Q. An ideal gas heat engine operates in Carnot cycle between $227℃$ and $127℃$ .It absorbs $6 \times 10^{4}$ cal of heat at higher temperature. Amount of heat converted into work is
A) $1.2 \times 10^{4}$ cal
B) $2.4 \times 10^{4}$ cal
C) $6 \times 10^{4}$ cal
D) $4.8 \times 10^{4}$ cal
Explanation
(a) As $\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}\ \ \therefore\ \ \frac{Q_{2}}{6 \times 10^{4}} = \frac{127 + 273}{227 + 273} = \frac{400}{500}$
$Q_{2} = \frac{4}{5} \times 6 \times 10^{4} = 4.8 \times 10^{4}$ cal
$\therefore\ W = Q_{1} - Q_{2} = 6 \times 10^{4} - 4.8 \times 10^{4} = 1.2 \times 10^{4}$ cal
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.