Q. An ideal gas heat engine operates in a Carnot’s cycle between $227℃\ and\ 127℃.$ It absorbs $6 \times 10^{4}J$ at high temperature. The amount of heat converted into work is

A) $1.6 \times 10^{4}J$
B) $1.2 \times 10^{4}J$
C) $4.8 \times 10^{4}J$
D) $3.5 \times 10^{4}J$
Explanation

(b) Using the relation
$\ \ \ \ \ \ \ \ \ \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}\ }{Q_{1}}$
$\ or\ \ \ \frac{W}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}}$
$or\ \ \ \ \frac{W}{Q_{1}} = 1 - \frac{T_{2}}{T_{1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} \right)$
$or\ \ \ \ \ \ W = Q_{1}\left( 1 - \frac{T_{2}}{T_{1}} \right)$
$\therefore\ \ \ W = 6 \times 10^{4}\left( 1 - \frac{(127 + 273)}{(227 + 273)} \right)$
$\ \ or\ \ \ \ \ \ \ W = 6 \times 10^{4}\left( 1 - \frac{400}{500} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6 \times 10^{4} \times \frac{100}{500}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.2 \times 10^{4}\ J$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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