Thermal conductivity of a solid is a measure of the ability of the solid to conduct heat through it.Greater is the thermal conductivity of a solid, the greater is its ability to conduct heat through it. If a solid has lower heat conductivity its ability to conduct heat through it would be lower.

The rate of heat conduction through a medium depends on following factors :-

- geometry of the medium,
- its thickness, and
- the material of the medium.
- It also depends on the temperature difference across the medium.

In this section we would have a look at how we can derive **Thermal Conductivity equation** or **Thermal conduction equation**.

Let us consider a piece of material made in the form of a bar of thickness$\Delta x$ and area of cross-section $A$ as shown below in the figure.

Now the heat that flows perpendicular to the faces for a time $t$ is measured. The experiment for measuring heat is repeated with other bars of same material but with different values of $\Delta x$ and $A$. the results of such experiments show that, for a given value of$\Delta T$

- The quantity of heat conducted is directly proportional to the area of each surface i.e., $Q\propto A$
- Quantity of heat conducted is directly proportional to the temperature gradient in direction of heat flow. That is

$Q\propto -\frac{\Delta T}{\Delta x}$

Provided both $\Delta T$ and $\Delta x$ is small. - Quantity of heat conducted is directly proportional to the time t i.e., $Q\propto t$

Here factor $K$ is the proportionality constant and is known coefficient thermal conductivity of the material. Its value depends on the nature of material used. Thermal conductivity is the property of the material and it is the ability of the substance to transfer heat.

If in above equation (1) we put

$A=1m^2$ , $\Delta x=1m$, $\Delta T=-1K$ and $t=1 \,sec$ then,

$Q=K$

The value of thermal conductivity determines the quantity of heat passing per unit area at a temperature drop of 1k per unit length in unit time.

Due to this limiting case rate of heat flow that is given by

$H= \frac{dQ}{dt}=-KA\frac{dT}{dx}$

Where $dQ$ is the amount heat crossing the thin layer of thickness $dx$ in time $dT$. $dT$ is the temperature difference across the layer of thickness $dx$.

$\text{K=}\frac{Q\Delta x}{A(\Delta T)t}$

SI unit of $K$ is $=\frac{J.m}{{{m}^{2}}K\cdot s}=J{{s}^{-1}}{{m}^{-1}}=w{{m}^{-1}}{{k}^{-1}}$

CGS unit of $K$ is $Cal\cdot {{s}^{-1}}c{{m}^{-1}}^{0}{{C}^{-1}}$

Dimensions of thermal conductivity $K$ are

$=\frac{[M{{L}^{2}}{{T}^{-2}}]\cdot [L]}{[{{L}^{2}}]\cdot [K]\cdot [T]}=[ML{{T}^{-3}}{{K}^{-1}}]$

$$Q=\frac{kA(T_1 -T_2)t}{x}$$ According to the conditions of the problem, $Q = mL$.

$$mL=\frac{kA(T_1 -T_2)t}{x}$$ Here it is given that

$m= 4.8 \, kg$ ; $L = 80 \, kcal/kg$ ; $A = 0.36 \, m^2$; $T_1 - T_2= 100 - 0 = 100^{\circ}C$

Now $t=1 \, hour = 3600 \, s \, , \, x=0.1 \, m \, k=?$

Therefore, $$4.8 \times 80 = \frac{k \times 0.36 \times 100 \times 3600}{0.1}$$ Or, thermal consuctivity $k=3 \times 10^{-4}\, kcal s^{-1}m{-1}^{\circ}C^{-1} $

$$\frac{K_1 A(100-T)}{x}=\frac{K_2 A (T-0)}{x}$$ $$\frac{300 A(100-T)}{x}=\frac{200 A (T-0)}{x} $$ or,

$$300-3T=2T$$ or, $$5T=300$$ $$T=60^{\circ}C$$

(a) $\frac{dQ}{dt}=\frac{K(T_1 - T_2)}{LA}$

(b) $\frac{dQ}{dt}=KLA(T_1 - T_2)$

(c) $\frac{dQ}{dt}=\frac{KA(T_1 - T_2)}{L}$

(d) $\frac{dQ}{dt}=\frac{KL(T_1 - T_2)}{A}$

- Heat Transfer
- Heat Conduction
- Thermal Conductivity
- Thermal resistance and conductance
- Convection heat transfer
- Black Body Radiation
- Stefan Boltzmann law
- Nature of thermal Radiation
- Kirchoff's law
- Newton's Law of Cooling
- Solved examples

Class 11 Maths Class 11 Physics Class 11 Chemistry

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