Question

A Carnot engine whose sink is at $300\,K$ 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. $275K$
• B. $325K$
• C. $250K$
• D. $380K$

Answer 1

Answer: (C)

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied, i.e.,
$\eta=\frac{ \text{Work done }}{\text{ Heat supplied }}$
$=\frac{W}{Q_{1}}=\frac{Q_{1} - Q_{2}}{Q_{1}}=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$
Here, $T_{1}$ is the temperature of source and $T_{2}$ is the temperature of sink.
As given, $\eta=40\%=\frac{40}{100}=0.4$
and $T_{2}=300K$
So, $0.4=1-\frac{300}{ T_{1}}$ .
$\Rightarrow T_{1}=\frac{300}{1 - 0 . 4}=\frac{300}{0 . 6}=500K$
Let, temperature of the source be increased by $xK,$ then efficiency becomes $\eta^{'}=40\%+50\%$ of $\eta$
$=\frac{40}{100}+\frac{50}{100}\times 0.4=0.4+0.5\times 0.4=0.6$
Hence, $0.6=1-\frac{300}{500 + x}$
$\Rightarrow \frac{300}{500 + x}=0.4$
$\Rightarrow 500+x=\frac{300}{0 . 4}=750$
$x=750-500=250\,K$