Q. 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 1 m thick. Given that the thermal conductivity of copper is $0.1 \; WK^{-1} \; m^{-1}$, the energy flux through it in the steady state is :

Solution:

$\left(\frac{dQ}{dt}\right) = \frac{kA\Delta T}{\ell} $
$ \Rightarrow \frac{1}{A} \left(\frac{dQ}{dt} \right) = \frac{\left(0.1\right)\left(900\right)}{1} = 90 W/m^{2} $

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