For a Carnot engine,the efficiency is given b | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Initially,$\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \qu

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$37^{\circ}C$ and $99^{\circ}C$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Initially,$\eta = 1 - \frac{T_2}{T_1} = \frac{1}{6} \quad \dots(i)$When the sink temperature $T_2$ is reduced by $62^{\circ}C$ (which is equivalent to $62 K$ in absolute scale),the new sink temperature is $T_2' = T_2 - 62$.The new efficiency is $\eta' = 1 - \frac{T_2 - 62}{T_1} = 2\eta$.Substituting $\eta = 1/6$,we get $\eta' = 2 	imes (1/6) = 1/3$.So,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3} \quad \dots(ii)$.From $(i)$,$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6} \implies T_2 = \frac{5}{6}T_1$.Substitute $T_2$ into $(ii)$: $\frac{\frac{5}{6}T_1 - 62}{T_1} = \frac{2}{3} \implies \frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$.$\frac{62}{T_1} = \frac{5}{6} - \frac{4}{6} = \frac{1}{6} \implies T_1 = 62 	imes 6 = 372 K$.Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ}C$.Now,$T_2 = \frac{5}{6} 	imes 372 = 310 K$.Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ}C$.Thus,the sink temperature is $37^{\circ}C$ and the source temperature is $99^{\circ}C$.

For a Carnot engine,the efficiency is given by $W/Q_1 = 1/6$. If the temperature of the sink is reduced by $62^{\circ}C$,the efficiency doubles. The initial temperatures of the sink and the source are,respectively:

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