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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperatu

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$99$

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.Given $\eta = 1/6$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6}$ (Equation $i$).When the sink temperature is reduced by $62^{\circ}C$ (which is equivalent to a change of $62 \ K$),the efficiency doubles to $2 	imes \frac{1}{6} = \frac{1}{3}$.Thus,$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$.Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3}$.Substituting $T_2 = \frac{5}{6}T_1$ from Equation $i$ into this expression:$\frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{2}{3}$.$\frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$.$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6}$.$T_1 = 62 	imes 6 = 372 \ K$.To convert Kelvin to Celsius: $T(^{\circ}C) = T(K) - 273$.$T_1 = 372 - 273 = 99^{\circ}C$.

An engine has an efficiency of $1/6$. When the temperature of the sink is reduced by $62^{\circ}C$,its efficiency is doubled. The temperature of the source is ....... $^{\circ}C$.

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