A Carnot engine whose heat sinks at 27 ° C , has an efficiency of 25 % . By how many degrees should the…

Efficiency of carnot engine is given by, $\eta =1-\frac{T_{\mathrm{sink}}}{T_{\mathrm{source}}}$.

For initial case: $T_{\mathrm{sink}}=\left(27+273\right) °\mathrm{C}=300 \mathrm{K}$

So we can write, $\frac{1}{4}=1-\frac{300}{T_1}$$\Rightarrow T_1=400 \mathrm{K}$

Now for the second case: $T_{\mathrm{sink}}=300 \mathrm{K}$

After increasing the previous efficiency by 100%, value of efficiency will get doubled.

Therefore, $\frac{1}{2}=1-\frac{300}{T_2}$ $\Rightarrow T_2=600 \mathrm{K}$

Increase in temperature required will be, $600 \mathrm{K}-400 \mathrm{K}=200 \mathrm{K}$