Assume that a homogeneous isotropic material is subjected to a temperature gradient dT/dx. The flow of heat will result in the direction opposite to the temperature gradient through the conducting medium.

The heat flux Q (heat flow per unit time per unit area) will be proportional to the temperature gradient i.e Q∞dT/dx

Or Q= -K dT/dx

Where K is the proportionality constant and is known as coefficient of thermal conductivity. if Q is expressed in W/m^{2} and dT/dx in K/m the units of K will be W/mK

As discussed ,the transformation of heat in solids is due to phonons and free electrons . Thus, the coefficient of thermal conductivity K can be written as

K=K_{phonon }+K_{electron}

In order to derive the expression for K, let us consider the heat flow from high temperature to low temperature in a metal slab having temperature gradient dT/dx.

Let c_{v }be the heat capacity, then the heat transfer per unit area per second will be

Q=mnv/3 c_{v} λdT/dx (1)

Where v is the velocity of electrons.

λ Is mean free path of collisions

Also ,heat flux Q=KdT/dx (2)

By comparing equations(1) and (2) ,we get

KdT/dx=mnv/3 C_{v}dT/dx

Or K=mnv/ C_{v}λ (3)

The energy of free electron is given by

M C_{v}T=3/2 K_{B}T

Or C_{v}=3/2m K_{b} (4)

Where K** _{B} is B**oltzmann Constant

By putting equation (4) in (3) ,we get

Thermal conductivity K=mnv/3(3/2 K_{B}/m)λ

Or K=K_{B}(nvλ/2) (5)

Specific heat at constant volume for an ideal gas is

C_{v}=3/2 n K_{B}

K_{B}=2/3 n C_{v} (6)

By putting equation (6) in (5) ,we get

K=1/3 C_{v}λv (7)

Expression (7) represents that the thermal conductivity of solid depends upon specific heat (C_{V}) ,mean free path of collisions (λ) and velocity of electrons (v)

Now consider the electrical conductivity σ

σ =ne^{2}r/m (8)

And relaxation time (collision time)

r=λ/v_{d} (9)

by putting equation (9) in (8) ,we get

σ =ne^{2}λ/mv_{d} (10)

Also ½ mv^{2}d=3/2 K_{B}T

Or m=3K_{B}T/v^{2}d (11)

By putting equation (11) in (10) ,we get

σ =ne^{2}λv_{d}/3K_{B}T (12)

Therefore, the ratio of thermal conductivity K to electrical conductivity σ is

K/ σ =K_{B}nvλ/2*3K_{B}T/ne^{2}λv_{d}[by dividing equation (5) by (12)]

=3/2 K^{2}B/e^{2}.T, if we assume v=v_{d}

Or K/ σ T=5.838*10^{-9} o cal K-sec

K/ σ T=2.44*10^{-8}oW/K^{2}=L

Which indicates that the ratio K/ σ is same for all metals and is a function of temperature only. This empirical law is known as **Weidemann –Franz Lorenz law .**Thus, we can say that best electrical conductor will be a best thermal conductor.

The L is known as the **Lorenz number. **

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Thanks Akshara for appreciating the blog.