26. (a) Eq. 21-11 leads to ε =1 − T L T H =1 − 333 K 373 K =0 . 107 . 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. 21-9, we obtain the heat input per unit time:
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.