\(\eta=1-\frac{ T _2}{ T _1}\)
\(\therefore \frac{1}{2}=1-\frac{ T _2}{600} \quad \Rightarrow T _2=300\,K\)
\(\Rightarrow \frac{ T _2}{600}=\frac{1}{2} \quad\)
Now efficiency is increased to \(70 \%\) and \(T _2=300\) \(K\), Let temp of source \(T _1= T\)
\(\Rightarrow \frac{7}{10}=1-\frac{300}{T}\)
\(\Rightarrow \frac{300}{ T }=1-\frac{7}{10}\)
\(\Rightarrow \frac{300}{ T }=\frac{3}{10} \quad \therefore T =1000\,K\)