A Carnot engine converts 1/6th of the heat in | vedclass

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(D) The efficiency $\eta$ 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 temp

Vedclass · Accepted Answer

$372 \ K, 310 \ K$

Vedclass · Answer

(D) The efficiency $\eta$ 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 $\eta_1 = 1/6$,so $1/6 = 1 - T_2/T_1$,which implies $T_2/T_1 = 5/6$ or $T_2 = \frac{5}{6}T_1$.When the sink temperature is reduced by $62 \ K$,the new efficiency $\eta_2$ becomes double the original,i.e.,$\eta_2 = 2 	imes (1/6) = 1/3$.So,$1/3 = 1 - \frac{T_2 - 62}{T_1}$.Substituting $T_2 = \frac{5}{6}T_1$ into the equation: $1/3 = 1 - \frac{\frac{5}{6}T_1 - 62}{T_1}$.$1/3 = 1 - (5/6 - 62/T_1) = 1 - 5/6 + 62/T_1 = 1/6 + 62/T_1$.$62/T_1 = 1/3 - 1/6 = 1/6$.$T_1 = 62 	imes 6 = 372 \ K$.Now,$T_2 = \frac{5}{6} 	imes 372 = 310 \ K$.Thus,the source temperature is $372 \ K$ and the sink temperature is $310 \ K$.

$A$ Carnot engine converts $1/6$th of the heat into work. When the temperature of the sink is reduced by $62 \ K$,the efficiency of the engine is doubled. Calculate the temperatures of the source and the sink.

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