Question

define coefficient of thermal conductivity. Derive its expression​

Answer 1

Coefficient of Thermal Conductivity:

If L be the length of the rod, A the area of cross section and $\sf{\theta_1\:and\:\theta_2}$ be the temperatures of its two faces, then the amount of heat flowing from one face to the another face in time t is

$\sf{Q\propto\:A}\\\sf{Q\propto\left(\theta_1-\theta_2\right)}\\\sf{Q\propto\:t}\\\sf{Q\propto\dfrac{1}{l}}$

In combined form,

$\sf{Q\propto\dfrac{A\left(\theta_1-\theta_2\right)t}{l}}$

$\sf{Q=\dfrac{KA\left(\theta_1-\theta_2\right)t}{l}}$

where K is coefficient of Thermal Conductivity of material of rod. It is the measure of the ability of a substance to conduct heat through it.

This relation can also be expressed as

$\sf{\dfrac{\Delta\:Q}{\Delta\:t}=\dfrac{KA\left(\theta_1-\theta_2\right)}{l}}$

If $\sf{A=1\:m^2,\:\left(\theta_1-\theta_2\right)=1^{\circ}C,\:t=1\:s\:and\:l=1\:m,}\\\sf{then,\:Q=K}$

Thus, Thermal Conductivity of a material is the amount of heat flowing per second during steady state through its rod of length 1 m and cross section 1 m² with a unit temperature difference between the opposite faces.

•Units: cal/cm-s°C (in CGS), kcal/m-sK (in MKS) and W/m-K (in SI)

•Dimensions: $\sf{\left[MLT^{-3}\theta^{-1}\right]}$

•The magnitude of K depends only on nature of material.

•For perfect conductors, $\sf{K=\infty}$ and for perfect insulators, K = 0

•The Thermal Conductivity of pure metals decreases with rise in temperature but for alloys thermal conductivity increases with increase of temperature.

Answer 2

If L be the length of the rod, A the area of cross section and be the temperatures of its two faces, then the amount of heat flowing from one face to the another face in time t