1. We use the first law of thermodynamics to find the change in internal energy: U = Q W = J ( J) = J.

Transcription

1 CHAPTER We use the first law of thermodynamics to find the change in internal energy: U = Q W = J ( J) = J. 2. (a) The internal energy of an ideal gas depends only on the temperature, U = 8nRT, so we have U = 8nR T =. (b) We use the first law of thermodynamics to find the heat absorbed: U = Q W; = Q J, which gives Q = J. 3. p Pressure p 1 Isobaric p 1/2 Isothermal V 1 2V 1 Volume V 4. For the isothermal process, we have P 1 V 1 = P 2 V 2 ; (6.5 atm)(1. L) = (1. atm)v 2, which gives V 2 = 6.5 L. P (atm) 1. B A.5 C V (L) P P B A 5. P /2 C V /2 V V Page 15 1

2 6. (a) Because there is no change in volume, we have W =. (b) We use the first law of thermodynamics to find the change in internal energy: U = Q W = 265 kj = 265 kj. 7. (a) In an adiabatic process there is no heat flow: Q =. (b) We use the first law of thermodynamics to find the change in internal energy: U = Q W = ( 135 J) = J. (c) The internal energy of an ideal gas depends only on the temperature, U = 8nRT, so we see that an increase in the internal energy means that the temperature must rise. 8. (a) Work is done only in the constant pressure process: W = p a (V b V a ) = (5. atm)( N/m 2 á atm)[(.66.4) 1 3 m 3 ] = J. (b) Because the initial and final temperatures are the same, we know that U =. We use the first law of thermodynamics to find the heat flow for the entire process: U = Q W; = Q ( J), which gives Q = J. P (atm) 5. a 4 b c 66 V (ml) 9. (a) Work is done only in the constant pressure process: W = p b (V c V b ) P (atm) = (1.4 atm)( N/m 2 á atm)[( ) 1 3 m 3 ] = J. (b) Because the initial and final temperatures are the same, we know that U =. (c) We use the first law of thermodynamics to find the heat flow for the entire process: U = Q W; = Q ( J), which gives Q = J (into the gas) a c b V (L) 1. (b) Work done in the constant pressure process is W = p 1 (V 2 V 1 ) = (45 N/m 2 )(8. m 3 2. m 3 ) = J. The internal energy of an ideal gas depends only on the temperature, so we have U = 8nR T = 8(nRT 2 nrt 1 ) = 8(p 2 V 2 p 1 V 1 ) = 8p 1 (V 2 V 1 ) = 8W = 8( J) = J. (d) Because both paths have the same initial and final states, the change in internal energy is the same: U = J. P (N/m 2 ) (A) (B) V (m 3 ) 1. Page 15 2

3 11. (a) We can find the internal energy change U c from the information for the curved path a c: U c = Q a c W a c = 63 J ( 35 J) = 28 J. For the path a b c, we have U c = Q a b c W a b c = Q a b c W a b ; 28 J = Q a b c ( 48 J), which gives Q a b c = 76 J. P b a (b) For the path c d a, work is done only during the constant pressure process, c d, so we have W c d a = P c (V d V c ) =!P b (V a V b ) =!W b a =!W a b =!( 48 J) = + 24 J. c d V (c) We use the first law of thermodynamics for the path c d a to find Q c d a : U a = (U c ) = Q c d a W c d a ; ( 28 J) = Q c d a (24 J), which gives Q c d a = + 52 J. (d) U a = (U c ) = ( 28 J) = + 28 J. (e) Because there is no work done for the path d a, we have U a U d = (U a ) + (U c U d ) = (U a ) (U d ) = Q d a W d a ; + 28 J (+ 5 J) = Q d a, which gives Q d a = + 23 J. 12. (a) We can find the internal energy change U a from the information for the curved path a c: (U c ) = (U a ) = Q a c W a c = 8 J ( 55 J) = 25 J, so U a = + 25 J. P b a (b) We use the first law of thermodynamics for the path c d a to find Q c d a : U a = Q c d a W c d a ; + 25 J = Q c d a (+ 38 J), which gives Q c d a = + 63 J. c d (c) For the path a b c, work is done only during the constant pressure process, a b, so we have W a b c = P a (V b V a ) = (2.5)P d (V c V d ) = (2.5)W d c = (2.5)W c d = (2.5)(+ 38 J) = 95 J. V (d) We use the first law of thermodynamics for the path a b c to find Q a b c : U c = Q a b c W a b c ; 25 J = Q a b c ( 95 J), which gives Q a b c = 12 J. (e) Because there is no work done for the path b c, we have U c U b = (U c ) + (U a U b ) = Q b c W b c ; 25 J + (+ 1 J) = Q b c, which gives Q b c = 15 J. Page 15 3

