Question

The motor in a refrigerator has a power output of 100 W. If the freezing compartment is at 271 K and the outside air is at 306 K, assuming ideal efficiency, what is the maximum amount of heat (in joules) that can be extracted from the freezing compartment in 30.0 minutes?

Answer 1

In refregerator, Q2 (source freezer at T2) energy is extracted and (W) work is done on it by motor whereby much larger energy Q1 is thrown out to air (at T1)
W = Q1-Q2
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efficiency = COP = output/input = Q1/W = useful work/expenditure = Q1/(Q1-Q2) = 1/[1 - (Q2/Q1)]
Q1/W = 1/[1 - (Q2/Q1)]
but Q1/T1 = Q2/T2 >> entropy loss=gain gives Q2/Q1=T2/T1
Q1/W = 1/[1 - (T2/T1)] -------- (1)
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W = power * time = 100(W) * 30*60(s) = 180000 Joule
Now

1 - (T2/T1) =1-271/306 = 35/306
Q1 = W [1/[1 - (T2/T1)] =180000{306/35] = 1573714.3 J
Q2 = from freezer = Q1 - W = 1573714.3 - 180000 =  1393714.3 Joule