A refrigerator removes heat from the freezing compartment at the rate of 20kj and ejects 24...
A refrigerator removes heat from the freezing compartment at the rate of 20kj and ejects 24 km into a rookper cycle. How much work is required in each cycle?
Homework Answers
Solution)
We know, Work done,
W = Q2 - Q1 / n
Where, n= 1 - Q1 / Q2
Efficiency, n= 1 - 20 / 24
So, n = 0.16
Now,
W = Q2 - Q1 / n
Hence,
W = 24 - 20 / 0.16
W = 24 k J
===========
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