ch20-p028 - 28. (a) Eq. 20-13 leads to = 1 TL 333 K = 1 =...

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L H 333 K =1 = 0.107. 373 K T T ε −− We recall that a Watt is Joule-per-second. Thus, the (net) work done by the cycle per unit time is the given value 500 J/s. Therefore, by Eq. 20-11, we obtain the heat input per unit
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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