JEE PHYSICS MCQ: THERMODYNAMICS

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Understanding +4 | -1

For an engine operating between

{t_1}^ \circ C

and

{t_2}^ \circ C

, the efficiency will be

View answer and explanation

Correct option: D

\frac{{{t_1} - {t_2}}}{{{t_1} + 273}}

Option D (

\frac{{{t_1} - {t_2}}}{{{t_1} + 273}}

) is correct because the efficiency of an engine is calculated using absolute temperatures (Kelvin).

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