**Thermal Conductivity Formula**

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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.

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 m

^{2}

**,**

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 m

^{2}if the temperature gradient is 40 K.

**Answer:**

Known:

Q (Heat flow)= 30 kW, L (length) = 4 m, A (Area )= 12 m

^{2}, Δ T (temperature difference) = 40 K.

The thermal conductivity is articulated as,

**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, T

_{i}(initial temperature)= 70 K, T

_{f}(Final temperature) = 100 K, ΔT (temperature gradient) = ?

The temperature gradient is articulated as

Δ T = Ti- Tf

_{}= 100 K – 70 K

= 30 K