A reversible heat engine converts one-fourth | vedclass

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(B) The efficiency of a heat 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.G

Vedclass · Accepted Answer

$208$

Vedclass · Answer

(B) The efficiency of a heat 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 $\eta = \frac{1}{4}$,we have $\frac{1}{4} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{3}{4}$,so $T_2 = 0.75 T_1$.When the sink temperature is reduced by $52 \, K$,the new efficiency $\eta' = 2 	imes \eta = 2 	imes \frac{1}{4} = \frac{1}{2}$.The new sink temperature is $T_2' = T_2 - 52$.Using the efficiency formula for the new state: $\frac{1}{2} = 1 - \frac{T_2 - 52}{T_1}$.This simplifies to $\frac{T_2 - 52}{T_1} = \frac{1}{2}$,or $T_2 - 52 = 0.5 T_1$.Substituting $T_2 = 0.75 T_1$ into the equation: $0.75 T_1 - 52 = 0.5 T_1$.$0.25 T_1 = 52$,which gives $T_1 = \frac{52}{0.25} = 208 \, K$.

$A$ reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by $52 \, K$,its efficiency is doubled. The temperature in Kelvin of the source will be ...... .

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