## Solutions to Physics: Principles with Applications, 5/E, Giancoli ...

Aug 12, 2011 ... Solutions to Physics: Principles with Applications, 5/E, Giancoli. Chapter 15. Page 15 – 1. CHAPTER 15. 1. We use the first law of ...

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

CHAPTER 15 1.

We use the first law of thermodynamics to find the change in internal energy: ®U = Q – W – 1.8 × 103 J. = – 3.42 × 103 J – (– 1.6 × 103 J) =

2.

(a) The internal energy of an ideal gas depends only on the temperature, U = 8nRT, so we have ®U = 8nR ®T = 0. (b) We use the first law of thermodynamics to find the heat absorbed: ®U = Q – W; 0 = Q – 4.40 × 103 J , which gives Q = 4.40 × 103 J.

3.

Pressure

p Isobaric

p1 p1/2

Isothermal

4.

V

V1 2V1 Volume

0

For the isothermal process, we have P1V1 = P2V2 ; (6.5 atm)(1.0 L) = (1.0 atm)V2 , which gives V2 = 6.5 L.

P (atm) B

1.0 0.5

C 0

5.

P P0

B

0

A

C

P0/2

V0/2

V0

A

V

Page 15 – 1

1.0

2.0

V (L)

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

6.

(a) Because there is no change in volume, we have W= 0. (b) We use the first law of thermodynamics to find the change in internal energy: ®U = Q – W – 265 kJ. = – 265 kJ – 0 =

7.

(a) In an adiabatic process there is no heat flow: Q= 0. (b) We use the first law of thermodynamics to find the change in internal energy: ®U = Q – W + 1350 J. = 0 – (– 1350 J) – 0 = (c) The internal energy of an ideal gas depends only on the temperature, U = 8nRT, so we see that an rise. increase in the internal energy means that the temperature must

8.

(a) Work is done only in the constant pressure process: W = pa(Vb – Va)

P (atm)

= (5.0 atm)(1.013 × – 0.400) × = 1.3 × 102 J. (b) Because the initial and final temperatures are the same, we know that ®U = 0. We use the first law of thermodynamics to find the heat flow for the entire process: ®U = Q – W; 0 = Q – (+ 1.3 × 102 J), which gives Q = + 1.3 × 102 J. 105 N/m2 á atm)[(0.660

9.

10–3

(a) Work is done only in the constant pressure process: W = pb(Vc – Vb) = (1.4 atm)(1.013 × 105 N/m2 á atm)[(9.3 – 6.8) × 10–3 m3] 3.5 × 102 J. = (b) Because the initial and final temperatures are the same, we know that ®U = 0. (c) We use the first law of thermodynamics to find the heat flow for the entire process: ®U = Q – W; 0 = Q – (+ 3.5 × 102 J), + 3.5 × 102 J (into the gas). which gives Q =

10. (b) Work done in the constant pressure process is W = p1(V2 – V1)

m3]

b

c

0

400

660

V (mL)

P (atm) a

2.2

c

1.4

b

0

6.8

9.3

V (L)

P (N/m2)

= (450 N/m2)(8.00 m3 – 2.00 m3) = 2.70 × 103 J. The internal energy of an ideal gas depends only on the temperature, so we have ®U = 8nR ®T = 8(nRT2 – nRT1) = 8(p2V2 – p1V1)

500

= 8p1(V2 – V1) = 8W = 8(2.70 × 103 J) = 4.05 × 103 J. (d) Because both paths have the same initial and final states, the change in internal energy is the same: 4.05 × 103 J. ®U =

100

Page 15 – 2

a

5.0

400 300 200

0

1

(A)

2

(B) 2.0 4.0 6.0 8.0 10.0

V (m3)

Solutions to Physics: Principles with Applications, 5/E, Giancoli

11. (a) We can find the internal energy change Uc – Ua from the information for the curved path a→c: Uc – Ua = Qa→c – Wa→c = – 63 J – (– 35 J) = – 28 J. For the path a→b→c, we have Uc – Ua = Qa→b→c – Wa→b→c = Qa→b→c – Wa→b ; – 28 J = Qa→b→c – (– 48 J), which gives Qa→b→c = – 76 J. (b) For the path c→d→a, work is done only during the constant pressure process, c→d, so we have Wc→d→a = Pc(Vd – Vc) = !Pb(Va – Vb)

Chapter 15

P b

a

c

d V

0

+ 24 J. = !Wb→a = – !Wa→b = – !(– 48 J) = (c) We use the first law of thermodynamics for the path c→d→a to find Qc→d→a : Ua – Uc = – (Uc – Ua) = Qc→d→a – Wc→d→a ; – (– 28 J) = Qc→d→a – (24 J), which gives Qc→d→a = + 52 J. + 28 J. (d) Ua – Uc = – (Uc – Ua) = – (– 28 J) = (e) Because there is no work done for the path d→a, we have Ua – Ud = (Ua – Uc) + (Uc – Ud) = (Ua – Uc) – (Ud – Uc) = Qd→a – Wd→a ; + 28 J – (+ 5 J) = Qd→a – 0, which gives Qd→a = + 23 J.

