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

Answer 1

Answer: (D)

Efficiency of a Carnot engine, $\eta=1-\frac{T_{2}}{T_{1}}$
or, $\frac{T_{2}}{T_{1}}=1-\eta=1-\frac{40}{100}=\frac{3}{5}$
$\therefore T_{1}=\frac{5}{3} \times T_{2}=\frac{5}{3} \times 300=500 K$
Increase in efficiency $=50 \%$ of $40 \%=20 \%$
New efficiency, $\eta^{\prime}=40 \%+20 \%=60 \%$
$\therefore \frac{T_{2}}{T_{1}^{\prime}}=1-\frac{60}{100}=\frac{2}{5} $
$ T_{1}^{\prime}=\frac{5}{2} \times T_{2}=\frac{5}{2} \times 300=750 K $
Increase in temperature of source $=T_{1}^{\prime}-T_{1}$
$=750-500=250 K$.