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Chapter 20

# Chapter 20 - 20.1 a 2200 J 20.2 a 9000 J 20.3 a c 3700...

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20.1: a) %. 8 . 33 338 . 0 b) J. 6500 J 4300 J 2200 6500 2200 = = = + 20.2: a) %. 9 . 28 289 . 0 b) J. 2600 J 6400 J 9000 J 9000 J 2600 = = = - 20.3: a) %. 0 . 23 230 . 0 100 , 16 3700 = = b) J. 12,400 J 3700 J 100 , 16 = - c) . g 350 . 0 kg J 10 60 . 4 J 100 , 16 4 = × d) hp. 298 kW 222 s) 60.0 J)( 3700 ( = = 20.4: a) J. 10 43 . 6 5 ) 280 . 0 ( s) W)(1.00 10 180 ( 1 3 × = = = × Pt Q e b) J. 10 4.63 s) W)(1.00 10 180 ( J 10 43 . 6 5 3 5 × = × - × = - Pt Q 20.5: a) MW. 970 MW 330 MW 1300 b) %. 25 25 . 0 MW 1300 MW 330 = - = = = e 20.6: Solving , for (20.6) Eq. r or ) 1 ln( ln ) 1 ( e r γ - = - . 8 . 13 ) 350 . 0 ( ) 1 ( 5 . 2 1 1 = = - = - - γ e r If the first equation is used (for instance, using a calculator without the y x function), note that the symbol “ e ” is the ideal efficiency , not the base of natural logarithms. 20.7: a) C. 453 K 726 K)(9.5) 15 . 295 ( 0.40 1 ° = = = = - γ a b r T T b) Pa. 10 99 . 1 Pa)(9.50) 10 50 . 8 ( 6 4 × = × = = γ γ a b r p p 20.8: a) From %. 58 58 . 0 ) 8 . 8 ( 1 1 (20.6), Eq. 40 . 0 1 = = - = - = - - γ r e b) %, 60 ) 6 . 9 ( 1 40 . 0 = - - an increase of 2%. If more figures are kept for the efficiencies, the difference is 1.4%. 20.9: a) J. 10 62 . 1 4 10 . 2 J 10 40 . 3 4 × = = = × K Q C W b) J. 10 02 . 5 ) 1 ( 4 1 C C H × = + = + = K Q W Q Q

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20.10: ) ( 1 f C T c L t m K t K Q t W P p + = = = ( 29 W. 128 K) K)(2.5 kg J (485 kg) J 10 60 . 1 ( s 3600 kg 0 . 8 8 . 2 1 5 = + × = 20.11: a) , b) W. 767 s 0 . 60 J 10 80 . 9 J 10 44 . 1 4 5 P H EER = = × - × or . 27 . 7 ) 413 . 3 ( W 767 W 1633 ) 413 . 3 ( ] s) (60 J) 10 8 . 9 ( s) (60 J) 10 44 . 1 ( [ s) (60 J) 10 8 . 9 ( 4 5 4 = = × - × × = EER 20.12: a) ) ( water water ice ice C T c T c L m Q f + + = ( 29 J. 10 90 . 8 K) K)(25.0 kg J 4190 ( K) K)(5.0 kg J 2100 ( kg J 10 334 kg) 80 . 1 ( 5 3 × = + + × = b) J. 10 37 . 3 5 40 . 2 J 10 08 . 8 | | 5 C × = = = × K Q W c) = × = × + × = + = | | that (note J 10 1.14 J 10 8.08 J 10 37 . 3 | | | | H 6 5 5 C H Q Q W Q ).) 1 ( | | 1 C K Q + 20.13: a) J. 215 J 335 J 550 | | | | C H = - = - Q Q b) K. 378 J) 550 J K)(335 620 ( |) | | | ( H C H C = = = Q Q T T c) %. 39 J) 550 J 335 ( 1 ) | | | | ( 1 H C = - = - Q Q 20.14: a) From Eq. (20.13), the rejected heat is J. 10 3.72 J) 6450 )( ( 3 K 520 K 300 × = b) J. 10 2.73 J 10 3.72 J 6450 3 3 × = × - c) From either Eq. (20.4) or Eq. (20.14), e= 0.423=42.3%. 20.15: a) C H f C H C H | | | | T T mL T T Q Q = = J, 10 088 . 3 K) (273.15 K) (287.15 kg) J 10 kg)(334 0 . 85 ( 7 3 × = × = or = - = - = × )) ( 1 ( | | | | | | | | b) figures. two to J 10 09 . 3 H H C H 7 T T Q Q Q W C × × J) 10 09 . 3 ( 7 J. 10 49 . 2 )) 15 . 297 15 . 273 ( 1 ( 6 × = -
20.16: a) From Eq. (20.13), J. 492 J) 415 )( ( K 270 K 320 = b) The work per cycle is J, 77 J 415 J 492 = - and W, 212 ) 75 . 2 ( s 1.00 J 77 = × = P keeping an extra figure. c) 5.4. K) (50 K) 270 ( ) ( C H C = = - T T T 20.17: For all cases, . | | | | | | C H Q Q W - = a) The heat is discarded at a higher temperature, and a refrigerator is required; × × = - = J) 10 00 . 5 ( ) 1 ) (( | | | | 3 C H C T T Q W J. 665 ) 1 ) 15 . 263 298.15 (( = - b) Again, the device is a refrigerator, and J. 190 ) 1 ) 15 . 263 / 15 . 273 (( | | | | C = - = Q W c) The device is an engine; the heat is taken form the hot reservoir, and the work done by the engine is × × = J) 10 00 . 5 ( | | 3 W J. 285 ) 1 ) 15 . 263 248.15 (( = - 20.18: For the smallest amount of electrical energy, use a Carnot cycle. ( 29 J 10 2.09 K) J 10 kg)(334 (5.00 K) 20 ( 4190 kg) (5.00 6 3 K kg J F water freeze C 0 water to Cool in × = × + = + = + = ° mL T mc Q Q Q Carnot cycle: K 293 K 268 J 10 09 . 2 out 6 hot out cold in Q T Q T Q = × = room) the J(into 10 28 . 2 6 out × = Q J 10 09 . 2 J 10 28 . 2 6 6 in out × - × = - = Q Q W energy) al J(electric 10 95 . 1 5 × = W 20.19: The total work that must be done is J 10 4.90 m) 100 )( s m kg)(9.80 500 ( 5 2 tot × = = = mgy W J 250 H = Q Find C Q so can calculate work W done each cycle: H T T Q Q C H C - = [ ] J 7 . 120 K) (773.15 K) (373.15 J) 250 ( ) ( H H C C - = - = - = Q T T Q J 3 . 129 H C = + = Q Q W The number of cycles required is cycles. 3790 J 3 . 129 J 10 09 . 4 5 tot = × = W W

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20.20: For a heat engine, ( 29 ( 29 J, 7500 600 . 0 1 ) J 3000 ( 1 / C H = - - - = - - = e Q Q J. 4500 ) J 7500 )( 600 . 0 ( then and H = = = eQ W This does not make use of the given value of ( 29 ( 29 ( 29 K 320 600 . 0 1 K 800 1 engine, Carnot a for then used, is If .
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