12. (a) We can find the internal energy change Ua – Uc from the information for the curved path a→c: P (Uc – Ua) = – (Ua – Uc) = Qa→c – Wa→c = – 80 J – (– 55 J) = – 25 J, b + 25 J. so Ua – Uc = (b) We use the first law of thermodynamics for the path c→d→a to find Qc→d→a : Ua – Uc = Qc→d→a – Wc→d→a ; c + 25 J = Qc→d→a – (+ 38 J), which gives Qc→d→a = + 63 J. (c) For the path a→b→c, work is done only during the constant 0 pressure process, a→b, so we have Wa→b→c = Pa(Vb – Va) = (2.5)Pd(Vc – Vd) = (2.5)Wd→c = – (2.5)Wc→d = – (2.5)(+ 38 J) = – 95 J. (d) We use the first law of thermodynamics for the path a→b→c to find Qa→b→c : Uc – Ua = Qa→b→c – Wa→b→c ; – 25 J = Qa→b→c – (– 95 J), which gives Qa→b→c = – 120 J. (e) Because there is no work done for the path b→c, we have Uc – Ub = (Uc – Ua) + (Ua – Ub) = Qb→c – Wb→c ; – 15 J. – 25 J + (+ 10 J) = Qb→c – 0, which gives Qb→c =

Page 15 – 3

a

d V

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

13. Because the directions along the path are opposite to the directions in Problem 12, all terms for Q and W will have the opposite sign. (a) For the work done around the cycle, we have Wcycle = Wc→b→a + Wa→d→c = – Wa→b→c – Wc→d→a = – (– 95 J) – (+ 38 J) = + 57 J. (b) For the heat flow during the cycle, we have Qcycle = Qc→b→a + Qa→d→c = – Qa→b→c – Qc→d→a = – (– 120 J) – (+ 63 J) = + 57 J. (c) For the internal energy change of the cycle, we have Ucycle = Qcycle – Wcycle = (+ 57 J) – (+ 57 J) = 0. Because the system returns to the initial state, this must always be true for a cycle. (d) The intake heat is Qc→b→a . For the efficiency, we have e = Wcycle/Qc→b→a = (57 J)/(120 J) = 0.475 = 48%.

14. Using the metabolic rates from Table 15–1, we have E = [(8.0 h)(70 W) + (1.0 h)(460 W) + (4.0 h)(230 W) + (10.0 h)(115 W) + (1.0 h)(1150 W)](3600 s/h) =

P b

a

c

d

0

1.5 × 107 J (3.6 × 103 kcal).

15. Using the metabolic rates from Table 15–1, we have Rate = E/t = [(8.0 h)(70 W) + (8.0 h)(115 W) + (4.0 h)(230 W) + (2.0 h)(115 W) + 1.6 × 102 W. (1.5 h)(460 W) + (0.5 h)(1150 W)]/(24 h) =

16. Using the metabolic rates from Table 15–1, we find the energy difference per year: ®E/t = [(1.0 h/day)(230 W – 70 W)(365 days/yr)(3600 s/h) = 2.1 × 108 J/yr. We find the equivalent amount of fat from 5.6 kg. ®m = (2.1 × 108 J/yr)/(4186 J/kcal)(9000 kcal/kg) =

17. For the heat input, we have QH = QL + W = 8200 J + 3200 J = 11,400 J. We find the efficiency from e = W/QH = (3200 J)/(11,400 J) = 0.28 =

28%.

18. We find the efficiency from e = W/QH = (7200 J)/(12.0 kcal)(4186 J/kcal) = 0.143 =

14.3%.

19. The maximum efficiency is the efficiency of the Carnot cycle: e = 1 – (TL/TH) = 1 – [(593 K)/(853 K)] = 0.305 = 30.5%.

