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

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- Solution 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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