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Q.
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 :
But $\eta=1-\frac{ T _2}{ T _1} $
$ \therefore \frac{1}{2}=1-\frac{ T _2}{600} $
$\Rightarrow \frac{ T _2}{600}=\frac{1}{2} \Rightarrow T _2=300\, K$
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} $
$ \therefore T =1000 \,K$