20. We find the temperature from the Carnot efficiency: e = 1 – (TL/TH); 0.28 = 1 – [(503 K)/TH], which gives TH = 699 K =

Page 15 – 4

426¡C.

V

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

21. The efficiency of the plant is e = 0.75eCarnot = 0.75[1 – (TL/TH)] = 0.75{1 – [(623 K)/(873 K)]} = 0.215. We find the intake heat flow rate from e = W/QH ; 0.215 = (1.3 × 109 J/s)(3600 s/h)/(QH/t), which gives QH/t = 2.18 × 1013 J/h. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 2.18 × 1013 J/h – (1.3 × 109 J/s)(3600 s/h) = 1.7 × 1013 J/h.

22. We find the efficiency from e = W/QH = (W/t)/(QH/t) = (440 × 103 J/s)/(680 kcal)(4186 J/kcal) = 0.155. We find the temperature from the Carnot efficiency: e = 1 – (TL/TH); 439¡C. 0.155 = 1 – [TL/(843 K)], which gives TL = 712 K =

23. (a) The efficiency of the engine is e = feCarnot ; 0.15 = f [1 – (TL/TH)] = f {1 – [(358 K)/(773 K)]}, which gives f = 0.28 = 28%. (b) The power output of the engine is P = (100 hp)(746 W/hp) = 7.46 × 104 W. We find the intake heat flow rate from e = P/(QH/t); 0.15 = (7.46 × 104 W)(3600 s/h)/(QH/t), which gives QH/t = 1.79 × 109 J/h. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 1.79 × 109 J/h – (7.46 × 104 J/s)(3600 s/h) = 1.52 × 109 J/h (3.64 × 105 kcal/h).

24. We find the low temperature from the Carnot efficiency: e1 = 1 – (TL/TH1); 0.30 = 1 – [TL/(823 K)], which gives TL = 576 K. Because the low temperature does not change, for the new efficiency we have e2 = 1 – (TL/TH2); 687¡C. 0.40 = 1 – [(576 K)/TH2], which gives TH2 = 960 K =

25. We find the high temperature from the Carnot efficiency: e1 = 1 – (TL1/TH); 0.39 = 1 – [(623 K)/TH], which gives TH = 1021 K. Because the high temperature does not change, for the new efficiency we have e2 = 1 – (TL2/TH); 0.50 = 1 – [TL2/(1021 K)], which gives TL2 = 511 K = 238¡C.

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Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

26. For the efficiencies of the engines, we have e1 = 0.60eCarnot = 0.60[1 – (TL1/TH1)] = 0.60{1 – [(713 K)/(943 K)]} = 0.146; e2 = 0.60eCarnot = 0.60[1 – (TL2/TH2)] = 0.60{1 – [(563 K)/(703 K)]} = 0.119. Because coal is burned to produce the input heat to the first engine, we need to find QH1. We relate W1 to QH1 from the efficiency: e1 = W1/QH1 , or W1 = 0.146QH1 . Because the exhaust heat from the first engine is the input heat to the second engine, we have e2 = W2/QH1 = W2/QL1 , or W2 = 0.119QL1 . For the first engine we know that QL1 = QH1 – W1 , so we get W2 = 0.119QL1 = 0.119(QH1 – 0.146QH1) = 0.102QH1 . For the total work, we have W = W1 + W2 = 0.146QH1 + 0.102QH1 = 0.248QH1 . When we use the rate at which this work is done, we get 1000 × 106 W = 0.248(QH1/t), which gives QH1/t = 4.03 × 109 J/s. We find the rate at which coal must be burned from m/t = (4.03 × 109 J/s)/(2.8 × 107 J/kg) = 144 kg/s.

TL2 Q2 Engine W2

29. The coefficient of performance for the refrigerator is CP = QL/W = TL/(TH – TL) = (258 K)/(295 K – 258 K) =

7.0.

