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# Thermal Conductivity Calculator

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This calculator helps to determine the thermal conductivity of a material. It is an intrinsic property of a material. This property shows the ability of a material to conduct heat. The formula used to find out the thermal conductivity is,
$\lambda$ = $\frac{QL}{A\Delta T}$
Where,
$\lambda$ = thermal conductivity,
Q = quantity of heat,
L = material thickness,
A = area of the material,
$\Delta$ T = temperature difference.

## Steps

Step 1 : Write down the parameters given in the question.

Step 2 : Plug these values in the formula and find out the required quantity

$\lambda$ = $\frac{QL}{A\Delta T}$.

## Problems

Solved problems related to Thermal conductivity are given below.

### Solved Examples

Question 1: Find out the thermal conductivity of a given material of area 5m2 and the thickness is 10mm. 1500W heat is passed through this material within a temperature gradient of 200K.

Solution:

Given that,
A = 5m2,
L = 10mm = 10$\times10^{-3}$m,
Q = 1500W,
$\Delta$T = 200K.

Thermal conductivity of a given material is,

$\lambda$ = $\frac{QL}{A\Delta T}$

$\lambda$ = $\frac{1500\times10\times10^{-3}}{5\times200}$ = 0.015W/mK

Question 2: Find out the thermal conductivity of a given material of area 2m2 and the thickness is 15cm. 2.5kW heat is passed through this material within a temperature gradient of 150K.

Solution:

Given that,
A = 2m2,
L = 15cm = 15$\times10^{-2}$m,
Q = 2.5kW = 2.5$\times10^{3}$W,
$\Delta$T = 150K.

Thermal conductivity of a given material is,

$\lambda$ = $\frac{QL}{A\Delta T}$

$\lambda$ = $\frac{2.5\times10^{3}\times15\times10^{-2}}{2\times150}$ = 1.25W/mK