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} $

Solition Image

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions

6. The spectrum of an oil flame is an example for ...........

KCET 2010 Dual Nature Of Radiation And Matter