4 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 W cycle = W c b a + W a d c = W a b c W c d a = ( 95 J) (+ 38 J) = + 57 J. (b) For the heat flow during the cycle, we have Q cycle = Q c b a + Q a d c = Q a b c Q c d a = ( 12 J) (+ 63 J) = + 57 J. (c) For the internal energy change of the cycle, we have U cycle = Q cycle W cycle = (+ 57 J) (+ 57 J) =. Because the system returns to the initial state, this must always be true for a cycle. (d) The intake heat is Q c b a. For the efficiency, we have e = W cycle /Q c b a = (57 J)/(12 J) =.475 = 48%. P b c a d V 14. Using the metabolic rates from Table 15 1, we have E = [(8. h)(7 W) + (1. h)(46 W) + (4. h)(23 W) + (1. h)(115 W) + (1. h)(115 W)](36 s/h) = J ( kcal). 15. Using the metabolic rates from Table 15 1, we have Rate = E/t = [(8. h)(7 W) + (8. h)(115 W) + (4. h)(23 W) + (2. h)(115 W) + (1.5 h)(46 W) + (.5 h)(115 W)]/(24 h) = W. 16. Using the metabolic rates from Table 15 1, we find the energy difference per year: E/t = [(1. h/day)(23 W 7 W)(365 days/yr)(36 s/h) = J/yr. We find the equivalent amount of fat from m = ( J/yr)/(4186 J/kcal)(9 kcal/kg) = 5.6 kg. 17. For the heat input, we have Q H = Q L + W = 82 J + 32 J = 11,4 J. We find the efficiency from e = W/Q H = (32 J)/(11,4 J) =.28 = 28%. 18. We find the efficiency from e = W/Q H = (72 J)/(12. kcal)(4186 J/kcal) =.143 = 14.3%. 19. The maximum efficiency is the efficiency of the Carnot cycle: ) = 1 [(593 K)/(853 K)] =.35 = 3.5%. 2. We find the temperature from the Carnot efficiency: );.28 = 1 [(53 K) ], which gives T H = 699 K = 426 C. Page 15 4

5 21. The efficiency of the plant is e =.75e Carnot =.75[1 (T L )] =.75{1 [(623 K)/(873 K)]} =.215. We find the intake heat flow rate from e = W/Q H ;.215 = ( J/s)(36 s/h)/(q H /t), which gives Q H /t = J/h. We find the discharge heat flow from Q L /t = (Q H /t) (W/t) = J/h ( J/s)(36 s/h) = J/h. 22. We find the efficiency from e = W/Q H = (W/t)/(Q H /t) = ( J/s)/(68 kcal)(4186 J/kcal) =.155. We find the temperature from the Carnot efficiency: );.155 = 1 [T L /(843 K)], which gives T L = 712 K = 439 C. 23. (a) The efficiency of the engine is e = fe Carnot ;.15 = f [1 (T L )] = f {1 [(358 K)/(773 K)]}, which gives f =.28 = 28%. (b) The power output of the engine is P = (1 hp)(746 W/hp) = W. We find the intake heat flow rate from e = P/(Q H /t);.15 = ( W)(36 s/h)/(q H /t), which gives Q H /t = J/h. We find the discharge heat flow from Q L /t = (Q H /t) (W/t) = J/h ( J/s)(36 s/h) = J/h ( kcal/h). 24. We find the low temperature from the Carnot efficiency: e 1 = 1 (T L 1 );.3 = 1 [T L /(823 K)], which gives T L = 576 K. Because the low temperature does not change, for the new efficiency we have e 2 = 1 (T L 2 );.4 = 1 [(576 K)2 ], which gives T H2 = 96 K = 687 C. 25. We find the high temperature from the Carnot efficiency: e 1 = 1 (T L1 );.39 = 1 [(623 K) ], which gives T H = 121 K. Because the high temperature does not change, for the new efficiency we have e 2 = 1 (T L2 );.5 = 1 [T L2 /(121 K)], which gives T L2 = 511 K = 238 C. Page 15 5

