**Thermal Conductivity**

Thermal Conductivity is the intrinsic property of a material which relates its ability to conduct heat. Heat transfer by conduction involves transfer of energy within a material without any motion of the material as a whole.

Conduction takes place when a temperature gradient exists in a solid medium. Conductive heat flow occurs in the direction of decreasing temperature because higher temperature equates to higher molecular energy or more molecular movement. Energy is transferred from the more energetic to the less energetic molecules when neighbouring molecules collide.

**What is Thermal Conductivity?**

Thermal Conductivity is defined as the quantity of heat (Q) transmitted through a unit thickness (L) in a direction normal to a surface of unit area due to a unit temperature gradient (ΔT) under steady state condition and when the heat transfer is dependent only on the temperature gradient.

Everybody has its own capacity to conduct heat. To determine how much, it is we use this term. Thermal Conductivity (λ) is the capacity of the body to conduct or transmit heat.

\(Thermal\,Conductivity(\lambda )=\left( \frac{\left( Heat(Q)\times Dis\tan ce(L) \right)}{\left( Area(A)\times Temperature\,Gradient(\Delta T) \right)} \right)\).

Where,

λ = Thermal Conductivity in W/ mK,

Q = Amount of heat transfer through the material in J/ sec or W,

A = Area of the body in m²,

ΔT = Temperature Difference in K.

**How to find the Thermal Conductivity?**

**Problem: **Calculate the Thermal Conductivity through a conductor when 30 KW of heat flows through it having length of 4m and area 12m², if the temperature gradient is 40K?

**Solution: **Given,

Heat flow (Q) = 30 KW

Length (L) = 4m

Area (A) = 12m²

Temperature difference (Δ) = 40 K

We know that:

\(Thermal\,Conductivity(\lambda)=\frac{QL}{A\times \Delta T}=\frac{30\times {{10}^{3}}\times 4}{12\times 40}=250\,W/mK\).

∴ Thermal Conductivity (λ) = 250 W/ mK.