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. 275 K
• B. 325 K
• C. 250 K
• D. 380 K

Answer 1

Answer: (C)

250 K

Answer 2

The efficiency of Cannot engine is defined as the ratio of work done to the heat supplied ie, $\eta=\frac{\text { work done }}{\text { heat supplied }}=\frac{W}{Q_{1}}$ $=\frac{Q_{1}-Q_{2}}{Q_{1}}=1-\frac{Q_{2}}{Q_{1}}$ $=1-\frac{T_{2}}{T_{1}}$ Here, $T_{1}$ is the temperature of source and $T_{2}$ is the temperature of sink. As given, $\eta=40$ and $T_{2}=300\, K$ So. $0.4=1-\frac{300}{T_{1}}$ $\Rightarrow T_{1}=\frac{300}{1-0.4}=\frac{300}{0.6}=500\, K$ Let temperature of the source be increased by $K$, then efficiency becomes $\eta=40$ of $\eta=\frac{400}{100}+\frac{50}{100} \times 0.4$ $=0.4+0.5 \times 0.4=0.6$ Hence $0.6=1-\frac{300}{500+x}$ $\frac{300}{500+x}=0.4$ $\Rightarrow 500+x=\frac{300}{0.4}=750$ $x=750-500=250\, K$