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(A) The efficiency $\eta$ of an ideal (Carnot) heat engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_H$ is the temperature of the source and $

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$-\,113$

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(A) The efficiency $\eta$ of an ideal (Carnot) heat engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_H$ is the temperature of the source and $T_L$ is the temperature of the sink in Kelvin.Given: $\eta = 60\, \% = 0.60$ and $T_H = 127\,^{\circ} C = 127 + 273 = 400\, K$.Substituting the values: $0.60 = 1 - \frac{T_L}{400}$.Rearranging the equation: $\frac{T_L}{400} = 1 - 0.60 = 0.40$.$T_L = 0.40 	imes 400 = 160\, K$.Converting back to Celsius: $T_L(^{\circ} C) = 160 - 273 = -113\,^{\circ} C$.

For an ideal heat engine,the temperature of the source is $127\,^{\circ} C$. In order to have $60\, \%$ efficiency,the temperature of the sink should be $........\,{ }^{\circ} C$. (Round off to the nearest integer)

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