Thermal Conductivity Calculator is a free online tool that displays the thermal conductivity of the given material. BYJU’S online thermal conductivity calculator tool performs the calculation faster, and it displays the thermal conductivity in a fraction of seconds.
How to Use the Thermal Conductivity Calculator?
The procedure to use the thermal conductivity calculator is as follows:
Step 1: Enter the area, heat transfer rate, the difference in temperature, distance, and x for the unknown in the input field
Step 2: Now click the button “Calculate x” to get the thermal conductivity
Step 3: Finally, the thermal conductivity of the material will be displayed in the output field
What is Meant by Thermal Conductivity?
The thermal conductivity is defined as the intrinsic ability of the material to transfer or conduct the heat. It is denoted by the symbol “λ” or “k”. There are three different methods of heat transfer. Thermal conductivity is one among them. The other two methods are radiation and convection. Generally, the process of heat transfer can be quantified in terms of the rate equations. The rate equation of thermal conductivity is based on Fourier’s law of heat conduction. The material with good thermal conductivity is used in heat sinks. The reciprocal of the thermal conductivity is the thermal resistivity. Each and every material has its own capacity to conduct or transfer heat. The formula to calculate the thermal conductivity of a material is given as:
Thermal Conductivity, λ = (QL)/(AΔT)
Here,
λ = Thermal conductivity
A = Area of the surface
Q = Amount of heat transferred
L = Distance between two isothermal planes
ΔT = Change in temperature
Example:
Determine the thermal conductivity of the metal, if one end of 0.25m long metal bar is placed in steam and the other end is placed in ice. Given that 15×10-3 kg of ice melts per minute, latent heat of ice is 80 cal/kg, and the cross-section of the metal bar is 7×10-4 m2.
Solution:
Given:
Length, L = 0.25 m
Cross-sectional area, A = 7×10-4 m2 (or) 0.0007 m2
Amount of heat transferred, Q = 15×10-3 x 80 x 1000 = 1200 cal
ΔT = 100 x 60
Now, substitute the values in the thermal conductivity formula, we get
λ = (QL)/(AΔT)
λ = (1200 x 0.25)/(0.0007 x 100x 60)
λ = 71.4 cal m-1s-1 °C-1 .
