Question

A Carnot has efficiency $40\%$ (heat sink ${27}^{o}C$). To increase efficiency by $10\%$ of previous efficiency, the temperature be increased by

• A. $15.7K$
• B. $25.7K$
• C. $50.7K$
• D. $35.7K$

Answer 1

The correct option is D $35.7K$

Efficiency is given by:
$\eta =1-\frac{T_{sink}}{T_{source}}$
Initially the efficiency is $40\%$ and $T_{sink}=300K$
$0.4=1-\frac{300}{T_{source}}$
i.e. $T_{source}=500K$

For efficiency to increase by $10\%$,
$\eta =0.44$ (44%)
i.e. $0.44=1-\frac{300}{T_{source}}$
$T_{source}=535.7K$

Hence temperature increases by $35.7K$