A Carnot engine,whose efficiency is 40%,takes | vedclass

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Question

(B) The efficiency $\eta$ 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 temp

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(B) The efficiency $\eta$ 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: $\eta_2 = 0.6$,$T_2 = 300\ K$ (same sink temperature).$0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.4 \implies T_1' = \frac{300}{0.4} = 750\ K$.Therefore,the required intake temperature is $750\ K$.

$A$ 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 ....... $K$.

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