(d)

Temperature of sink; \(T_2=300 K\)

Original efficiency; \(\eta=50 \%=0.5\)

Let initial temperature \(: \rightarrow T_1\)

we know; \(\eta=1-\frac{T_2}{T_1}\)

Substituting the values we get:

\(0.5=1-\frac{300}{T_1}\)

\(\Rightarrow T_1=600\,K\)

Now; new efficiency; \(\eta^{\prime}=70 \%=.7\)

New initial temperature \(=T_1^{\prime}\)

\(\Rightarrow 0.7=1-\frac{T_2}{\left(T_1\right)^{\prime}}\)

\(\Rightarrow 0.7=1-\frac{300}{\left(T_1\right)^{\prime}}\)

\(\Rightarrow\left(T_1\right)^{\prime}=1000\,K\)

\(\therefore\) Increase in source temperature is:

\(\Delta T =(1000-600)\,K\)

\(\Rightarrow \Delta T =400\,K\)

So the temperature of the source should be increased by \(400\,K\).