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Thermal Conductivity Calculator
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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

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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

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  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





    Answer  :  

    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



    Answer  :  

    The thermal conductivity of iron is,

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



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