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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. Enter the thickness, conductivity, and area to find the thermal resistance.

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