA 84.3 84.3 0

(c) The input energy is Qh = 149 J , the waste is Qc = 65 0. J , and Weng J= 84 3. .

(d) The efficiency is: eW

Qh= = =

eng J149 J84 3

0 565.

. .

(e) Let f represent the angular speed of the crankshaft. Then f2

is the frequency at which we

obtain work in the amount of 84.3 J/cycle:

1 0002

84 3

2 00084 3

23 7 1 42 103

J s J cycle

J s J cycle

rev s rev min

= FHGIKJ

= = = ×

f

f

.

.. .

b g

Section 22.6 Entropy

P22.35 For a freezing process,

∆∆

SQ

T= =

− ×= −

0 500 3 33 10

273610

5. . kg J kg

K J K

b ge j.