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Q.

A refrigerator with coefficient of performance 1/3 releases 200 J of heat to a hot reservoir, then the work done on the working substance is

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a

1003Joule

b

100 Joule

c

2003Joule

d

150 Joule

answer is D.

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Detailed Solution

Coefficient of performanceβ=Q2Q1−Q2=133Q2=Q1−Q24Q2=Q1=200Q2=50W=Q1−Q2=200−50=150J
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