A Carnot engine with efficiency 50 % takes he | vedclass

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.Given initial efficiency $\eta_1 = 50 \% = 0.5$ and source temperature

Vedclass · Accepted Answer

$1000$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.Given initial efficiency $\eta_1 = 50 \% = 0.5$ and source temperature $T_{H1} = 600 \ K$.$0.5 = 1 - \frac{T_C}{600} \implies \frac{T_C}{600} = 0.5 \implies T_C = 300 \ K$.Now,the efficiency is increased to $\eta_2 = 70 \% = 0.7$ while keeping the sink temperature $T_C = 300 \ K$ constant.$0.7 = 1 - \frac{300}{T_{H2}}$.$\frac{300}{T_{H2}} = 1 - 0.7 = 0.3$.$T_{H2} = \frac{300}{0.3} = 1000 \ K$.Thus,the new temperature of the source is $1000 \ K$.

$A$ Carnot engine with efficiency $50 \%$ takes heat from a source at $600 \ K$. To increase the efficiency to $70 \%$,keeping the temperature of the sink same,the new temperature of the source will be (in $K$)

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