A refrigerator with a coefficient of performance $\frac{1}{3}$ releases $200 J$ of heat to a hot reservoir. Then the work done on the working substance is

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$\frac{100}{3} J$

$100 J$

$\frac{200}{3} J$

$150 J$

answer is D.

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Detailed Solution

The coefficient of performance of a refrigerator is given by

$β = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}}$

Substituting the given values, we get,

$\frac{1}{3} = \frac{Q_{2}}{200 - Q_{2}} o r, 200 - Q_{2} = 3 Q_{2} 4 Q_{2} = 200 Q_{2} = 50 J W = Q_{1} - Q_{2} = 200 J - 50 J = 150 J$

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