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Heat Conduction Calculator (Fourier’s law)

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Understanding the heat conduction calculator — concept and principle

This calculator applies Fourier's law to find how much heat flows through a flat layer of material, such as an insulation board, a wall or a window pane, when one side is warmer than the other. It also returns the layer's thermal resistance and its U-value, the numbers used to compare insulation and estimate heat loss.

Heat conducting through an insulation layerA 100 mm layer with k = 0.04 W/m·K, 10 m² and a 20 K difference passes 0.04 × 10 × 20 ÷ 0.1 = 80 W.Heat leaks through the layerwarmcoldk = 0.04ΔT = 20 KA = 10 m²d = 100 mm = 0.1 mQ = k × A × ΔT ÷ d0.04 × 10 × 20 ÷ 0.1Q = 80 W flows throughR = d ÷ k = 2.5 m²K/W · U = k ÷ d = 0.42× thicker → 40 W2× area → 160 W
Heat flow grows with k, area and temperature difference and shrinks with thickness: 80 W here.

How it is calculated

For steady one-dimensional conduction through a flat slab: Q = k · A · ΔT / d.

The layer's thermal resistance is R = d/k (m²K/W), and its conductance per unit area is U = k/d = 1/R (W/m²K), so Q = U · A · ΔT.

The principle

Fourier's law states that heat flux is proportional to the temperature gradient: q = −k dT/dx. In a flat, uniform slab with fixed face temperatures and no internal heat sources, the temperature falls linearly through the thickness, so the gradient is simply ΔT/d. Doubling the thickness halves the gradient and therefore halves the heat flow; doubling k or the area doubles it. As with Ohm's law, layers in series simply add their resistances.

Worked example

  1. Inputs: k = 0.04 W/m·K (typical of mineral wool or polystyrene), A = 10 m², ΔT = 20 K, t = 100 mm.
  2. d = 100/1000 = 0.1 m.
  3. Q = 0.04 × 10 × 20 / 0.1 = 80 W.
  4. R = 0.1 / 0.04 = 2.5 m²K/W.
  5. U = 0.04 / 0.1 = 0.4 W/m²K (check: 0.4 × 10 × 20 = 80 W).
How Heat flow (W) changes with Thickness (mm)How Heat flow changes with Thickness0100200300400255075100125150200Example: 80Thickness (mm)
Each point is the calculator re-run with only “Thickness (mm)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Thickness (mm)Heat flow (W)
25320
50160
75106.6666667
10080
12564
15053.33333333
20040

How to read the result

Q is the continuous heat flow in watts, so 80 W running for 24 hours is 1.92 kWh. A larger R or a smaller U means better insulation. Typical conductivities: still air about 0.026, insulation foams and wools roughly 0.02–0.04, timber around 0.1–0.2, brick about 0.6–1, concrete about 1–2, steel around 50 and copper about 400 W/m·K.

Common mistakes

Keep in mind: The formula assumes steady state, one-dimensional flow and uniform material. Thermal bridges, edges, moisture and air leaks change real heat loss; building designs must follow local energy codes and be checked by a qualified professional.

Frequently asked questions

What is the difference between thermal conductivity and U-value?

Conductivity k is a material property independent of thickness. The U-value describes a specific layer or assembly of a given thickness, including all its layers.

How do I combine several layers?

Add the R values (d/k) of each layer plus the surface resistances, then take U = 1/R_total.

Sources: Incropera, DeWitt, Bergman and Lavine, Fundamentals of Heat and Mass Transfer, Wiley · ISO 6946 — Building components and building elements: Thermal resistance and thermal transmittance

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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