Top

Thermal Conductivity Calculator
Top
Thermal conductivity calculator is an online tool, which helps to find out the thermal conductivity of a material. The material property governing the flow of heat through a material at steady state is known as thermal conductivity. It is an intrinsic property of a material. This measurement reveals the ability of a material for the conduction of heat. So it is defined as the amount of the heat energy transmitted through a unit area and thickness due to its temperature difference. It is denoted by λ or k. The mathematical expression for the thermal conductivity is given by,

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

Where $\lambda$ is thermal conductivity
Q is the amount of heat flow through the material
L is the thickness of the material
A is the area of the material
$\Delta$ T is the temperature difference

## Steps for Thermal Conductivity Calculator

Step 1 :

Identify the given parameters from the question.

Step 2 :

Sustitute the values in the below equation and find out the thermal conductivity.

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

## Problems on Thermal Conductivity Calculator

1. ### Calculate the thermal conductivity of iron, when 10kW heat is flowing through the thickness of 5 cm and area 2m2. The temperature gradient is given as 100K.

Step 1 :

The given quantities are,

Q=10kW=10000W

L=5cm=0.05m

A=2m2

$\Delta$ T=100K

Step 2 :

We know the equation of thermal conductivity,

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

$\lambda$ =$\frac{10000\times 0.05}{2\times 100}$=2.5W/mK

The thermal conductivity of iron is,

$\lambda$ =$\frac{10000\times 0.05}{2\times 100}$=2.5W/mK

2. ### Determine the thermal conductivity of a material whose thickness and area be 10cm and 4m2. The quantity of heat which is flowing through the material is given by 9 kW and the temperature gradient is 150K.

Step 1 :

Q=9kW=9000W

L=10cm=0.1m

A=4m2

$\Delta$ T=150K

Step 2 :

We know the equation of thermal conductivity,

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

$\lambda$ =$\frac{9000\times 0.1}{4\times 150}$=1.5W/mK

$\lambda$ =$\frac{9000\times 0.1}{4\times 150}$=1.5W/mK