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26. (a) Eq. 2111 leads to
ε
=1
−
T
L
T
H
=1
−
333 K
373 K
=0
.
107
.
We recall that a Watt is Joulepersecond. Thus, the (net) work done by the cycle per unit time is
the given value 500 J/s. Therefore, by Eq. 219, 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.
 Fall '08
 SPRUNGER
 Physics, Work, Heat

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