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(D) 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

$232$

Vedclass · Answer

(D) 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 $\eta_1 = 0.25$,we have $0.25 = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = 0.75 = \frac{3}{4}$.When the sink temperature is reduced by $58\,^{\circ}C$,the new sink temperature is $T_2' = T_2 - 58$. The new efficiency is $\eta_2 = 2 	imes 0.25 = 0.50$.Thus,$0.50 = 1 - \frac{T_2 - 58}{T_1}$.Substituting $\frac{T_2}{T_1} = \frac{3}{4}$,we get $0.50 = 1 - (\frac{T_2}{T_1} - \frac{58}{T_1})$.$0.50 = 1 - \frac{3}{4} + \frac{58}{T_1}$.$0.50 = 0.25 + \frac{58}{T_1}$.$0.25 = \frac{58}{T_1} \Rightarrow T_1 = \frac{58}{0.25} = 232\,K$.Since the change in temperature $\Delta T$ is the same in Celsius and Kelvin,the source temperature $T_1$ is $232\,^{\circ}C$.

An engine has an efficiency of $0.25$ when the temperature of the sink is reduced by $58\,^{\circ}C$,its efficiency is doubled. The temperature of the source is ..... $^{\circ}C$.

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