Heat Conduction (Fourier) Calculator
Heat conduction is the steady flow of thermal energy through a solid layer, and Fourier's law puts a number on it. The rate of heat flow through a flat slab depends on four things: how well the material conducts, set by its thermal conductivity, how large the area is, how big the temperature difference is across the layer, and how thick the layer is. Heat always moves from the hot side to the cold side, and the rate grows with a larger area or a wider temperature gap while falling as the slab gets thicker. This is the calculation behind sizing insulation, designing heat sinks and estimating heat loss through a wall or a window. This calculator takes the thermal conductivity in watts per meter-kelvin, the area in square meters, the temperature difference in kelvin and the thickness in meters, then returns the heat conduction rate in watts. Because the law uses a temperature difference rather than an absolute temperature, you can enter the gap in kelvin or in degrees Celsius and get the same answer. Every figure here is computed deterministically from the standard physics formula shown below, with a worked example that reconciles exactly to the calculator so you can follow each step.
Fourier's law gives the conduction rate as thermal conductivity times area times the temperature difference, divided by thickness: Q = k A dT / d. With k = 0.5 W/(m K), an area of 2 m^2, a difference of 20 K and a thickness of 0.1 m, the heat conduction rate is 200.00 W.
Fourier's law formula
Q = k A dT / d
Q = heat conduction rate in watts
k = thermal conductivity in W/(m K)
A = cross-sectional area in square meters
dT = temperature difference in kelvin
d = layer thickness in meters
Multiply the thermal conductivity by the area and by the temperature difference, then divide by the thickness. The temperature difference can be entered in kelvin or degrees Celsius because the two scales share the same step size.
Worked example
A layer has a thermal conductivity of 0.5 W/(m K), an area of 2 square meters, a temperature difference of 20 kelvin across it, and a thickness of 0.1 meters.
- k x A = 0.5 x 2 = 1
- k x A x dT = 1 x 20 = 20
- Q = 20 / 0.1 = 200.00 W
The heat conduction rate is 200.00 watts. These are the calculator's default inputs, so the result above matches the widget exactly.
Sample conductivity values
| Material type | Conductivity k (W/(m K)) | Heat flow tendency |
|---|---|---|
| Insulating foam | 0.03 to 0.05 | Low |
| Brick or glass | 0.7 to 1.0 | Moderate |
| Metals | 50 to 400 | High |
Ranges are typical published values for broad material classes. Use a specific sourced value for engineering work.
Heat conduction calculator: frequently asked questions
What is Fourier's law of heat conduction?
Fourier's law states that the rate of heat conduction through a flat layer equals the thermal conductivity multiplied by the cross-sectional area and the temperature difference, divided by the thickness. Heat flows from the hot side to the cold side, and the rate rises with a bigger area or temperature gap and falls as the layer gets thicker. It is the steady-state basis for sizing insulation and heat sinks.
What units does this calculator expect?
Thermal conductivity k is in watts per meter-kelvin, W/(m K). Area A is in square meters. The temperature difference dT is in kelvin, which is numerically the same as a difference in degrees Celsius. Thickness d is in meters. The output, the heat conduction rate Q, is in watts, the rate of heat energy flow.
Why is the result the same in kelvin and Celsius?
Fourier's law uses a temperature difference, not an absolute temperature. A difference of 20 kelvin is exactly the same size as a difference of 20 degrees Celsius, because the two scales share the same step size. So you can enter dT as a Celsius difference and the watts come out identical. Only absolute-temperature laws require kelvin.
What do the default inputs give?
The defaults are a thermal conductivity of 0.5 W/(m K), an area of 2 square meters, a temperature difference of 20 kelvin and a thickness of 0.1 meters. Fourier's law gives 0.5 times 2 times 20 divided by 0.1, which is 200.00 watts. That is the steady-state rate of heat flow through the layer described.
Does a thicker layer always reduce heat flow?
Yes, all else equal. Thickness sits in the denominator, so doubling the thickness halves the conduction rate. That is why thicker insulation slows heat loss. Likewise a lower thermal conductivity, a smaller area or a smaller temperature difference each reduce the watts conducted through the layer.
Official sources
- Thermal conductivity and heat transfer references: US National Institute of Standards and Technology (NIST). As at 25 June 2026.
Reviewed by the CalculatorHub team, edited by James Graham, 25 June 2026. See our methodology. This is general information, not financial, tax, legal or investment advice.