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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1003Joule

100 Joule

2003Joule

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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