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 sink same, the new temperature of the source will be $.........\,K$
JEE MAIN 2023, Medium
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$\text { Initially } \eta=\frac{1}{2}$
$\eta=1-\frac{ T _2}{ T _1}$
$\therefore \frac{1}{2}=1-\frac{ T _2}{600} \quad \Rightarrow T _2=300\,K$
$\Rightarrow \frac{ T _2}{600}=\frac{1}{2} \quad$
Now efficiency is increased to $70 \%$ and $T _2=300$ $K$, Let temp of source $T _1= T$
$\Rightarrow \frac{7}{10}=1-\frac{300}{T}$
$\Rightarrow \frac{300}{ T }=1-\frac{7}{10}$
$\Rightarrow \frac{300}{ T }=\frac{3}{10} \quad \therefore T =1000\,K$
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