A refrigerator extracts 65 J of heat from a heat reservoir at 0.00°C. If the coefficient of performance of the refrigerator is 2.5, then how much work is needed to run the refrigerator?

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) of a refrigerator is given by $COP = \frac{Q_L}{W}$, where $Q_L$ is the heat extracted from the cold reservoir and $W$ is the work input. Rearranging for $W$, we get $W = \frac{Q_L}{COP}$. Substituting $Q_L = 65 \, \text{J}$ and $COP = 2.5$, we find $W = \frac{65 \, \text{J}}{2.5} = 26 \, \text{J}$.

A Refrigerator Extracts 65 J of Heat from a Heat