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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

Vedclass · Accepted Answer

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

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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 in Kelvin.Given $\eta = \frac{1}{6}$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_1 = 1.2 T_2$.When the sink temperature is reduced by $62^{\circ} C$,the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is doubled,so $\eta' = 2 	imes \frac{1}{6} = \frac{1}{3}$.Thus,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$.Substituting $T_1 = 1.2 T_2$ into the equation: $\frac{T_2 - 62}{1.2 T_2} = \frac{2}{3}$.$3(T_2 - 62) = 2.4 T_2 \implies 3 T_2 - 186 = 2.4 T_2 \implies 0.6 T_2 = 186 \implies T_2 = 310 \ K$.Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ} C$.Then $T_1 = 1.2 	imes 310 = 372 \ K$. Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ} C$.Therefore,the temperatures are $99^{\circ} C$ and $37^{\circ} C$.

$A$ reversible engine converts one-sixth of the heat supplied into work. When the temperature of the sink is reduced by $62^{\circ} C$,the efficiency of the engine is doubled. The temperatures of the source and sink are

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