P21 026

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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:

ε =

W

|Q

H

|

=

0.500 kJ/s

0.107

= 4.66 kJ/s .

(b) Considering Eq. 21-6 on a per unit time basis,we find 4.66

0.500 = 4.16 kJ/s for the rate of heat

exhaust.


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