A Carnot engine operating between two reservo | vedclass

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.Given $T_H = 99^{\circ} C = 99 + 273 = 372 K$.For the first case,$\eta_

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$62$

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.Given $T_H = 99^{\circ} C = 99 + 273 = 372 K$.For the first case,$\eta_1 = \frac{1}{3} = 1 - \frac{T_C}{372}$.$\frac{T_C}{372} = 1 - \frac{1}{3} = \frac{2}{3}$.$T_C = \frac{2}{3} 	imes 372 = 248 K$.For the second case,the cold reservoir temperature is increased by $x$,so $T_C' = T_C + x = 248 + x$.The new efficiency is $\eta_2 = \frac{1}{6} = 1 - \frac{248 + x}{372}$.$\frac{248 + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$.$248 + x = \frac{5}{6} 	imes 372 = 5 	imes 62 = 310$.$x = 310 - 248 = 62 K$.

$A$ Carnot engine operating between two reservoirs has an efficiency of $\frac{1}{3}$. When the temperature of the cold reservoir is raised by $x$,its efficiency decreases to $\frac{1}{6}$. The value of $x$,if the temperature of the hot reservoir is $99^{\circ} C$,will be $........ K$.

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