If the sink and source temperatures of a refr | vedclass

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(C) The efficiency (or coefficient of performance) of a refrigerator is defined by the temperatures of the sink $(T_2)$ and the source $(T_1)$.Note: I

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$0.0382$

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(C) The efficiency (or coefficient of performance) of a refrigerator is defined by the temperatures of the sink $(T_2)$ and the source $(T_1)$.Note: In thermodynamics,the efficiency of a refrigerator is often represented by the coefficient of performance $(COP = T_2 / (T_1 - T_2))$,but if the question asks for the Carnot efficiency limit $\eta = 1 - (T_2 / T_1)$,we calculate it as follows:First,convert temperatures to Kelvin:$T_2 = 273 + 4 = 277 \ K$$T_1 = 273 + 15 = 288 \ K$Using the formula $\eta = 1 - (T_2 / T_1)$:$\eta = 1 - (277 / 288)$$\eta = (288 - 277) / 288$$\eta = 11 / 288$$\eta \approx 0.0382$

If the sink and source temperatures of a refrigerator are $4^{\circ} C$ and $15^{\circ} C$ respectively,then the efficiency of the refrigerator is:

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