The efficiency of a Carnot engine is 50 % whe | 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

$400$

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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.Given $\eta = 0.5$ and $T_2 = 500 \ K$,we have $0.5 = 1 - \frac{500}{T_1}$,which gives $\frac{500}{T_1} = 0.5$,so $T_1 = 1000 \ K$.Now,keeping the source temperature $T_1 = 1000 \ K$ constant,we want to achieve an efficiency $\eta' = 0.6$ by changing the sink temperature to $T_2'$.Using the formula $\eta' = 1 - \frac{T_2'}{T_1}$,we get $0.6 = 1 - \frac{T_2'}{1000}$.Rearranging the terms,$\frac{T_2'}{1000} = 1 - 0.6 = 0.4$.Therefore,$T_2' = 0.4 	imes 1000 = 400 \ K$.

The efficiency of a Carnot engine is $50 \%$ when the temperature of the sink is $500 \ K$. If the efficiency of the Carnot engine is to be increased to $60 \%$,what should be the temperature of the sink,keeping the temperature of the source constant (in $K$)?

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