How much work will need to be input to extract 5.0 J of heat from the cold reservoir of a Carnot refrigerator operated between 20 $\circ$ C and 0 $\circ$ C?

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The efficiency of a Carnot refrigerator is given by the formula: $COP = \frac{Q_C}{W}$, where $Q_C$ is the heat extracted from the cold reservoir, and $W$ is the work input. The Coefficient of Performance (COP) for a Carnot refrigerator can also be expressed in terms of temperatures: $COP = \frac{T_C}{T_H - T_C}$, where $T_C$ and $T_H$ are the absolute temperatures (in Kelvin) of the cold and hot reservoirs, respectively.First, convert the temperatures from Celsius to Kelvin:- $20^\circ C = 293 K$- $0^\circ C = 273 K$Then, calculate the COP using the temperature-based formula:$COP = \frac{273}{293 - 273} = \frac{273}{20} = 13.65$Now, use the first formula to find the work input ($W$):$13.65 = \frac{5.0 J}{W}$$W = \frac{5.0 J}{13.65} \approx 0.37 J$Therefore, the work needed to extract 5.0 J of heat from the cold reservoir is approximately 0.37 J.

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