Question

The efficiency of carnot's engine is 50%. The temperature of its sink is $7^{o}C$. To increase its efficience 70%. What is the increase in temperature of the source ?

Answer 1

$\begin{array}{c}n=\frac{T_1-T_2}{T_1}=\frac{50}{100}=\frac{T-260}{T}=\frac{1}{2}=1-\frac{266}{T}\\ \Rightarrow \frac{266}{T}=\frac{1}{2}\\ T_s=532k\end{array}$

Now,

$\begin{array}{c}\Delta T_{source}=866.66-532k\\ =354.66\end{array}$