Question

A heat source at $T={{10}^{3}}K$ is connected to another heat reservoir at $T={{10}^{2}}K$ by a copper slab which is $1m$ thick. Given that the thermal conductivity of copper is $0.1K{{W}^{-1}}{{m}^{-1}}$, the energy flux through it in the steady state is
A. $90W{{m}^{-2}}$
B. $200W{{m}^{-2}}$
C. $65W{{m}^{-2}}$
D. $120W{{m}^{-2}}$

Answer 1

We know that in the heat transfer by the conduction the heat flux is directly proportional to the temperature gradient, that is
${{q}_{x}}=-K\dfrac{dT}{dx}$
Where $K$ is the thermal conductivity of the material and $\dfrac{dT}{dx}$ is the temperature gradient

Complete step by step answer:
We know that in the heat transfer by the conduction the heat flux is directly proportional to the temperature gradient, that is
${{q}_{x}}=-K\dfrac{dT}{dx}$
Where $K$ is the thermal conductivity of the material and $\dfrac{dT}{dx}$is the temperature gradient.
Here the negative sign indicates the heat flow from the hot surface to the cold surface.
Given data is as follows
Temperature of the heat source is ${{T}_{s}}={{10}^{3}}K$
Temperature of the reservoir is ${{T}_{r}}={{10}^{2}}K$
Thickness of the copper slab is $\Delta x=1m$
Thermal conductivity of the copper is $K=0.1K{{W}^{-1}}{{m}^{-1}}$
Substitute all the in the above formula
${{q}_{x}}=0.1\dfrac{{{10}^{3}}-{{10}^{2}}}{1}$
${{q}_{x}}=90W{{m}^{-2}}$
Therefore the required heat flux or the energy flux is ${{q}_{x}}=90W{{m}^{-2}}$
Hence the option $\left( A \right)$ is the correct answer.

Note:
The law that is involved here is the Fourier’s law of Heat conduction in the mode of heat transfer involved here is the conduction where the heat transfer takes place by the atomic vibrations of the material of the heat transfer. In this mode of heat transfer there is no actual motion of the medium of the heat flow.