JEE Main 2019PhysicsThermal Properties of MatterEasyMCQ

JEE Main 2019 — Thermal Properties of Matter Question with Solution

JEE Main 2019 (10 Jan Shift 1)

Question

A heat source at T=103 K is connected to another heat reservoir at T=102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 W K-1 m-1, the energy flux through it in the steady-state is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C90 W m-2

Step-by-step explanation

Given,

Temperature of heat source, TH=103 K,

Temperature of heat reservoir, TL=102 K,

Conductivity of copper K=0.1 W K-1m-1

From the equation of steady-state heat flow,

dQdt=KA∆TL

Heat energy flux can be defined as the rate of heat energy transfer through a given surface.

1AdQdt=K×1000-1001=90 W m-2

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About this question

This is a previous-year question from JEE Main 2019, covering the Thermal Properties of Matter 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.