Thermal Properties of Matter

JEE Main level, single correct, medium

103 of 126

Question 103 of 126

A heat source at T=103KT={10}^{3}K is connected to another heat reservoir at T=102KT={10}^{2}K by a copper slab which is 1m1m thick. Given that the thermal conductivity of copper is 0.1WK−1m−10.1W{K}^{-1}{m}^{-1}, the energy flux through it in the steady-state is:

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Answer: (C) 90Wm−290W{m}^{-2}

Given,
Temperature of heat source, TH=103K{T}_{H}={10}^{3}K,
Temperature of heat reservoir, TL=102K{T}_{L}={10}^{2}K,
Conductivity of copper K=0.1WK−1m−1K=0.1W{K}^{-1}{m}^{-1}
From the equation of steady-state heat flow,
dQdt=KA∆TL\dfrac{dQ}{dt}=\dfrac{KA∆T}{L}
Heat energy flux can be defined as the rate of heat energy transfer through a given surface.
1A(dQdt)=K×1000−1001=90Wm−2\dfrac{1}{A}\left(\dfrac{dQ}{dt}\right)=K\times \dfrac{1000-100}{1}=90W{m}^{-2}

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