6 26. For the efficiencies of the engines, we have e 1 =.6e Carnot =.6[1 (T L1 1 )] =.6{1 [(713 K)/(943 K)]} =.146; e 2 =.6e Carnot =.6[1 (T L2 2 )] =.6{1 [(563 K)/(73 K)]} =.119. Because coal is burned to produce the input heat to the first engine, we need to find Q H1. We relate W 1 to Q H1 from the efficiency: e 1 = W 1 /Q H1, or W 1 =.146Q H1. Because the exhaust heat from the first engine is the input heat to the second engine, we have e 2 = W 2 /Q H1 = W 2 /Q L1, or W 2 =.119Q L1. For the first engine we know that Q L1 = Q H1 W 1, so we get W 2 =.119Q L1 =.119(Q H1.146Q H1 ) =.12Q H1. For the total work, we have W = W 1 + W 2 =.146Q H1 +.12Q H1 =.248Q H1. When we use the rate at which this work is done, we get W =.248(Q H1 /t), which gives Q H1 /t = J/s. We find the rate at which coal must be burned from m/t = ( J/s)/( J/kg) = 144 kg/s. T L2 Q 2 Engine W 2 T Q H2 i T L1 Engine W 1 Q 1 T H1 27. We find the discharge heat flow for the plant from Q L2 /t = (Q H1 /t) (W/t) = J/s J/s = J/s. We find the rate at which water must pass through the plant from Q L2 /t = (m/t )c T; ( J/s)(36 s/h) = (m/t )(4186 J/kg á C )(6. C ), which gives m/t = kg/h. 28. The maximum coefficient of performance for a cooling coil is CP = Q L /W = T L /(T H T L ) = (258 K)/(33 K 258 K) = The coefficient of performance for the refrigerator is CP = Q L /W = T L /(T H T L ) = (258 K)/(295 K 258 K) = We find the low temperature from the coefficient of performance: CP = Q L /W = T L /(T H T L ); 5. = T L /(32 K T L ), which gives = 252 K = 21 C. 31. (a) The coefficient of performance for the heat pump is CP 1 = T H /(T H T L1 ) = (295 K)/(295 K 273 K) = We find the work from CP 1 = Q H /W 1 ; 13.4 = (28 J)/W 1, which gives W 1 = J. (b) The coefficient of performance for the heat pump is now CP 2 = T H /(T H T L2 ) = (295 K)/(295 K 258 K) = We find the work from CP 2 = Q H /W 2 ; 7.97 = (28 J)/W 2, which gives W 2 = J. Page 15 6