30. We find the low temperature from the coefficient of performance: CP = QL/W = TL/(TH – TL); – 21¡C. 5.0 = TL/(302 K – TL), which gives = 252 K =

31. (a) The coefficient of performance for the heat pump is CP1 = TH/(TH – TL1) = (295 K)/(295 K – 273 K) = 13.4. We find the work from CP1 = QH/W1 ; 13.4 = (2800 J)/W1 , which gives W1 = 2.1 × 102 J. (b) The coefficient of performance for the heat pump is now CP2 = TH/(TH – TL2) = (295 K)/(295 K – 258 K) = 7.97. We find the work from CP2 = QH/W2 ; 7.97 = (2800 J)/W2 , which gives W2 = 3.5 × 102 J. Page 15 – 6

TL1 Engine W1

Q1

27. We find the discharge heat flow for the plant from QL2/t = (QH1/t) – (W/t) = 4.03 × 109 J/s – 1000 × 106 J/s = 3.03 × 109 J/s. We find the rate at which water must pass through the plant from QL2/t = (m/t )c ®T; (3.03 × 109 J/s)(3600 s/h) = (m/t )(4186 J/kg á C¡)(6.0 C¡), which gives m/t =

28. The maximum coefficient of performance for a cooling coil is CP = QL/W = TL/(TH – TL) = (258 K)/(303 K – 258 K) = 5.7.

TH2

Qi

TH1

4.3 × 108 kg/h.

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Page 15 – 7

Chapter 15

Solutions to Physics: Principles with Applications, 5/E, Giancoli

32. The efficiency of the engine is e = 1 – (TL/TH). For the coefficient of performance, we have CP = TH/(TH – TL) = 1/[1 – (TL/TH)] = 1/e = 1/0.35 =

Chapter 15

2.9.

33. We assume a room temperature of 20¡C. The efficiency of the engine is e = 1 – (TL/TH). We find the coefficient of performance of the refrigerator from the efficiency of the engine: CP = QL/W = TL/(TH – TL) = (TL/TH)/[1 – (TL/TH)] = (1 – e)/e = (1 – 0.30)/(0.30) = 2.33. The heat flow required to cool and freeze the water is QL = m(c ®T + L). We find the time from t = W/P = (QL/CP)/P = m(c ®T + L)/(CP)P = (2.33)(12)(0.040 kg){[(4186 J/kg á C¡)(20¡C – 0¡C)] + 3.33 × 105 J/kg}/(450 W) = 190 s = 3.2 min.

34. The condensation occurs at constant temperature. Because there is a heat flow out of the steam, we have – 0.145 kcal/K. ®S = Q/T = – mL/T = – (0.100 kg)(539 kcal/kg)/(373 K) =

35. The heating does not occur at constant temperature. To estimate the entropy change, we will use the average temperature of 50¡C. Because there is a heat flow into the water, we have ®S ≈ Q/T = mc ®T/Tav = (1.00 kg)(1.00 kcal/kg á C¡)(100¡C – 0¡C)/(323 K) = + 0.31 kcal/K.

36. The freezing occurs at constant temperature. Because there is a heat flow out of the water, we have ®S = Q/T = – mL/T = – (1.00 m3)(1000 kg/m3)(79.7 kcal/kg)/(273 K) = – 292 kcal/K.

37. We assume that when the ice is formed at 0¡C, it is removed, so its entropy change will be the same as in Problem 36. The heat flow from the water went into the great deal of ice at – 10¡C. Because there is a great deal of ice, its temperature will not change. We find the entropy change of the ice from ®Sice = Q/Tice = + mL/T = (1.00 m3)(1000 kg/m3)(79.7 kcal/kg)/(263 K) = + 303 kcal/K. Thus the total entropy change is ®Sice = ®Sice + ®S = + 303 kcal/K + (– 292 kcal/K) = + 11 kcal/K. If the new ice were not removed, we would include an additional heat term which would be negative for cooling the new ice and positive for the great deal of ice. The net additional entropy change would be small.

38. The total rate of the entropy change is ®Stotal/t = ®Ssource/t + ®Swater/t = (– Q/t)/Tsource + (+ Q/t)/Twater = (– 7.50 cal/s)/(513 K) + (+ 7.50 cal/s)/(300 K) = + 0.0104 cal/K á s.

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Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

39. We find the final temperature from heat lost = heat gained; mhotcwater ®Thot = mcoldcwater ®Tcold ; (1.0 kg)(1000 cal/kg á C¡)(60¡C – T) = (1.0 kg)(1000 cal/kg á C¡)(T – 30¡C), which gives T = 45¡C. The heat flow from the hot water to the cold water is Q = mcoldcwater ®Tcold = (1.0 kg)(1000 cal/kg á C¡)(45¡C – 30¡C) = 1.5 × 104 cal. The heating and cooling do not occur at constant temperature. To estimate the entropy changes, we will use the average temperatures: Tcoldav = !(Tcold + T) = !(30¡C + 45¡C) = 37.5¡C; Thotav = !(Thot + T) = !(60¡C + 45¡C) = 52.5¡C. The total entropy change is ®S = ®Shot + ®Scold = (– Q/Thotav) + (+ Q/Tcoldav) = [(– 1.5 × 104 cal)/(325.7 K)] + [(+ 1.5 × 104 cal)/(310.7 K)] =

+ 2.2 cal/K.

