JEE Main 2024PhysicsThermodynamicsMediumMCQ

JEE Main 2024 — Thermodynamics Question with Solution

JEE Main 2024 (27 Jan Shift 1)

Question

0.08 kg air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal kg-1 °C-1 and 1 J=4.18 joule cal-1. The change in its internal energy is approximately.

Choose an option

Show full solutionCorrect option: C
Correct answer
C284 J

Step-by-step explanation

The amount of work done for the given process should be zero, as the process is an isochoric process.

Thus, the change in internal energy will be equal to the heat exchanged in the process. Mathematically,

Q=∆U

Now,

∆U=ms∆T   ...1

From equation (1), it follows that

∆U=0.08×(170×4.18)×5 J≈ 284 J

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About this question

This is a previous-year question from JEE Main 2024, 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.