A cornot engine whose heat sinks at 27°C has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

0.25 = 1 - $\frac{T_2}{T_1}$ $\frac{{}_{}}{{}_{}}$

0.25 = 1 - $\frac{\left(27+273\right)}{T_1}\Rightarrow \frac{300}{T_1}=1-0.25$ $\frac{}{{}_{}}\frac{}{{}_{}}$

$\Rightarrow \frac{300}{T_1}=0.75$

$T_1=\frac{300*100}{75}$

T1 = 400 k

Now, efficiency increases by 100%

${\eta }_2=100\mathrm{\%}{\eta }_1+{\eta }_1$

$=2{\eta }_1$

= 0.25 * 2

= 0.50

0.50 = 1 - $\frac{T_2\text{'}}{T_1}$ $\frac{{}_{}}{{}_{}}$

$\Rightarrow \frac{T_2\text{'}}{T_1}=0.50$

$T_1\text{'}=\frac{300}{0.5}$

$T_1\text{'}=600k$

$$ $T_1\text{'}-T_1=$  (600 – 400) = 200 k or 200° C