An ideal Carnot engine whose efficiency is 50 | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the intake temperature and $T_2$ is the exhaust tempera

Vedclass · Accepted Answer

$625$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the intake temperature and $T_2$ is the exhaust temperature.For the first case, $\eta_1 = 50 \% = 0.5$ and $T_1 = 500 \ K$.$0.5 = 1 - \frac{T_2}{500} \implies \frac{T_2}{500} = 0.5 \implies T_2 = 250 \ K$.For the second case, $\eta_2 = 60 \% = 0.6$ and the exhaust temperature $T_2$ remains $250 \ K$. Let the new intake temperature be $T_1'$.$0.6 = 1 - \frac{250}{T_1'} \implies \frac{250}{T_1'} = 1 - 0.6 = 0.4$.$T_1' = \frac{250}{0.4} = \frac{2500}{4} = 625 \ K$.

An ideal Carnot engine whose efficiency is $50 \%$ receives heat at $500 \ K$. If the efficiency is to be $60 \%$, then the intake temperature for the same exhaust temperature is: (in $K$)

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