A reversible engine converts 1/6 of its input | vedclass

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q}$.Initially,$\eta = 1/6$,so $1 - \frac{T_2}{T_1} = \frac{1}

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

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q}$.Initially,$\eta = 1/6$,so $1 - \frac{T_2}{T_1} = \frac{1}{6} \implies \frac{T_2}{T_1} = \frac{5}{6} \implies T_1 = \frac{6}{5}T_2$.When the sink temperature is reduced by $62^{\circ}C$,the new sink temperature is $T_2' = T_2 - 62$.The new efficiency is $\eta' = 2\eta = 2(1/6) = 1/3$.Thus,$1 - \frac{T_2 - 62}{T_1} = \frac{1}{3} \implies \frac{T_2 - 62}{T_1} = \frac{2}{3}$.Substituting $T_1 = \frac{6}{5}T_2$ into the equation: $\frac{T_2 - 62}{(6/5)T_2} = \frac{2}{3}$.$\frac{5(T_2 - 62)}{6T_2} = \frac{2}{3} \implies 15(T_2 - 62) = 12T_2 \implies 15T_2 - 930 = 12T_2$.$3T_2 = 930 \implies T_2 = 310 	ext{ K}$.Converting to Celsius: $T_2 = 310 - 273 = 37^{\circ}C$.Now,$T_1 = \frac{6}{5} 	imes 310 = 6 	imes 62 = 372 	ext{ K}$.Converting to Celsius: $T_1 = 372 - 273 = 99^{\circ}C$.

$A$ reversible engine converts $1/6$ of its input heat into work. When the temperature of the sink is reduced by $62^{\circ}C$,the efficiency of the engine is doubled. Find the temperatures of the source and the sink.

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