Solution:

Coefficient of thermal conductivity:

• Let us consider a metallic rod of cross sectional area A.

• Let the two ends be separated by a distance d, maintained at temperatures $θ1$ and $θ2$ Let $θ1$ be greater than $θ2$.

• Heat flows from the end at the higher temperature to the end at the lower temperature.

When the steady state is reached, the quantity of heat Q conducted is,

• i) directly proportional to the area of cross section A.

• ii) directly proportional to the difference in temperature between the ends $(θ1 - θ2)$.

• iii) directly proportional to the time for which the heat is conducted (t).

• iv) and inversely proportional to the distance between the two ends.

Hence, $$ \mathrm{Q} \alpha \mathrm{A} \frac{\left(\theta_{1}-\theta_{2}\right) \mathrm{t}}{\mathrm{d}} \\ $$

$$or$$

$$ Q=\lambda A \frac{\left(\theta_{1}-\theta_{2}\right) t}{d} \\ $$

• where l is a constant known as the coefficient of thermal conductivity of the material and $(θ1-θ2) / d$ is known as temperature gradient.

when, $A=1$, $\frac{\left(\theta_{1}-\theta_{2}\right)}{d}=1 \quad$, and, $t=1$ then, $Q=\lambda$

• Hence, Coefficient of thermal conductivity (l) of the material of a conductor is defined as the quantity of heat conducted per second per unit area per unit temperature gradient at the steady state.

unit of $\lambda$ : $$ \begin{aligned} Q &=\lambda A \frac{\left(\theta_{1}-\theta_{2}\right) t}{d} \quad \text { or } \\\\ \lambda &=\frac{Q d}{A\left(\theta_{1}-\theta_{2}\right) t} \\\\ \end{aligned} $$

Substituting the units, Unit of, $ \lambda=\frac{\mathrm{Jm}}{\mathrm{m}^{2} \mathrm{Ks}}=\mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}=\mathrm{Wm}^{-1} \mathrm{~K}^{-1} \\ $

$ \left(\right.$ as $\mathrm{Js}^{-1} \\ $ is W)

Good and poor thermal conductors:-

• Materials having higher values of coefficient of thermal conductivity are termed as good thermal conductors and the materials having lower values of coefficient of thermal conductivity are called as poor thermal conductors or insulators.

• The presence of free electrons in metals helps the easy transfer of heat from one part of the metal to the other part.

• So, all metals are good thermal conductors of heat. Some important materials and their values of coefficient of thermal conductivity are listed above.