Question

A Carnot's engine operates with an efficiency of 40% with its sink at $27°C$. The amount by which the temperature of the source be increased with an aim to increase the efficiency by 10% is K.

Answer 1

Let $T_1$ be the initial temperature of source, then
Using $\eta =1-\frac{T_2}{T_1}$
We have $\frac{40}{100}=1-\left(\frac{273+27K}{T_1}\right)$ or $T_1=500K$

For the efficiency to be 10 % more, i.e. 50 %, let $T_1^{{}^{\prime }}$ be the new temperature of the sink.
then, $\frac{50}{100}=1-\frac{(273+27K)}{T_1^{{}^{\prime }}}$ or $T_1^{{}^{\prime }}=600K$
The required increase in the temperature of the source is $T_1^{{}^{\prime }}-T=600K-500K=100K$