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8 32. The efficiency of the engine is ). For the coefficient of performance, we have CP = T H /(T H T L ) = 1/[1 (T L )] = 1/e = 1/.35 = We assume a room temperature of 2 C. The efficiency of the engine is ). We find the coefficient of performance of the refrigerator from the efficiency of the engine: CP = Q L /W = T L /(T H T L ) = (T L )/[1 (T L )] = (1 e)/e = (1.3)/(.3) = The heat flow required to cool and freeze the water is Q L = m(c T + L). We find the time from t = W/P = (Q L /CP)/P = m(c T + L)/(CP)P = (2.33)(12)(.4 kg){[(4186 J/kg á C )(2 C C)] J/kg}/(45 W) = 19 s = 3.2 min. 34. The condensation occurs at constant temperature. Because there is a heat flow out of the steam, we have S = Q/T = ml/t = (.1 kg)(539 kcal/kg)/(373 K) =.145 kcal/k. 35. The heating does not occur at constant temperature. To estimate the entropy change, we will use the average temperature of 5 C. Because there is a heat flow into the water, we have S Q/T = mc T/T av = (1. kg)(1. kcal/kg á C )(1 C C)/(323 K) = +.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. m 3 )(1 kg/m 3 )(79.7 kcal/kg)/(273 K) = 292 kcal/k. 37. We assume that when the ice is formed at 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 1 C. Because there is a great deal of ice, its temperature will not change. We find the entropy change of the ice from S ice = Q/T ice = + ml/t = (1. m 3 )(1 kg/m 3 )(79.7 kcal/kg)/(263 K) = + 33 kcal/k. Thus the total entropy change is S ice = S ice + S = + 33 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 S total /t = S source /t + S water /t = ( Q/t)/T source + (+ Q/t)/T water = ( 7.5 cal/s)/(513 K) + (+ 7.5 cal/s)/(3 K) = +.14 cal/k á s. Page 15 8

9 39. We find the final temperature from heat lost = heat gained; m hot c water T hot = m cold c water T cold ; (1. kg)(1 cal/kg á C )(6 C T) = (1. kg)(1 cal/kg á C )(T 3 C), which gives T = 45 C. The heat flow from the hot water to the cold water is Q = m cold c water T cold = (1. kg)(1 cal/kg á C )(45 C 3 C) = cal. The heating and cooling do not occur at constant temperature. To estimate the entropy changes, we will use the average temperatures: T coldav =!(T cold + T) =!(3 C + 45 C) = 37.5 C; T hotav =!(T hot + T) =!(6 C + 45 C) = 52.5 C. The total entropy change is S = S hot + S cold = ( Q/T hotav ) + (+ Q/T coldav ) = [( cal)/(325.7 K)] + [( cal)/(31.7 K)] = cal/k. 4. The aluminum and water are isolated, so we find the final temperature from heat lost = heat gained; m Al c Al T Al = m water c water T water ; (3.8 kg)(.22 kcal/kg á C )(3 C T) = (1. kg)(1. kcal/kg á C )(T 2 C), which gives T = C. The heat flow from the aluminum to the water is Q = m Al c Al T Al = (3.8 kg)(.22 kcal/kg á C )(3 C 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: T waterav =!(T water + T) =!(2 C C) = 22.3 C; T Alav =!(T Al + T) =!(3 C C) = 27.3 C. The total entropy change is S = S Al + S water = ( Q/T Alav ) + (+ Q/T waterav ) = [( cal)/(3.5 K)] + [( cal)/(295.5 K)] = +.26 cal/k. 41. (a) The actual efficiency of the engine is e actual = W/Q H = (55 J)/(22 J) =.25 = 25.%. The ideal efficiency is e ideal = 1 (T L ) = 1 [(65 K)/(97 K)] =.33 = 33.%. Thus the engine is running at e actual /e ideal = (.25)/(.33) =.758 = 75.8% of ideal. (b) We find the heat exhausted in one cycle from Q L = Q H W = 22 J 55 J = 165 J. In one cycle of the engine there is no entropy change for the engine. The input heat is taken from the universe at T H and the exhaust heat is added to the universe at T L. The total entropy change is S total = ( Q H ) + (Q L /T L ) = [ (22 J)/(97 K)] + [(165 J)/(65 K)] = +.27 J/K. (c) For a Carnot engine we have W = eq H =.33Q H ; Q L = Q H W = (1 e)q H = (1.33)Q H =.667 Q H. The total entropy change is S total = ( Q H ) + (Q L /T L ) = [ Q H /(97 K)] + [+.667 Q H /(65 K)] =, as expected for an ideal engine. Page 15 9

