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Heat conduction: how much heat flows through a layer?

Calculate the heat flowing through a layer of copper, steel, concrete, wood or insulation with Fourier's law, along with its thermal resistance, R-value and daily energy loss.

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W/(m·K)

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°F

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Results

Heat flow through the layer

101.6

Conduction through this layer only, without surface films or thermal bridges

Thermal resistance R (SI)

2.54 m²·K/W

Heat flux (per unit area)
10.94 W/m²
US R-value (ft²·°F·h/BTU)
14.4
Energy lost per day at this temperature difference
2.44 kWh
Thermal conductivity used
0.035 W/(m·K)

How heat moves through solids

Heat flows from warm to cold in three ways: conduction through solids, convection in moving fluids and radiation. In walls, windows, pans and heat sinks, conduction is often the main route. Joseph Fourier described it in 1822 with a law that is still the basis of insulation design.

Fourier’s law for a flat layer

Q = k × A × ΔT ÷ d

  • Q is the heat flow in watts.
  • k is the thermal conductivity of the material, in W/(m·K).
  • A is the area, d the thickness and ΔT the temperature difference between the two faces.

The thermal resistance of the layer is R = d ÷ k, in m²·K/W. Multiply it by 5.678 to get the US R-value printed on insulation.

Thermal conductivity of common materials

Material k, W/(m·K)
Copper about 400
Aluminum about 237
Carbon steel about 50
Stainless steel about 16
Concrete 1.4 to 2
Brick 0.6 to 1
Glass about 1
Softwood about 0.13
Mineral wool, EPS about 0.035
Polyurethane foam about 0.023

Example

A 100 ft² (9.29 m²) wall section with 3½ inches (8.9 cm) of mineral wool, 70 °F inside and 20 °F outside (a difference of 27.8 K):

  • R = 0.0889 ÷ 0.035 = 2.54 m²·K/W, or about R-14.4
  • Q = 0.035 × 9.29 × 27.8 ÷ 0.0889 ≈ 102 W (about 347 BTU/h)
  • Over a day: 102 × 24 ≈ 2.4 kWh

Replace the insulation with the same thickness of concrete and the heat flow would be about 50 times larger.

Several layers

For a wall made of several layers, add the resistances: R_total = d₁/k₁ + d₂/k₂ + … plus the small resistances of the air films on each side (about 0.13 m²·K/W inside and 0.04 outside). The heat flow is then ΔT × A ÷ R_total. Use this calculator layer by layer, then sum the R values.

Limits

The formula assumes steady conditions, a uniform flat layer and heat flowing straight through. It ignores thermal bridges such as studs, fasteners and window frames, which can raise the real heat loss of a framed wall by 10 to 20% or more. Wet insulation conducts much more heat than dry insulation, and conductivity changes with temperature.

Frequently asked questions

What is Fourier's law?

The rate of heat flow through a material is proportional to its thermal conductivity, the area and the temperature difference, and inversely proportional to its thickness: Q = k × A × ΔT ÷ d. Joseph Fourier published it in 1822 in his Analytical Theory of Heat.

What is thermal conductivity?

How easily heat passes through a material, in watts per meter per kelvin. Copper conducts about 400 W/(m·K), steel about 50, concrete 1 to 2, wood about 0.13 and mineral wool about 0.035. Insulation works by trapping still air or gas, which conducts heat very poorly.

How do I convert R-values?

The SI R-value (m²·K/W) times 5.678 gives the US R-value (ft²·°F·h/BTU). A layer with RSI 2.54 is about R-14.4. The US Department of Energy recommends about R-49 to R-60 for attics in cold climates.

Why is the real heat loss of a wall higher?

Real walls have several layers, air films on each surface and thermal bridges such as studs, which conduct more heat than the insulation between them. Wood studs can lower the effective R-value of a framed wall by 10 to 20%. For a whole house, use the heat loss calculator.

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Results are estimates provided for information purposes. They do not replace professional advice.

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