Case 1
- Thermal conductivity k
- 0.04W/(m·K)
- Cross-sectional area A
- 10m²
- Temperature difference ΔT
- 20K
- Thickness d
- 0.1m
80.00W
Open with these valuesHeat Conduction Calculator
Change any value to update the result.
80.00W
Heat flow rate: 80.00 WExamples
How the result moves
Thickness d (m) → Heat flow rate (W)Your current valid inputs, with one value varied across the range.
How the result moves
Held fixed: Thermal conductivity k 0.0400 W/(m·K), Cross-sectional area A 10.0000 m², Temperature difference ΔT 20.00 K.
| Thickness d (m) | Heat flow rate (W) |
|---|---|
| 0.0800 | 100.00 |
| 0.1000Your value | 80.00 |
| 0.1200 | 66.67 |
Fourier's law in one line: Q = k × A × ΔT ÷ d. A 10 m² wall of mineral wool 100 mm thick with 20 K across it passes 80 watts. Double the thickness and the loss halves; the temperature difference may be negative, and then heat flows the other way.
Worked examples
Case 2
- Thermal conductivity k
- 1W/(m·K)
- Cross-sectional area A
- 4m²
- Temperature difference ΔT
- 15K
- Thickness d
- 0.005m
12,000.00W
Open with these valuesCase 3
- Thermal conductivity k
- 400W/(m·K)
- Cross-sectional area A
- 2m²
- Temperature difference ΔT
- 50K
- Thickness d
- 0.01m
4,000,000.00W
Open with these valuesHow it's calculated
Q = k × A × ΔT ÷ d
- StepEnter the thermal conductivity of the material, in W/(m·K).Thermal conductivity k=0.04W/(m·K)
- StepEnter the area the heat crosses, in square metres.Cross-sectional area A=10m²
- StepEnter the temperature difference between the two faces, in kelvin.Temperature difference ΔT=20K
- StepEnter the thickness along the heat path, in metres.Thickness d=0.1m
- ResultRead the heat flow in watts — joules per second.80.00W
Reference table
| k, A, ΔT, d | What that could be | Heat flow in W |
|---|---|---|
| 0.04, 10, -20, 0.1 | same wall, heat flowing inward | -80.00 |
| 1, 1, 1, 1 | unit check of the formula | 1.00 |
| 0.025, 5, 30, 0.05 | a 50 mm still-air gap | 75.00 |
| 0.04, 10, 20, 0.1 | 100 mm of mineral wool | 80.00 |
| 1, 4, 15, 0.005 | 5 mm of window glass | 12000.00 |
| 400, 2, 50, 0.01 | 10 mm of copper | 4000000.00 |
Questions
How do I calculate the rate of heat conduction?
Use Fourier's law: multiply the conductivity by the area and the temperature difference, then divide by the thickness. A 10 m² panel with k = 0.04, ΔT = 20 K and d = 0.1 m passes 0.04 × 10 × 20 ÷ 0.1 = 80 W.
What is thermal conductivity k?
It is how readily a material passes heat, in W/(m·K). Insulators are low — air about 0.026, mineral wool about 0.04 — while glass sits near 1 and copper near 400.
Why does a thicker layer conduct less?
The thickness is in the denominator, so the heat flow is inversely proportional to it. Doubling an insulation layer halves the loss at the same temperature difference.
How does this relate to thermal resistance?
Thermal resistance is R = d ÷ (k × A) in K/W, and the heat flow is simply Q = ΔT ÷ R. The two calculators use the same three inputs and answer opposite questions: how much heat gets through, and how hard the layer makes it.
What does steady state mean here?
It means both faces sit at constant temperatures and the flow no longer changes with time. While the wall is still warming up, the flow varies and this simple form does not apply.