646_Physics ProblemsTechnical Physics

646_Physics ProblemsTechnical Physics - 648*P22.39 Heat...

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648 Heat Engines, Entropy, and the Second Law of Thermodynamics *P22.39 We take data from the description of Figure 20.2 in section 20.3, and we assume a constant specific heat for each phase. As the ice is warmed from –12°C to 0°C, its entropy increases by S dQ T mc dT T mc T dT mc T S S i f == = = =⋅ ° ° F H G I K J F H G I K J = zz z ice K 273 K ice K 273 K ice K 273 K kg 2 090 J kg C K K kg 2 090 J kg C JK 261 1 261 261 0 027 0 273 261 0 027 0 273 261 254 ln .l n l n n . bg af As the ice melts its entropy change is S Q T mL T f = × = 00270 273 32 9 . . kg 3.33 10 J kg K 5 ej As liquid water warms from 273 K to 373 K, S mc dT T mc T T i f f i F H G I K J ° F H G I K J = z liquid liquid kg 4186 J kg C J K ln . ln . 373 273 35 3 As the water boils and the steam warms, S mL T mc T T S v f i =+ F H G I K J = × +⋅ ° F H G I K J steam 6 kg 2.26 10 J kg K kg2010 Jkg C ln . n . 373 388 373 164 2 14 The total entropy change is 2 54 32 9 35 3 164 2 14 236 .. . . +++ + = a f JK . We could equally well have taken the values for specific heats and latent heats from Tables 20.1 and
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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