Question

A carnot engine, efficiency $40$% and temperature of sink $300K$ to increase efficiency up to $60$% calculate change in temperature of source.

Answer 1

Given : $T_L=300K$
Efficiency of carnot engine $\eta =1-\frac{T_L}{T_H}$
Case 1 : $\eta =0.4$
$\therefore$ $0.4=1-\frac{300}{T_H}$
$⟹T_H=500K$
Case 2 : $\eta =0.6$
$\therefore$ $0.6=1-\frac{300}{T_H^{\prime }}$
$⟹T_H^{\prime }=750K$
Thus change in temperature of source $\Delta T_H=750-500=250K$