40. The aluminum and water are isolated, so we find the final temperature from heat lost = heat gained; mAlcAl ®TAl = mwatercwater ®Twater ; (3.8 kg)(0.22 kcal/kg á C¡)(30¡C – T) = (1.0 kg)(1.00 kcal/kg á C¡)(T – 20¡C), which gives T = 24.55¡C. The heat flow from the aluminum to the water is Q = mAlcAl ®TAl = (3.8 kg)(0.22 kcal/kg á C¡)(30¡C – 24.55¡C) = 4.55 kcal. The heating and cooling do not occur at constant temperature. To estimate the entropy changes, we will use the average temperatures: Twaterav = !(Twater + T) = !(20¡C + 24.55¡C) = 22.3¡C; TAlav = !(TAl + T) = !(30¡C + 24.55¡C) = 27.3¡C. The total entropy change is ®S = ®SAl + ®Swater = (– Q/TAlav) + (+ Q/Twaterav) = [(– 4.55 × 103 cal)/(300.5 K)] + [(+ 4.55 × 103 cal)/(295.5 K)] =

+ 0.26 cal/K.

41. (a) The actual efficiency of the engine is eactual = W/QH = (550 J)/(2200 J) = 0.250 = 25.0%. The ideal efficiency is eideal = 1 – (TL/TH) = 1 – [(650 K)/(970 K)] = 0.330 = 33.0%. Thus the engine is running at 75.8% of ideal. eactual /eideal = (0.250)/(0.330) = 0.758 = (b) We find the heat exhausted in one cycle from QL = QH – W = 2200 J – 550 J = 1650 J. In one cycle of the engine there is no entropy change for the engine. The input heat is taken from the universe at TH and the exhaust heat is added to the universe at TL. The total entropy change is ®Stotal = (– QH/TH) + (QL/TL) = [– (2200 J)/(970 K)] + [(1650 J)/(650 K)] = + 0.270 J/K. (c) For a Carnot engine we have W = eQH = 0.330QH ; QL = QH – W = (1 – e)QH = (1 – 0.330)QH = 0.667 QH . The total entropy change is ®Stotal = (– QH/TH) + (QL/TL) 0, as expected for an ideal engine. = [– QH/(970 K)] + [+ 0.667 QH/(650 K)] =

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Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

42. Because each die has six possible results, there are (6)(6) = 36 possible microstates. (a) We find the number of microstates that give a 5 by listing all the possibilities: (1, 4), (2, 3), (3, 2), and (4, 1), so there are 4 microstates. Thus the probability of obtaining a 5 is 1/9. P5 = (4 microstates)/(36 microstates) = (b) We find the number of microstates that give an 11 by listing all the possibilities: (5, 6), and (6, 5), so there are 2 microstates. Thus the probability of obtaining an 11 is P11 = (2 microstates)/(36 microstates) = 1/18.

43. We are not concerned with the order that the cards are placed in the hand. (a) One of the possible microstates is the four aces and one of the four kings. Because the suit of the king is not specified, there are four different possibilities, and thus the macrostate of four aces and one king has 4 microstates. (b) Because the deck contains only one of each card specified, there is only 1 microstate for the macrostate of six of hearts, eight of diamonds, queen of clubs, three of hearts, jack of spades. (c) If we call the four jacks J1, J2, J3, J4, without regard to order we have the following possible pairs: J1J2, J1J3, J1J4, J2J3, J2J4, J3J4, so there are 6 combinations for the jacks. Similarly, there will be 6 combinations for the queens, but only 4 combinations for the ace. Thus the total number of microstates for the macrostate of two jacks, two queens, and an ace is (6)(6)(4) = 144. (d) We construct the hand by considering the number of ways we can draw each card. Because we are not concerned with any specific values, there will be 52 possibilities for the first card. For the second draw, because we cannot have any of the cards equal in value to the first one, there will be only 48 possibilities. Similarly, there will be 44 possibilities for the third draw, 40 possibilities for the fourth draw, and 36 possibilities for the fifth draw. If we were concerned with order, the total number of possibilities would be the product of these. The number of microstates must be reduced by dividing by the number of ways of arranging five cards, which is (5)(4)(3)(2)(1) = 120. Thus the number of microstates for a hand with no two equal-value cards is (52)(48)(44)(40)(36)/120 = 1.32 × 106. The probability will increase with the number of microstates, so the order is (b), (a), (c), (d).

