An ideal Carnot engine,whose efficiency is 40 | vedclass

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Question

(C) 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

$600$

Vedclass · Answer

(C) 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.For the first case: $\eta_1 = 0.4$,$T_1 = 500 \; K$.$0.4 = 1 - \frac{T_2}{500} \implies \frac{T_2}{500} = 0.6 \implies T_2 = 300 \; K$.For the second case: $\eta_2 = 0.5$,$T_2 = 300 \; K$ (same exhaust temperature).$0.5 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.5 \implies T_1 = \frac{300}{0.5} = 600 \; K$.

An ideal Carnot engine,whose efficiency is $40 \%$ receives heat at $500 \; K$. If its efficiency is $50 \%$ then the intake temperature for the same exhaust temperature is ......... $K$.

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