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JEE Main 20238 AprilEvening Shift

Exam: JEE Main
Year: 2023
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Thermodynamics

JEE Main 2023 · 8 April · Evening Shift · Physics · Thermodynamics

Work done by a Carnot engine operating between temperatures 127 ° C...

Work done by a Carnot engine operating between temperatures

127 ° C

and

27 ° C

is

2kJ

. The amount of heat transferred to the engine by the reservoir is:

Options

A $8kJ$
B $2 . 67kJ$
C $2kJ$
D $4kJ$

Correct answer

A. $8kJ$

Step-by-step solution

The temperatures in Kelvin scales are
${T}_{2} = 27 ° C + 273 = 300K{T}_{1} = 127 ° C + 273 = 400K$
The efficiency of the Carnot engine is
$η = 1 - \dfrac{{T}_{2}}{{T}_{1}} = 1 - \dfrac{300}{400} = \dfrac{1}{4}$
Using the relation, $η = \dfrac{W}{Q}$
$⇒ \dfrac{2kJ}{Q} = \dfrac{1}{4} ⇒ Q = 8kJ$

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