44. We use H for head, and T for tail. For the microstates we construct the following table: Macrostate Microstates Number 6 heads HHHHHH 1 5 heads, 1 tail H H H H H T, H H H H T H, H H H T H H, H H T H H H, H T H H H H, T H H H H H 6 4 heads, 2 tails H H H H T T, H H H T H T, H H H T T H, H H T H H T, H H T H T H, H H T T H H, 15 H T H H H T, H T H H T H, H T H T H H, H T T H H H, T H H H H T, T H H H T H, T H H T H H, T H T H H H,T T H H H H 3 heads, 3 tails H H H T T T, H H T H T T, H H T T H T, H H T T T H, H T H H T T, H T H T H T, H T H T T H, 20 H T T H H T, H T T H T H, H T T T H H, T H H H T T, T H H T H T, T H H T T H, T H T H H T, T H T H T H, T H T T H H, T T H H H T, T T H H T H, T T H T H H, T T TH H H 2 heads, 4 tails H H T T T T, H T H T T T, H T T H T T, H T T T H T, H T T T T H, T H H T T T, 15 T H T H T T, T H T T H T, T H T T T H, T T H H T T, T T H T H T, T T H T T H, T T T H H T, T T T H T H, T T T T H H 1 head, 5 tails H T T T T T, T H T T T T, T T H T T T, T T T H T T, T T T T H T, T T T T T H 6 6 tails TTTTTT 1

There are a total of 64 microstates. (a) The probability of obtaining three heads and three tails is P33 = (20 microstates)/(64 microstates) = 5/16. (b) The probability of obtaining six heads is Page 15 – 10

Solutions to Physics: Principles with Applications, 5/E, Giancoli

P60 = (1 microstate)/(64 microstates) =

1/64.

Page 15 – 11

Chapter 15

Solutions to Physics: Principles with Applications, 5/E, Giancoli

Chapter 15

45. (a) We assume that our hand is the first dealt and we receive the aces in the first four cards. The probability of the first card being an ace is 4/52. For the next three cards dealt to the other players, there must be no aces, so the probabilities are 48/51, 47/50, 46/49. The next card is ours; the probability of it being an ace is 3/48. For the next three cards, there must be no aces, so the probabilities are 45/47, 44/46, 43/45. For the next cards, we have 2/44, 42/43, 41/42, 40/41, 1/40, 39/39, 38/38,… . For the product of all these, we have P1 = (4/52)(48/51)(47/50)(46/49)(3/48)(45/47)(44/46)(43/45)(2/44)(42/43)(41/42)(40/41)(1/40) = (4)(3)(2)(1)/(52)(51)(50)(49). Because we do not have to receive the aces as the first four cards, we multiply this probability by the number of ways 4 cards can be drawn from a total of 13, which is 13!/4!9! = (13)(12)(11)(10)/(4)(3)(2)(1). Thus the probability of being dealt four aces is P = [(4)(3)(2)(1)/(52)(51)(50)(49)][(13)(12)(11)(10)/(4)(3)(2)(1)] = (13)(12)(11)(10)/(52)(51)(50)(49) = 0.00264 = 1/379. [Note that this is (48! 4!/52!)(13!/4! 9!).] (b) We assume that our hand is the first dealt. Because the suit is not specified, any first card is acceptable, so the probability is (52/52) = 1. For the next three cards dealt to the other players, there must be no cards in the suit we were dealt, so the probabilities are 39/51, 38/50, 37/49. The next card is ours; the probability of it being in the same suit is 12/48. For the next three cards, there must be no cards in the suit we were dealt, so the probabilities are 36/47, 35/46, 34/45. For the next cards, we have 11/44, 33/43, 32/42, 31/41, 10/40, … . For the product of all these, we have P1 = (1)(39/51)(38/50)(37/49)(12/48)(36/47)(35/46)(34/45)(11/44))(33/43)(32/42)(31/41)(10/40)… = 12! 39!/51! = 6.3 × 10–12 = 1/1.59 × 1011.

46. We find the area from E = (IA)t; 3 22x10 Wh/day = (40 W/m2)A(12 h/day), which gives A = 46 m2 ≈ 490 ft2. would probably fit, it would have to be larger because it is Although this area (about 20 ft × 25 ft) not directly facing the sun for the 12 hours.

