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. 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, $\frac{T_2}{T_1}=1-\eta =1-\frac{40}{100}=\frac{3}{5}$
$\therefore T_1=\frac{5}{3}\times T_2=\frac{5}{3}\times 300=500K$
Increase in efficiency $=50\%$ of $40\%=20\%$
New efficiency, ${\eta }^{\prime }=40\%+20\%=60\%$
$\therefore \frac{T_2}{T_1^{\prime }}=1-\frac{60}{100}=\frac{2}{5}$
$T_1^{\prime }=\frac{5}{2}\times T_2=\frac{5}{2}\times 300=750K$
Increase in temperature of source $=T_1^{\prime }-T_1$
$=750-500=250K$.