Question

One end A of a metallic rod length $10\text{cm}$ is inserted in a furnace whose temperature is ${827}^{\circ }C$. The curved surface of the rod is insulated. The room temperature is ${27}^{\circ }C$. When the steady state is attained, the temperature of the other end B of the rod is ${702}^{\circ }C$. Find the thermal conductivity of the metal $\left(\text{in SI units}\right)$

Stefan's constant $=5.67\times {10}^{-8}{\text{W m}}^{-2}{\text{K}}^{-4}$.

Answer 1

Heat lost only from the end B of the rod is by radiation.
If ${\theta }_B$ is the absolute temperature of end B, then the energy radiated per unit area per time from end B from Stefan's law is
$Q_1=\sigma ({\theta }_B^4-{\theta }_0^4)....\left(1\right)$
where ${\theta }_0$ is the room temperature constant.
Also, energy received at end B by conduction through the rod per unit area unit time is
$Q_2=\frac{K({\theta }_A-{\theta }_B)}{l}....\left(2\right)$
Where ${\theta }_A$ = temperature of end A of the rod,
$l$ is the lenght of the rod and
$k$ its thermal conductivity.
In the steady state $Q_1=Q_2$.
Equating equation (1) and (2) we have
$\frac{K({\theta }_A-{\theta }_B)}{l}=\sigma ({\theta }_B^4-{\theta }_0^4)$ or $K=\frac{\sigma l({\theta }_B^4-{\theta }_0^4)}{({\theta }_A-{\theta }_B)}....\left(3\right)$
Given
$\sigma =5.67\times {10}^{-10}{\text{Wm}}^{-2}{K}^{-4},l=0.1m,{\theta }_A=1100\text{K}$
${\theta }_B=925\text{K}{\theta }_0=300\text{K}$
$\therefore K=36.6{\text{Js}}^{-1}{\text{m}}^{-1}{\text{K}}^{-4}$.

Why this question ? Key concept: Solids can conduct and radiate heat at the same time. Importance in JEE: Combination of types of heat transfer mechanisms are noteworthy for JEE.