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(A) The efficiency of a reversible heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = 0.5$. From this, we find $\frac{T_2}{T_1} = 1 - 0.5 = 0.5$. T

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(A) The efficiency of a reversible heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = 0.5$. From this, we find $\frac{T_2}{T_1} = 1 - 0.5 = 0.5$. The coefficient of performance $(COP)$ of a refrigerator working between the same temperatures is given by $\beta = \frac{T_2}{T_1 - T_2}$. Dividing the numerator and denominator by $T_1$, we get $\beta = \frac{T_2/T_1}{1 - T_2/T_1}$. Substituting the value $\frac{T_2}{T_1} = 0.5$, we get $\beta = \frac{0.5}{1 - 0.5} = \frac{0.5}{0.5} = 1$.

The efficiency of a reversible heat engine working between two temperatures is $50 \%$. The coefficient of performance of a refrigerator working between the same two temperatures but in reverse direction is

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