Question

How much heat will be conducted through a slab of area 90 x 10-4 m2 and
thickness 1.2 x 10-3 m in one second when its opposite faces are maintained at
difference in temperature of 20 K. The coefficient of thermal conductivity of that
material is 0.04 Wm-1K-1.​

Answer 1

Given:-

$\text{Area, }A=90\times10^{-4}m^{2}\\\\\text{thickness, }t=1.2\times10^{-3}m\\\\\text{difference in temperatures, }\Delta T=20K\\\\\text{thermal conductivity, }k=0.04W/m-K$

To Find:-

Heat conducted(Q)

Explanation:-

According to Fourier's law, heat conducted across the rectangular slab is given by:-

$Q=\dfrac{\Delta T}{\frac{t}{kA}}=\dfrac{kA\Delta T}{t}$

Put all the values in above equation:-

$\Rightarrow Q=\dfrac{0.04\times90\times10^{-4}\times20}{1.2\times10^{-3}}=6W$

Therefore, heat conducted across the slab is 6W