JEE PHYSICS MCQ: THERMODYNAMICS

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An ideal heat engine exhausting heat at

{27^ \circ }C

is to have 25% efficiency. It must take heat at

View answer and explanation

Correct option: A

{127^ \circ }C

Option A (

{127^ \circ }C

) is correct because it is calculated using the efficiency formula for a heat engine and converting temperatures to Kelvin.

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0.25\,\,{m^3}

{10^3}N/{m^2},

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P

V

\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)