10 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 P 5 = (4 microstates)/(36 microstates) = 1/9. (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 P 11 = (2 microstates)/(36 microstates) = 1/ 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, 4 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) = 12. Thus the number of microstates for a hand with no two equal-value cards is (52)(48)(44)(4)(36)/12 = 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 H H H H H H 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, 2 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 T T T T T T 1 There are a total of 64 microstates. (a) The probability of obtaining three heads and three tails is P 33 = (2 microstates)/(64 microstates) = 5/16. (b) The probability of obtaining six heads is Page 15 1

11 P 6 = (1 microstate)/(64 microstates) = 1/64. Page 15 11

12 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, 47/5, 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, 4/41, 1/4, 39/39, 38/38,. For the product of all these, we have P 1 = (4/52)(48/51)(47/5)(46/49)(3/48)(45/47)(44/46)(43/45)(2/44)(42/43)(41/42)(4/41)(1/4) = (4)(3)(2)(1)/(52)(51)(5)(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)(1)/(4)(3)(2)(1). Thus the probability of being dealt four aces is P = [(4)(3)(2)(1)/(52)(51)(5)(49)][(13)(12)(11)(1)/(4)(3)(2)(1)] = (13)(12)(11)(1)/(52)(51)(5)(49) =.264 = 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, 38/5, 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, 1/4,. For the product of all these, we have P 1 = (1)(39/51)(38/5)(37/49)(12/48)(36/47)(35/46)(34/45)(11/44))(33/43)(32/42)(31/41)(1/4) = 12! 39!/51! = = 1/ We find the area from E = (IA)t; 22x1 3 Wh/day = (4 W/m 2 )A(12 h/day), which gives A = 46 m 2 49 ft 2. Although this area (about 2 ft 25 ft) would probably fit, it would have to be larger because it is not directly facing the sun for the 12 hours. 47. If we assume 1% efficiency, the electrical power is the rate at which the gravitational energy of the water is changed: P = ( m/ t)gh = (35 m 3 /s)(1 kg/m 3 )(9.8 m/s 2 )(45 m) = W = 15 MW. 48. (a) The energy required to increase the gravitational potential energy of the water is E = ( m/ t)ght = ( kg/s)(9.8 m/s 2 )(135 m)(1. h) = Wh = kwh. (b) If the stored energy is released at 75% efficiency, we have P = (.75)E/t = (.75)( kwh)/(14 h) = kw. Page 15 12

13 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 W 1 = mgh av = 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 = 4W 1 = 2ρAgh 2 = 2(1 kg/m 3 )( m 2 )(9.8 m/s 2 )(8.5 m) 2 = J. 5. We find the heat released from Q = mc T = ρvc T = (1 kg/m 3 )(1 m) 3 (4186 J/kg á C )(1 K) = 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= (75 J), which gives T = 41 K = 41 C. 52. First we check to see if energy is conserved: Q L + W = 1.5 MW MW = 3. MW = Q H, so energy is conserved. The efficiency of the engine is e = W/Q H = (1.5 MW)/(3. MW) =.5 = 5.%. The maximum possible efficiency is the efficiency of the Carnot cycle: ) = 1 [(215 K)/(425 K)] =.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(v 2 V 1 ) P (atm) = ( N/m 2 )(4.1 m m 3 ) = J. (b) We use the first law of thermodynamics to find the change in internal energy: U = Q W = J J = J V (m 3 ) 54. (a) We find the rate at which work is done from P = W/t = (2 J/cycle/cyl)(4 cyl)(25 cycles/s) = J/s. (b) We find the heat input rate from e = W/Q H ;.25 = ( J/s)/(Q H /t), which gives Q H /t = J/s. (c) We find the time to use one gallon from t = E/(Q H /t) = ( J)/( J/s) = s = 27 min. 55. The maximum possible efficiency is the efficiency of the Carnot cycle: ) = 1 [(9 K)/(293 K)] =.69 = 69%. Page 15 13

