A Carnot engine with 40% efficiency takes hea | vedclass

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(C) 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 temperat

Vedclass · Accepted Answer

$600$

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(C) 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: $\eta_1 = 0.4$,$T_1 = 500 \ K$.$0.4 = 1 - \frac{T_2}{500} \implies \frac{T_2}{500} = 0.6 \implies T_2 = 300 \ K$.Since the exhaust temperature $T_2$ remains the same for the second case:For the second case: $\eta_2 = 0.5$,$T_2 = 300 \ K$.$0.5 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.5 \implies T_1' = \frac{300}{0.5} = 600 \ K$.Thus,the new intake temperature is $600 \ K$.

$A$ Carnot engine with $40\%$ efficiency takes heat at $500 \ K$. If its efficiency is to be $50\%$,then for the same exhaust temperature,the intake temperature must be ..... $K$.

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