Q. In a Carnot engine, the temperature of reservoir is $972℃$ and that of sink is $27℃.$ If the work done by the engine when it transfers heat from reservoir to sink is $12.6 \times 10^{6}J,$ the quantity of heat absorbed y the engine from the reservoir is
A) $16.8 \times 10^{6}J$
B) $4 \times 10^{6}J$
C) $7.6 \times 10^{6}J$
D) $4.25 \times 10^{6}J$
Explanation
(a) Efficiency of engine $\eta = \frac{Work\ done}{Heat\ in\ put}$
Also, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = 1 - \frac{T_{2}}{T_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{W}{Q} = 1 - \frac{T_{2}}{T_{1}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{12.6 \times 10^{6}}{Q} = 1 - \frac{27 + 273}{927 + 273}$
$Q = 16.8 \times 10^{6}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.