A Carnot engine of efficiency 40 % takes heat | vedclass

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(B) The efficiency $\eta$ of a Carnot engine operating between a source temperature $T_1$ and a sink temperature $T_2$ is given by $\eta = (1 - \frac{

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$750$

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(B) The efficiency $\eta$ of a Carnot engine operating between a source temperature $T_1$ and a sink temperature $T_2$ is given by $\eta = (1 - \frac{T_2}{T_1}) 	imes 100$. For the first case,$\eta_1 = 40 \% = 0.4$ and $T_1 = 500 \ K$. $0.4 = 1 - \frac{T_2}{500} \implies \frac{T_2}{500} = 0.6 \implies T_2 = 300 \ K$. For the second case,we want $\eta_2 = 60 \% = 0.6$ with the same sink temperature $T_2 = 300 \ K$. $0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 1 - 0.6 = 0.4$. $T_1' = \frac{300}{0.4} = 750 \ K$. Thus,the required source temperature is $750 \ K$.

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

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