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

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$372 \,K$ and $310 \,K$

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(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.Given $\eta_1 = \frac{1}{6}$, we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$, which implies $\frac{T_2}{T_1} = \frac{5}{6}$ or $T_2 = \frac{5}{6} T_1$ (Eq. $1$).When the sink temperature is reduced by $62^{\circ} C$, the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is $\eta_2 = 2 	imes \eta_1 = 2 	imes \frac{1}{6} = \frac{1}{3}$.Thus, $\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$.Substituting $T_2 = \frac{5}{6} T_1$ into the equation: $\frac{1}{3} = 1 - \frac{\frac{5}{6} T_1 - 62}{T_1} = 1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{6} + \frac{62}{T_1}$.Rearranging gives $\frac{1}{3} - \frac{1}{6} = \frac{62}{T_1}$, so $\frac{1}{6} = \frac{62}{T_1}$, which means $T_1 = 372 \,K$.Using Eq. $1$, $T_2 = \frac{5}{6} 	imes 372 = 310 \,K$.

Consider a reversible engine of efficiency $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{\circ} C$, its efficiency gets doubled. The temperature of the source and sink respectively are

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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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