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Thermal Resistance Calculator

Find the thermal resistance of a material layer.

How it is calculated

For heat conducting through a flat layer the tool uses R = L / (k × A). L is the thickness in metres, k is the thermal conductivity in W/(m·K) and A is the area facing the heat flow in m². The result is in K/W, shown with 4 decimals. All three must be above zero. A thicker layer or a lower conductivity gives a higher resistance. Layers in series add their resistances, and the heat flow through a wall is then Q = ΔT / R in watts. Surface films and air gaps are not included.

Worked example

A 0.2 m insulation layer with k = 0.04 W/(m·K) over a 10 m² wall. Multiply k by area: 0.04 × 10 = 0.4. Divide the thickness: 0.2 / 0.4 = 0.5000 K/W. With 20 K between inside and outside, heat flow is 20 / 0.5 = 40 W.

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Common mistakes

  • Entering thickness in millimetres or centimetres. It must be metres, so 20 cm is 0.2.
  • Confusing R in K/W with the R-value used for building products, which is per unit area. Here the area is included.
  • Using the conductivity from a different unit system, such as BTU·in/(h·ft²·°F).
  • Forgetting that a stack of materials adds up. Compute each layer and add them, not average them.

About this calculator

Thermal resistance quantifies how strongly a material layer resists conductive heat flow: R = L / (k·A), where L is thickness, k is the material's thermal conductivity, and A is the cross-sectional area. It behaves exactly like electrical resistance in Ohm's law — heat flow Q equals the temperature difference divided by R (Q = ΔT/R), and layers in series simply add their resistances.

HVAC engineers and building scientists use thermal resistance to evaluate walls, roofs, and windows for energy efficiency (it's the physical basis behind R-values on insulation), while electronics engineers use the identical formula to size heat sinks and predict junction temperatures in semiconductor packages.

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Comments (1)

  • Editorial team

    Insulation labels usually quote an R-value per unit area, not K/W. To compare, multiply this result by the area: 0.5 K/W over 10 m² is 5 m²K/W. The K/W figure itself changes with the size of the surface, so it suits a specific wall, pipe jacket or heatsink plate, while the area-based one suits comparing products. Tell us whether your material data gives conductivity or an R-value, and we may consider supporting the other.

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