Q. An ideal gas heat engine is operating between $227℃$ and $127℃$. It absorbs $10^{4}$ J of heat at the higher temperature. The amount of heat converted into work is
A) 2000 J
B) 4000 J
C) 8000 J
D) 5600 J
Explanation
(a) $\eta = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{(127 + 273)}{(227 + 273)} = \frac{1}{5}$
$W = \eta Q_{1} = \frac{1}{5} \times 10^{4}\ J = 2000\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.