 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/section-a/a6/solution

For a reversible refrigerator, zero net entropy production gives

Q_{h} / T_{h} = Q_{c} / T_{c}

. Thus

Q_{h}^{m i n} = 1 J \times \frac{300}{4} = 75 J .

The necessary work is

W = Q_{h} - Q_{c} = 74 J

. Irreversibility increases the heat dumped to the surroundings;

74 J

is the minimum work, not the requested rejected heat.

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