Thermal Conductivity Formula

Every object has its own capacity to conduct heat.  To determine how much it is we use this term. Thermal conductivity (λ or k) is the capacity of the body to conduct or spread heat.

\(K=\frac{QL}{A\Delta T}\)

Where,
thermal conductivity is k in W/m K,
the amount of heat transfer through the material is Q in J/S or W,
the area of the body is A in m2,
the difference in temperature is ΔT  in K.

Thermal Conductivity Solved Examples

Let’sexplore some numerical on thermal conductivity :

Problem 1: Compute the thermal conductivity through a conductor when 30 kW of heat flows through it having length of 4 m and area of 12 m2 if the temperature gradient is 40 K.
Answer:

Known:
Q (Heat flow)= 30 kW, L (length) = 4 m, A (Area )= 12 m2, Δ T (temperature difference) = 40 K.

The thermal conductivity is articulated as,

\(K=\frac{QL}{A\Delta T}=\frac{30\times 10^{3}W\times 4m}{12m^{2}\times 40K}=250\;W/mK\)

Problem  2: What is the temperature gradient if a conductor at 70 K after a flow of heat of 20 kW reaches a temperature of 100 K?
Answer:

Known:
Q (Heat flow) = 20 kW, L (length) = 4 m, Ti (initial temperature)= 70 K, Tf (Final temperature) = 100 K, ΔT (temperature gradient) = ?

The temperature gradient is articulated as
Δ T = Ti- Tf
= 100 K – 70 K
= 30 K


Practise This Question

If a body describes a circular motion under inverse square field, the time taken to complete one revolution, T is related to the radius of the circular orbit as