This is the mechanism by which heat is transferred from one part of an object to another part through vibrations without net movement of material. If one part of an object is hotter than its neighboring part, the molecules or atoms in the hotter part have more energy and vibrate more vigorously than their neighbours. When they collide with their neighbouring molecules or atoms which vibrate less vigorously, energy is transferred to the latter and the temperature of the colder part increases.

Free electrons also play an important part in the thermal conduction as they provide an effective mechanism for carrying heat energy from one part of an object to another part. Metals are good electrical conductors because they have a high concentration of free electrons. Hence, they are also good conductors of heat. The ratio of thermal conductivity to electrical conductivity in metals is given by the Wiedemann-Franz law.

$$\frac{k}{\sigma} = LT$$

,where

- k is thermal conductivity
- $\sigma$ is electrical conductivity
- L is Lorentz number ($2.45 \times 10^{-8} \, W \Omega \, K^{-2}$)

### Thermal Conductivity

The rate of heat transfer by thermal conduction is proportional to the cross-sectional area (A) and the temperature gradient ($\frac{dT}{dx}$).

$$\frac{dQ}{dt} = \, – k A \frac{dT}{dx}$$

– The proportionality constant k is the thermal conductivity. The negative sign means that heat flows down the temperature gradient. (In the direction of decreasing temperature.

For an insulated uniform rod of length L, in steady state, the rate of heat flow (i.e. heat current) must be constant along the whole length and given by,

$$\frac{dQ}{dt} = k \frac{A}{L} \, \Delta T$$

where $\Delta T$ is the difference in temperature between the hot end and the cold end. We dropped the minus sign with the understanding that $\frac{dQ}{dt}$ flows from the hot end to the cold end.

We can see that $\frac{kA}{L}$ is the thermal conductance of the rod.

### Comparison with electricity

You can compare the $k \frac{A}{L}$ (thermal conductance of the rod) with the electrical conductance, where electrical conductance = $\frac{1}{R} = \frac{1}{\rho} \frac{A}{L}$. To see this,

$$\begin{aligned} I &= \frac{V}{R} \\ &= \frac{V}{\rho \frac{l}{A}} \\ &= \frac{1}{\rho} \frac{A}{l} V\end{aligned}$$

where V is the potential difference (which is similar to $\Delta T$ in the heat conductance equation.)

### Heat Conductance Equation

Consider an infinitesimally thin slab with different temperatures $T_{1}$ on one side and $T_{2}$ on the other side. The rate of heat transfer by thermal conduction across the infinitesimally thin slab is proportional to the cross-sectional area (A) of that slab and the temperature gradient (Definition of temperature gradient: $\frac{dT}{dx} = \frac{T_{2}-T_{1}}{x_{2}-x_{1}}$ across that slab.

$$\frac{dQ}{dt} = \, – k A \frac{dT}{dx}$$

, where

- $\frac{dQ}{dt}$ is the heat flow across the slab
- negative sign is that heat flows down the temperature gradient
- k is the thermal conductivity of slab

**Meaning of the negative sign:** If the positive x-axis is to the right and the lower temperature is also to the right, then the temperature gradient is negative (in the direction of positive x-axis), which makes the direction of heat flow across the slab also positive.