Q. 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) 280 K
B) 275 K
C) 325 K
D) 250 K
Explanation
(d) $\frac{T_{2}}{T_{1}} = 1 - \eta = 1 - \frac{40}{100} = \frac{3}{5}$
$\therefore\ T_{1} = \frac{5}{3}T_{2} = \frac{5}{3} \times 300 = 500K$
Increase in efficiency $= 50\%$ of $40\% = 20\%$
$\therefore$ New efficiency $\eta^{'} = 40 + 20 = 60\%$
$\therefore\ \frac{T_{2}}{{T'}_{1}} = 1 - \eta^{'} = 1 - \frac{60}{100} = \frac{2}{5}$
$T_{1}^{'} = \frac{5}{2} \times 300 = 750\ K$
Increase in temperature of source $= T_{1}^{'} - T_{1}$
$= 750 - 500 = 250\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.