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(C) The coefficient of performance $(\beta)$ of a refrigerator is defined as $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold

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$200 \, J$

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(C) The coefficient of performance $(\beta)$ of a refrigerator is defined as $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.Given $\beta = 0.25 = \frac{1}{4}$, so $Q_2 = \frac{W}{4}$.The heat released to the hot reservoir is $Q_1 = Q_2 + W$.Substituting $Q_2 = \frac{W}{4}$, we get $Q_1 = \frac{W}{4} + W = \frac{5W}{4}$.Given $Q_1 = 250 \, J$, we have $250 = \frac{5W}{4}$.Solving for $W$, $W = 250 	imes \frac{4}{5} = 200 \, J$.

$A$ refrigerator with a coefficient of performance $0.25$ releases $250 \, J$ of heat to a hot reservoir. The work done on the working substance is

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