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A refrigerator with coefficient of performance 13releases 200 J of heat to a hot reservoir. Then the work done on the working substance is

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a

1003J

b

100 J

c

2003J

d

150 J

answer is D.

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Detailed Solution

The coefficient of performance of a refrigerator is given by β=Q2W=Q2Q1−Q2Substituting the given values, we get 13=Q2200−Q2 or 200−Q2=3Q2or 4Q2=200 or Q2=2004J=50J∴ W=Q1−Q2=200J−50J=150J

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