A Carnot engine whose efficiency is 40% takes | 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 source temperature and $T_2$ is the sink temperature

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(B) 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: $\eta_2 = 0.6$,$T_2 = 300 \ K$,and we need to find the new source temperature $T_1'$.$0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.4$.$T_1' = \frac{300}{0.4} = 750 \ K$.

$A$ Carnot engine whose efficiency is $40\%$ takes in heat from a source maintained at a temperature of $500 \ K$. It is desired to have an engine of efficiency $60\%$. Then,the intake temperature for the same exhaust (sink) temperature must be (in $K$)

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