Efficiency of a heat engine whose sink is at | vedclass

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(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

$250$

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 $\eta_1 = 40 \% = 0.4$ and $T_2 = 300 ~K$:$0.4 = 1 - \frac{300}{T_1} \Rightarrow \frac{300}{T_1} = 0.6 \Rightarrow T_1 = \frac{300}{0.6} = 500 ~K$.Now,to increase the efficiency to $\eta_2 = 60 \% = 0.6$ with the same sink temperature $T_2 = 300 ~K$:$0.6 = 1 - \frac{300}{T_1^{\prime}} \Rightarrow \frac{300}{T_1^{\prime}} = 0.4 \Rightarrow T_1^{\prime} = \frac{300}{0.4} = 750 ~K$.The increase in source temperature is $\Delta T = T_1^{\prime} - T_1 = 750 ~K - 500 ~K = 250 ~K$.

Efficiency of a heat engine whose sink is at a temperature of $300 ~K$ is $40 \%$. To increase the efficiency to $60 \%$,keeping the sink temperature constant,the source temperature must be increased by (in $~K$)

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