Q. An ideal heat engine exhausting heat at $27℃$ is to have 25% efficiency. It must take heat at

A) $127℃$
B) $227℃$
C) $327℃$
D) None of these
Explanation

(a) $T_{2} = 27℃ = (27 + 273)K = 300\ K,\ \ \ \eta = 25\% = \frac{1}{4}$
We know that, $\eta = 1 - \frac{T_{2}}{T_{1}}$
$\Rightarrow \ \ \ \ \ \frac{1}{4} = 1 - \frac{300}{T_{1}}$
$or\ \ \frac{300}{T_{1}} = 1 - \frac{1}{4}$
$\Rightarrow \ \ \frac{300}{T_{1}} = \frac{3}{4}$
$or\ \ T_{1} = \frac{300 \times 4}{3}$
$\Rightarrow {\ \ T}_{1} = 400\ k$
$or\ \ \ T_{1} = (400 - 273)℃ = 127℃$

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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