The efficiency of a Carnot engine is 50% and | vedclass

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(D) Given,temperature of sink,$T_2 = 500\,K$.Efficiency,$\eta_1 = 50\% = 0.5$.The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) Given,temperature of sink,$T_2 = 500\,K$.Efficiency,$\eta_1 = 50\% = 0.5$.The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Substituting the values: $0.5 = 1 - \frac{500}{T_1} \implies \frac{500}{T_1} = 0.5 \implies T_1 = 1000\,K$.Now,the temperature of the source $T_1$ is kept constant at $1000\,K$.New efficiency,$\eta_2 = 60\% = 0.6$.Let the new temperature of the sink be $T_2'$.Using the formula: $\eta_2 = 1 - \frac{T_2'}{T_1}$.$0.6 = 1 - \frac{T_2'}{1000}$.$\frac{T_2'}{1000} = 1 - 0.6 = 0.4$.$T_2' = 0.4 	imes 1000 = 400\,K$.

The efficiency of a Carnot engine is $50\%$ and the temperature of the sink is $500\,K$. If the temperature of the source is kept constant and its efficiency is to be raised to $60\%$,then the required temperature of the sink will be ........... $K$.

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