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 $0.1W{K}^{-1}{m}^{-1}$, the energy flux through it in a steady state is :

• A. $90W/m^2$
• B. $120W/m^2$
• C. $65W/m^2$
• D. $200W/m^2$

Answer 1

The correct option is A

$90W/m^2$

Explanation for the correct option:

Step 1: Given data:

Heat source, $T_1={10}^{3}K$

Heat reservoir, $T_2={10}^{2}K$

Thus, $∆T=T_1-T_2=1000-100=900K$

Thickness of the copper slab which is $l=1m$

Thermal conductivity of copper is $k=0.1W{K}^{-1}{m}^{-1}$

Step 2: Finding Energy flux in a steady-state:

Energy Flux $\left(\frac{1}{A}\right)\left(\frac{dQ}{dt}\right)=\frac{k\Delta T}{l}$

$$\begin{gathered} =\frac{0.1x900}{1} \\ =90W/m^2 \end{gathered}$$

Hence, option A is correct.