14 56. The maximum possible efficiency is the efficiency of the Carnot cycle: ) = 1 [(277 K)/(3 K)] =.77 = 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(!mv 2 )/T = mv 2 /T = (11 kg)[(1 km/h)/(3.6 ks/h)] 2 /(293 K) = J/K. 58. (a) We find the final temperature from heat lost = heat gained; m water c water T water = m Al c Al T Al ; (.14 kg)(4186 J/kg á C )(5 C T) = (.12 kg)(9 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 = m Al c Al T Al = (.12 kg)(9 J/kg á C )(44.6 C 15 C) = J. The heating and cooling do not occur at constant temperature. To estimate the entropy changes, we will use the average temperatures: T waterav =!(T water + T) =!(5 C C) = 47.3 C = 32.5 K; T Alav =!(T Al + T) =!(15 C C) = 29.8 C = 33. K. The total entropy change is S = S Al + S water = (+ Q/T Alav ) + ( Q/T waterav ) = [( J)/(33. K)] + [( J)/(32.5 K)] = +.58 J/K. 59. (a) The coefficient of performance for the heat pump is CP = T H /(T H T L ) = (297 K)/(297 K 279 K) = 17. (b) We find the heat delivered at the high temperature from CP = Q H /W ; 17 = Q H /(1 J/s)(36 s/h), which gives Q H = J/h. 6. We use the units to help us find the rate of heat input to the engine from the burning of the gasoline: Q H /t = [(3. kcal/gal)/(41 km/gal)](9 km/h)(4186 J/kcal) = J/h. The horsepower is the work done by the engine. We find the efficiency from e = W/Q H = (25 hp)(746 W/hp)(36 s/h)/( J/h) =.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 =!mv 2 /T =!(15 kg)(2.4 m/s) 2 /(293 K) =.15 J/K. Page 15 14

15 62. We find the change in length of the steel I-beam from L = Lα T = (7.5 m)[ (C ) 1 ]( 6. C ) = m. The heat flow from the steel beam is Q = mc T = (3 kg)(.11 cal/kg á C )( 6. C )(4186 J/kcal) = 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 = [ N +!(3 kg)(9.8 m/s 2 )]( m) = 214 J. We use the first law of thermodynamics to find the change in internal energy: U = Q W = J ( 214 J) = J. 63. We find the heat input rate from e = W/Q H ;.33 = (8 MW)/(Q H /t), which gives Q H /t = 2424 MW. We find the discharge heat flow from Q L /t = (Q H /t) (W/t) = 2424 MW 8 MW = 1624 MW. If this heat flow warms the air, we have Q L /t = (n/t)c T; ( W)(36 s/h)(24 h/day) = (n/t)(7. cal/mol á C )(7. C )(4.186 J/cal), which gives n/t = mol/day. To find the volume rate, we use the ideal gas law: P(V/t) = (n/t)rt; ( Pa)(V/t) = ( mol/day)(8.315 J/mol á K)(293 K), which gives V/t = m 3 /day = 17 km 3 /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 km 3 /day)(1 day)/(.2 km) = 83 km (a) The efficiency of the plant is ) = 1 [(285 K)/(6 K)] =.525 = 52.5%. We find the heat input rate from e = W/Q H ;.525 = (9 MW)/(Q H /t), which gives Q H /t = 1714 MW. We find the discharge heat flow from Q L /t = (Q H /t) (W/t) = 1714 MW 9 MW = 814 MW. If this heat flow warms some river water, which mixes with the rest of the river water, we have Q L /t = (n/t)c T; ( W) = (37 m 3 /s)(1 kg/m 3 )(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: T waterav = T water +! T = 285 K +!(5.3 C ) = K; The entropy change is S/m = (Q L /t )/(m/t)t waterav = ( W)/(37 m 3 /s)(1 kg/m 3 )(287.6 K) = + 77 J/kg á K. Page 15 15