JEE Main 2023PhysicsThermodynamicsMediumMCQ

JEE Main 2023 — Thermodynamics Question with Solution

JEE Main 2023 (08 Apr Shift 2)

Question

Work done by a Carnot engine operating between temperatures 127°C and 27°C is 2 kJ. The amount of heat transferred to the engine by the reservoir is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A8 kJ

Step-by-step explanation

The temperatures in Kelvin scales are 

T2=27 °C+273=300 KT1=127 °C+273=400 K

The efficiency of the Carnot engine is

 η=1-T2T1=1-300400=14
Using the relation, η=WQ
⇒2 kJQ=14⇒Q=8 kJ

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About this question

This is a previous-year question from JEE Main 2023, covering the Thermodynamics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.