Q 11-10-108JEE MainJEE Main 2019 (10 Jan, Shift 1)Easy
A heat source at $T = 10^3\ \text{K}$ is connected to another heat reservoir at $T = 10^2\ \text{K}$ by a copper slab which is $1\ \text{m}$ thick. Given that the thermal conductivity of copper is $0.1\ \text{W K}^{-1}\text{m}^{-1}$, the energy flux through it in the steady state is:
Answer: (C) $90\ \text{W m}^{-2}$
Heat flux $= \dfrac{1}{A}\dfrac{dQ}{dt} = k\dfrac{\Delta T}{L}$:
$$\text{flux} = 0.1\times\frac{1000-100}{1} = 90\ \text{W m}^{-2}$$
Solution by Sreeraj P, M.Sc Physics