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

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Question

(C) 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

$400$

Vedclass · Answer

(C) 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.Given $\eta_1 = 50\% = 0.5$ and $T_2 = 500\, K$.$0.5 = 1 - \frac{500}{T_1} \Rightarrow \frac{500}{T_1} = 0.5 \Rightarrow T_1 = 1000\, K$.Now,the source temperature $T_1$ is kept constant at $1000\, K$ and the efficiency is to be increased to $\eta_2 = 60\% = 0.6$.$0.6 = 1 - \frac{T_2'}{1000}$.$\frac{T_2'}{1000} = 1 - 0.6 = 0.4$.$T_2' = 0.4 	imes 1000 = 400\, K$.Thus,the required temperature of the sink is $400\, K$.

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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