Thermal Resistance Calculator
Enter a layer's thickness, the material's thermal conductivity, and the area to get the conductive thermal resistance in kelvin per watt — and see why thicker, low-conductivity layers insulate best.
Resistance of the whole layer
Enter thickness, conductivity, and area and the calculator returns the conductive thermal resistance R = L ÷ (k × A) in K/W for the whole component.
Use SI units
Thickness in metres, conductivity in W/(m·K), and area in square metres give the resistance in K/W — convert millimetres to metres before you start.
What is thermal resistance?
How strongly a layer opposes heat flow
Thermal resistance is a measure of how strongly a material or layer opposes the flow of heat passing through it. The thermal resistance calculator turns three measurements — the thickness of the layer in metres, the material's thermal conductivity in watts per metre-kelvin, and the cross-sectional area in square metres — into a single figure in kelvin per watt. The higher that figure, the better the layer insulates: it is the number behind how warm a wall keeps a room, how well a window holds back the cold, and how slowly a cooler lets heat leak in.
Enter a thickness in metres, a thermal conductivity in W/(m·K), and an area in square metres to get the thermal resistance in K/W instantly.
The conductive thermal resistance of a flat layer is its thickness divided by the product of the material's thermal conductivity and the area through which heat flows.
R = L ÷ (k × A)Thickness sits on top of the fraction, so a thicker layer resists heat more, while conductivity and area sit underneath, so a more conductive material or a larger surface lowers the resistance. Keep thickness in metres, conductivity in W/(m·K), and area in square metres and the resistance comes back in kelvin per watt.
Suppose you have an insulation panel 0.1 m (100 mm) thick, made from a material with a thermal conductivity of 0.04 W/(m·K), covering an area of 10 m².
Multiply conductivity by area
0.04 × 10 = 0.4 — the conductance of the layer per unit thickness.
Divide the thickness by that product
0.1 ÷ 0.4 = 0.25 — thickness over conductance.
Read the result
R = 0.25 K/W — it takes a 0.25 K temperature difference to drive one watt of heat through the panel.
The formula is exact for steady-state conduction, but a few practical points are worth keeping in mind.
Conduction only, steady state, consistent units
This calculator gives the conductive thermal resistance of a single uniform layer once temperatures have settled. It does not include the surface (convective and radiative) resistances at the faces of the layer, nor heat carried by air leaks or moisture, and it assumes the material conducts heat evenly. Keep your units consistent — metres, W/(m·K), and square metres — or the kelvin per watt will be wrong: convert millimetres to metres by dividing by 1000 before you enter the thickness.