16_InstSolManual_PDF_Part11

16_InstSolManual_PDF_Part11 - Q C 5 2 206 J. Q C 5 Q H 2 W...

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(c) (d) apart from a small difference due to rounding. (e) Reflect: The net for the cycle is This is zero, except for a small difference due to rounding. For any cycle, 16.46. Set Up: The work done by the engine each cycle is with and For a Carnot engine, Solve: (a) The pV diagram is sketched in Figure 16.46. Figure 16.46 (b) and (c) (d) (e) The maximum pressure is for state a . This is also where the volume is a minimum, so 16.47. Set Up: The least possible and the smallest is for a Carnot cycle. For water, and Solve: The heat that must be removed from the water is Q 5 mc D T 2 mL f 5 1 5.00 kg 213 4190 J / kg # K 43 2 20.0 C° 4 2 3.34 3 10 5 J / kg 2 52 2.09 3 10 6 J. Q C Q H 52 T C T H . W , 0. Q C . 0, Q H , 0, 0 Q H 0 5 0 W 0 1 0 Q C 0 . L f 5 3.34 3 10 5 J / kg. c w 5 4190 J / kg # K T H 5 20.0°C 5 293 K. T C 52 5.0°C 5 268 K. 0 Q H 0 0 W 0 p a 5 nRT a V a 5 1 2.00 mol 21 8.315 J / mol # K 21 773 K 2 5.00 3 10 2 3 m 3 5 2.57 3 10 6 Pa. T a 5 T H 5 773 K. 5.00 3 10 2 3 m 3 . V a 5 5.00 L 5 0 Q C 0 5 206 J. e 5 1 2 T C T H 5 1 2 318 K 773 K 5 0.589 5 58.9%. T C 52 T H 1 Q C Q H 2 52 1 773 K 2 1 2 206 J 500 J 2 5 318 K 5 45°C.
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Unformatted text preview: Q C 5 2 206 J. Q C 5 Q H 2 W 5 500 J 2 294 J 5 206 J, W 5 mg D y 5 1 15.0 kg 21 9.80 m / s 2 21 2.00 m 2 5 294 J. a Q H Q C b c d p V Q C , 0. Q H . 0, W 5 Q H 2 Q C . e Carnot 5 1 2 T C T H . Q C Q H 5 2 T C T H . D y 5 2.00 m. m 5 15.0 kg mg D y , D U 5 0. D U tot 5 2183 J 1 1 2 786 J 2 1 1 2 1396 J 2 . D U e , e Carnot . e Carnot 5 1 2 T C T H 5 1 2 300 K 600 K 5 0.500 5 50.0%. e 5 W Q H 5 W Q 1 S 2 5 227 J 2183 J 5 0.104 5 10.4%. Q tot 5 W tot , Q tot 5 Q 1 S 2 1 Q 2 S 3 1 Q 3 S 1 5 2183 J 1 1 1 2 1955 J 2 5 228 J. W tot 5 W 1 S 2 1 W 2 S 3 1 W 3 S 1 5 1 786 J 1 1 2 559 J 2 5 1 227 J The Second Law of Thermodynamics 16-11...
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