Question

A Carnot engine whose sink is at $300K$ has an efficiency of $40\%$. By how much should the temperature of source be increased so as to increase its efficiency by $50\%$ of original efficiency?

• A. $280K$
• B. $275K$
• C. $325K$
• D. $250K$

Answer 1

The correct option is D $250K$

Temperature of sink $T_L=300$ $K$

Original efficiency $\eta =40$ $\%$ $=0.4$

Let the Initial temperature of the source be $T_H$

Using $\eta =1-\frac{T_L}{T_H}$

$\therefore$ $0.4=1-\frac{300}{T_H}$ $⟹T_H=500$ $K$

Now the efficiency of the engine is increased by $50$ $\%$ of original efficiency,

$\therefore$ New efficiency ${\eta }^{\prime }=40$ $\%$ $+20$ $\%$ $=60$ $\%$

$\therefore$ ${\eta }^{\prime }=1-\frac{T_L}{T_H^{\prime }}$

OR $0.6=1-\frac{300}{T_H^{\prime }}$

OR $\frac{300}{T_H^{\prime }}=0.4$ $⟹T_H^{\prime }=750$ $K$

Increase in source temperature $\Delta T_H=T_H^{\prime }-T_H=750-500=250$ $K$