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Heat Conduction Calculator

Find the heat conduction rate through a material.

How it is calculated

The tool applies Fourier's law for steady conduction through a flat layer: Q/t = k × A × ΔT / d. Use SI units: k, the thermal conductivity, in W/(m·K); A, the area facing the heat flow, in m²; ΔT, the temperature difference between the two faces, in K or °C (a difference is the same in both); and d, the thickness in m. The result is heat flow in watts. Thickness must be above zero, and k and area cannot be negative. ΔT is not restricted, so a negative difference gives a negative heat flow, meaning heat moves in the opposite direction. The answer has two decimals. It covers one uniform layer only.

Worked example

An insulation layer has k = 0.04 W/(m·K), an area of 20 m², a temperature difference of 20 °C between inside and outside, and a thickness of 0.1 m. Step 1: k × A = 0.04 × 20 = 0.8. Step 2: 0.8 × ΔT = 0.8 × 20 = 16. Step 3: 16 / 0.1 = 160. The heat flow through the layer is 160.00 W.

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Common mistakes

  • Entering thickness in centimetres or millimetres; it must be in metres, so 10 cm is 0.1.
  • Entering the two surface temperatures instead of their difference; the field wants ΔT only.
  • Using k in the wrong units such as BTU·in/(h·ft²·°F); the tool assumes W/(m·K).
  • Using the result for a wall made of several layers; each layer needs its own calculation, or the resistances must be added.
  • Using the air temperatures rather than the surface temperatures, which ignores the film resistance at the surfaces and overstates the heat flow.

About this calculator

This calculator applies Fourier's law of heat conduction, Q/t = k·A·ΔT/d, to find the rate of heat flow through a material, where k is the material's thermal conductivity (a property specific to each substance — copper conducts far better than wood or fiberglass insulation), A is the cross-sectional area, ΔT is the temperature difference across the material, and d is its thickness. A higher conductivity, larger area, bigger temperature gap, or thinner material all increase the rate of heat transfer.

Building engineers and architects calculating heat loss through walls, windows, or roofing to size heating and cooling systems, materials scientists comparing insulation products, and physics and engineering students working through heat-transfer problem sets use this.

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Comments (1)

  • Editorial team

    The tool handles one uniform layer, which is useful for comparing options: with k = 0.04 W/(m·K), 20 m² and a 20 K difference, a 0.1 m layer passes 160 W and a 0.2 m layer passes 80 W. A real wall is a stack of layers plus surface air films, so entering its total thickness with a single k is only a rough guess. Which material or build-up are you comparing, and over what area?

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