47. If we assume 100% efficiency, the electrical power is the rate at which the gravitational energy of the water is changed: P = (®m/®t)gh = (35 m3/s)(1000 kg/m3)(9.80 m/s2)(45 m) = 1.5 × 107 W = 15 MW.

48. (a) The energy required to increase the gravitational potential energy of the water is E = (®m/®t)ght 1.32 × 106 kWh. = (1.00 × 105 kg/s)(9.80 m/s2)(135 m)(10.0 h) = 1.32 × 109 Wh = (b) If the stored energy is released at 75% efficiency, we have P = (0.75)E/t = (0.75)(1.32 × 106 kWh)/(14 h) = 7.09 × 104 kW.

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49. Because the change in elevation of the water is not constant during the tidal flow, we take half the difference in height as the average change. In one pass through the turbines, for the work done by the falling water, we have W1 = mghav = ρAhg(!h). If there are two high and two low tides each day, there will be four passes through the turbines, so the total work in one day is W = 4W1 = 2ρAgh2 = 2(1000 kg/m3)(23 × 106 m2)(9.80 m/s2)(8.5 m)2 = 3.3 × 1013 J.

50. We find the heat released from Q = mc ®T = ρVc ®T = (1000 kg/m3)(1000 m)3(4186 J/kg á C¡)(1 K) =

4.2 × 1015 J.

51. The internal energy of an ideal gas depends only on the temperature, U = 8nRT. For an adiabatic process there is no heat flow. For the first law of thermodynamics we have ®U = 8nR ®T = Q – W; 8(1.5 mol)(8.31 J/mol á K) ®T= 0 – (7500 J), which gives ®T = – 401 K = – 401 C¡.

52. First we check to see if energy is conserved: QL + W = 1.50 MW + 1.50 MW = 3.00 MW = QH, so energy is conserved. The efficiency of the engine is e = W/QH = (1.50 MW)/(3.00 MW) = 0.500 = 50.0%. The maximum possible efficiency is the efficiency of the Carnot cycle: e = 1 – (TL/TH) = 1 – [(215 K)/(425 K)] = 0.494 = 49.4%. Yes, there is something fishy, because his claimed efficiency is not possible.

53. (a) Because the pressure is constant, we find the work from (c) W = p(V2 – V1) 2.2 × 105 J. = (1.013 × 105 N/m2)(4.1 m3 – 1.9 m3) = (b) We use the first law of thermodynamics to find the change in internal energy: ®U = Q – W = + 5.30 × 104 J – 2.2 × 105 J = – 1.7 × 105 J.

P (atm)

1.0

0

54. (a) We find the rate at which work is done from 2.0 × 104 J/s. P = W/t = (200 J/cycle/cyl)(4 cyl)(25 cycles/s) = (b) We find the heat input rate from e = W/QH ; 0.25 = (2.0 × 104 J/s)/(QH/t), which gives QH/t = 8.0 × 104 J/s. (c) We find the time to use one gallon from t = E/(QH/t) = (130 × 106 J)/(8.0 × 104 J/s) = 1.6 × 103 s = 27 min.

55. The maximum possible efficiency is the efficiency of the Carnot cycle: e = 1 – (TL/TH) = 1 – [(90 K)/(293 K)] = 0.69 = 69%.

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2.0

4.0

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56. The maximum possible efficiency is the efficiency of the Carnot cycle: e = 1 – (TL/TH) = 1 – [(277 K)/(300 K)] = 0.077 = 7.7%. The engine might be feasible because the great amount of water in the ocean could allow a large flow rate through the engine. Possible adverse environmental effects would be that mixing the waters on a large scale would change the environment for those creatures that live in the cooler water, and change in the surface temperature of the ocean over a large area could cause atmospheric changes.

57. We assume that the loss in kinetic energy is transferred to the environment as a heat flow. For the entropy change we have ®S = Q/T = 2(!mv2)/T = mv2/T 2.9 × 103 J/K. = (1100 kg)[(100 km/h)/(3.6 ks/h)]2/(293 K) =

58. (a) We find the final temperature from heat lost = heat gained; mwatercwater ®Twater = mAlcAl ®TAl ; (0.140 kg)(4186 J/kg á C¡)(50¡C – T) = (0.120 kg)(900 J/kg á C¡)(T – 15¡C), which gives T = 44.6¡C. (b) The heat flow from the water to the aluminum is Q = mAlcAl ®TAl = (0.120 kg)(900 J/kg á C¡)(44.6¡C – 15¡C) = 3.20 × 103 J. The heating and cooling do not occur at constant temperature. To estimate the entropy changes, we will use the average temperatures: Twaterav = !(Twater + T) = !(50¡C + 44.6¡C) = 47.3¡C = 320.5 K; TAlav = !(TAl + T) = !(15¡C + 44.6¡C) = 29.8¡C = 303.0 K. The total entropy change is ®S = ®SAl + ®Swater = (+ Q/TAlav) + (– Q/Twaterav) = [(+ 3.20 × 103 J)/(303.0 K)] + [(– 3.20 × 103 J)/(320.5 K)] =

+ 0.58 J/K.

