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(A) The coefficient of performance $(COP)$ of a refrigerator is given by the formula: $COP = \frac{T_2}{T_1 - T_2}$, where $T_2$ is the temperature of

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$5$

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(A) The coefficient of performance $(COP)$ of a refrigerator is given by the formula: $COP = \frac{T_2}{T_1 - T_2}$, where $T_2$ is the temperature of the low-temperature reservoir and $T_1$ is the temperature of the high-temperature reservoir.Given:$T_2 = -23^{\circ}C = (-23 + 273) \, K = 250 \, K$$T_1 = 27^{\circ}C = (27 + 273) \, K = 300 \, K$Substituting these values into the formula:$COP = \frac{250}{300 - 250}$$COP = \frac{250}{50}$$COP = 5$Thus, the theoretical coefficient of performance is $5$.

In a mechanical refrigerator, the low temperature coils are at a temperature of $-23^{\circ}C$ and the compressed gas in the condenser has a temperature of $27^{\circ}C$. The theoretical coefficient of performance is

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