Case 1
- Thickness d
- 0.1m
- Thermal conductivity k
- 0.04W/(m·K)
- Area A
- 10m²
0.250000K/W
Open with these valuesThermal Resistance Calculator
0.250000K/W
Result: 0.250000 K/WThermal resistance is how many kelvin of difference it takes to push one watt through a layer: R = d ÷ (k × A). A 10 m² wall of mineral wool 100 mm thick comes to 0.25 K/W. Higher is better insulation — and the heat flow that follows is simply ΔT ÷ R.
Worked examples
Case 2
- Thickness d
- 0.012m
- Thermal conductivity k
- 1W/(m·K)
- Area A
- 1m²
0.012000K/W
Open with these valuesCase 3
- Thickness d
- 0.05m
- Thermal conductivity k
- 400W/(m·K)
- Area A
- 1m²
0.000125K/W
Open with these valuesHow it's calculated
R = d ÷ (k × A)
- StepEnter the thickness of the layer along the heat path, in metres.
- StepEnter the thermal conductivity of the material, in W/(m·K).
- StepEnter the area the heat passes through, in square metres.
- ResultRead the resistance in kelvin per watt; divide ΔT by it for the heat flow.
Reference table
| d, k, A | What that could be | Resistance in K/W |
|---|---|---|
| 0, 0.04, 10 | no layer at all | 0.000000 |
| 0.05, 400, 1 | 50 mm of copper over 1 m² | 0.000125 |
| 0.012, 1, 1 | 12 mm of glass over 1 m² | 0.012000 |
| 0.1, 0.04, 10 | 100 mm of mineral wool over 10 m² | 0.250000 |
| 0.15, 0.04, 5 | 150 mm of mineral wool over 5 m² | 0.750000 |
| 0.2, 0.025, 1 | 200 mm of still air over 1 m² | 8.000000 |
Questions
How do I calculate thermal resistance?
Divide the thickness by the conductivity times the area: R = d ÷ (k × A). A 0.1 m panel of 0.04 W/(m·K) over 10 m² gives 0.1 ÷ 0.4 = 0.25 K/W.
What is thermal resistance?
It is how strongly a layer opposes heat passing through it, measured in kelvin per watt. The value is the temperature difference needed to drive exactly one watt across the layer.
How does it relate to the R-value?
The R-value used in construction is area-specific, R = d ÷ k in m²·K/W, and describes one square metre of material. This calculator divides by the area as well and gives the resistance of the whole component in K/W.
Is a higher or a lower value better?
For insulation, higher is better — less heat escapes through the wall, window or roof. For a heat sink or a pan you want it low, so heat leaves quickly.
How is this different from the heat conduction calculator?
This one answers how hard the layer makes it, in K/W, using only its geometry and material. The conduction calculator adds a temperature difference and answers how many watts actually get through, and the two are linked by Q = ΔT ÷ R.
Sources and last check
Information, not professional advice.