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(D) The coefficient of performance $(COP)$ of a refrigerator is given by $\alpha = \frac{Q_2}{W}$,where $Q_2$ is the heat extracted from the cold rese

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$150$

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(D) The coefficient of performance $(COP)$ of a refrigerator is given by $\alpha = \frac{Q_2}{W}$,where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.Also,$W = Q_1 - Q_2$,where $Q_1$ is the heat released to the hot reservoir.Given: $\alpha = 1/3$ and $Q_1 = 200 \ J$.Substituting these into the formula $\alpha = \frac{Q_2}{Q_1 - Q_2}$:$\frac{1}{3} = \frac{Q_2}{200 - Q_2}$$200 - Q_2 = 3Q_2$$4Q_2 = 200 \implies Q_2 = 50 \ J$.Now,calculate the work done $W$:$W = Q_1 - Q_2 = 200 \ J - 50 \ J = 150 \ J$.

$A$ refrigerator with a coefficient of performance of $1/3$ releases $200 \ J$ of heat. What is the work done on the working substance in Joules?

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