An ideally efficient freezer cools food to 210K. If room temperature is 300K, and the machine consumes 75 J of work, how much heat was removed from the food?
A) 210 J
B) 75 J
C) 175 J
D) 300 J

Question

Quizplus · Accepted Answer

The efficiency of an ideal refrigerator or freezer is given by the coefficient of performance (COP), which for a refrigerator is defined as $COP = \frac{Q_C}{W}$, where $Q_C$ is the heat removed from the cold reservoir (in this case, the food), and $W$ is the work input. The COP for an ideal refrigerator can also be expressed in terms of temperatures as $COP = \frac{T_C}{T_H - T_C}$, where $T_C$ is the temperature of the cold reservoir (210K in this case) and $T_H$ is the temperature of the hot reservoir (300K in this case). Rearranging the formula to solve for $Q_C$, we get $Q_C = COP \times W$. Substituting the values for $T_C$ and $T_H$ into the COP formula gives us $COP = \frac{210}{300 - 210} = \frac{210}{90}$. Therefore, $Q_C = \frac{210}{90} \times 75 = 2.333 \times 75 = 175 J$.

An Ideally Efficient Freezer Cools Food to 210K