59. (a) The coefficient of performance for the heat pump is CP = TH/(TH – TL) = (297 K)/(297 K – 279 K) = 17. (b) We find the heat delivered at the high temperature from CP = QH/W ; 6.1 × 107 J/h. 17 = QH/(1000 J/s)(3600 s/h) , which gives QH =

60. We use the units to help us find the rate of heat input to the engine from the burning of the gasoline: QH/t = [(3.0 kcal/gal)/(41 km/gal)](90 km/h)(4186 J/kcal) = 2.76 × 108 J/h. The horsepower is the work done by the engine. We find the efficiency from e = W/QH = (25 hp)(746 W/hp)(3600 s/h)/(2.76 × 108 J/h) = 0.24 = 24%.

61. We assume that the loss in kinetic energy is transferred to the environment as a heat flow. For the entropy change we have ®S = Q/T = !mv2/T 0.15 J/K. = !(15 kg)(2.4 m/s)2/(293 K) =

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62. We find the change in length of the steel I-beam from ®L = Lα ®T = (7.5 m)[11 × 10–6 (C¡)–1](– 6.0 C¡) = – 4.95 × 10–4 m. The heat flow from the steel beam is Q = mc ®T = (300 kg)(0.11 cal/kg á C¡)(– 6.0 C¡)(4186 J/kcal) = – 8.29 × 105 J. Work is done on the beam as the load at the top moves down a distance ®L and the center of gravity moves down a distance !®L. If we take the beam as our system, the work done by the system is negative: W = F ®L + mg!®L = (F + !mg) ®L = [4.3 × 105 N + !(300 kg)(9.80 m/s2)](– 4.95 × 10–4 m) = – 214 J. We use the first law of thermodynamics to find the change in internal energy: ®U = Q – W = – 8.29 × 105 J – (– 214 J) = – 8.3 × 105 J.

63. We find the heat input rate from e = W/QH ; 0.33 = (800 MW)/(QH/t), which gives QH/t = 2424 MW. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 2424 MW – 800 MW = 1624 MW. If this heat flow warms the air, we have QL/t = (n/t)c ®T; (1624 × 106 W)(3600 s/h)(24 h/day) = (n/t)(7.0 cal/mol á C¡)(7.0 C¡)(4.186 J/cal), which gives n/t = 6.84 × 1011 mol/day. To find the volume rate, we use the ideal gas law: P(V/t) = (n/t)RT; (1.013 × 105 Pa)(V/t) = (6.84 × 1011 mol/day)(8.315 J/mol á K)(293 K), 17 km3/day. which gives V/t = 1.65 × 1010 m3/day = Depending on the dispersal by the winds, the local climate could be heated significantly. We find the area from A = (V/t)t/h = (16.5 km3/day)(1 day)/(0.200 km) = 83 km2.

64. (a) The efficiency of the plant is e = 1 – (TL/TH) = 1 – [(285 K)/(600 K)] = 0.525 = 52.5%. We find the heat input rate from e = W/QH ; 0.525 = (900 MW)/(QH/t), which gives QH/t = 1714 MW. We find the discharge heat flow from QL/t = (QH/t) – (W/t) = 1714 MW – 900 MW = 814 MW. If this heat flow warms some river water, which mixes with the rest of the river water, we have QL/t = (n/t)c ®T; (814 × 106 W) = (37 m3/s)(1000 kg/m3)(4186 J/kg á C¡) ®T, which gives ®T = 5.3 C¡. (b) The heat flow does not occur at constant temperature. To estimate the entropy changes, we will use the average temperatures: Twaterav = Twater + ! ®T = 285 K + !(5.3 C¡) = 287.6 K; The entropy change is ®S/m = (QL/t )/(m/t)Twaterav + 77 J/kg á K. = (814 × 106 W)/(37 m3/s)(1000 kg/m3)(287.6 K) =

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