23. (a) The eﬃciency is
ε
=
T
H
−
T
L
T
H
=
(235
−
115) K
(235 + 273) K
=0
.
236 = 23
.
6%
.
We note that a temperature diference has the same value on the Kelvin and Celsius scales. Since the
temperatures in the equation must be in Kelvins, the temperature in the denominator is converted
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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

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