A Carnot engine with efficiency 50,% takes he | vedclass

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(B) Initially,the efficiency $\eta_1 = 50\,\% = 0.5$. The source temperature $T_1 = 600\,K$.Using the Carnot efficiency formula $\eta = 1 - \frac{T_2}

Vedclass · Accepted Answer

$1000$

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(B) Initially,the efficiency $\eta_1 = 50\,\% = 0.5$. The source temperature $T_1 = 600\,K$.Using the Carnot efficiency formula $\eta = 1 - \frac{T_2}{T_1}$:$0.5 = 1 - \frac{T_2}{600}$$\frac{T_2}{600} = 0.5$$T_2 = 300\,K$ (This is the temperature of the sink).Now,the efficiency is increased to $\eta_2 = 70\,\% = 0.7$,while the sink temperature $T_2$ remains $300\,K$. Let the new source temperature be $T_1'$.Using the formula again:$0.7 = 1 - \frac{300}{T_1'}$$\frac{300}{T_1'} = 1 - 0.7 = 0.3$$T_1' = \frac{300}{0.3} = 1000\,K$.

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

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