A Carnot engine with efficiency 50 % takes heat from a source at 600   K . In order to increase the…

Efficiency of Carnot engine is given by $\eta =1-\frac{T_{\mathrm{sink}}}{ T_{\mathrm{source}}}=1-\frac{T_2}{T_1}$

Given: Initial efficiency $\eta =\frac{1}{2}$

$\therefore \frac{1}{2}=1-\frac{T_2}{600}$

$\Rightarrow \frac{T_2}{600}=\frac{1}{2}$

$\Rightarrow T_2=300 \mathrm{K}$

When efficiency is increased to $70\%$ and $T_2=300 \mathrm{K}$,

Let $T^{\text{'}}$ be new temperature of source 

$\Rightarrow \frac{7}{10}=1-\frac{300}{T^{\text{'}}}$

$\Rightarrow \frac{300}{T^{\text{'}}}=1-\frac{7}{10}$

$\therefore T^{\text{'}}=1000 \mathrm{K}$.