Heat Conduction Calculator
Enter a material's conductivity, the area, the temperature difference and the thickness to get the steady-state heat transfer rate in watts using Fourier's law.
Fourier's law in one step
Enter the four inputs and the calculator returns the conductive heat transfer rate Q = k·A·ΔT/d in watts.
Use SI units
Conductivity in W/(m·K), area in m², temperature difference in K and thickness in m give the heat rate in watts — convert cm to m first.
What is heat conduction?
Thermal energy flowing through a material
Heat conduction is the flow of thermal energy through a material from its warmer side to its cooler side, without the material itself moving. This heat conduction calculator applies Fourier's law to a flat layer — a wall, a window pane, an insulation panel — and returns the steady-state rate at which energy passes through it, measured in watts. It takes four inputs: the material's thermal conductivity, the area the heat crosses, the temperature difference between the two faces, and the thickness of the layer. The result is the number behind heating bills, insulation choices, and how quickly a hot object loses heat to its surroundings.
Enter the conductivity, area, temperature difference and thickness to get the conductive heat transfer rate in watts instantly.
Fourier's law for a flat layer in steady state multiplies the thermal conductivity by the area and the temperature difference, then divides by the thickness.
Q = k × A × ΔT ÷ dConductivity (k) and area (A) sit on top, so a more conductive material or a larger surface lets more heat through. The temperature difference (ΔT) is the driving force — no gap, no flow. Thickness (d) is in the denominator, so a thicker layer slows the flow. Use SI units and the answer comes back in watts: joules of heat per second.
Suppose a 10 m² insulation panel is 0.1 m (10 cm) thick, has a conductivity of 0.04 W/(m·K), and separates spaces 20 K apart.
Multiply conductivity by area
0.04 × 10 = 0.4 — the conductance per unit thickness across the panel.
Multiply by the temperature difference
0.4 × 20 = 8 — the temperature gap drives the heat flow.
Divide by the thickness
8 ÷ 0.1 = 80 W — the steady-state heat transfer rate through the panel.
The heat transfer rate (80 W for the panel above) is how much thermal energy crosses the layer every second — the same as an 80 W light bulb's power, leaking continuously through that wall. That tells you directly what a heater must replace to keep the warm side at a steady temperature, and over a day it adds up to 80 W × 86,400 s ≈ 6.9 megajoules of energy. Two levers dominate the outcome. Lowering the conductivity (switching from still air to a dedicated insulator) or adding thickness both cut the rate, and because thickness is in the denominator, doubling it from 0.1 to 0.2 m halves the loss to 40 W. The temperature difference scales the result in direct proportion: on a colder day with a 40 K gap the same panel would lose 160 W. This is why insulation is rated by its conductivity and thickness together, and why a bigger temperature contrast between inside and outside always costs more to maintain.
The formula is exact for the case it describes, but a few assumptions are worth keeping in mind.
Steady state, one material, consistent units
This calculator gives the steady-state conduction through a single uniform flat layer. It does not cover the transient warm-up phase, convection or radiation, layered walls with several materials, or curved shapes, all of which need a fuller model. Keep your units consistent — W/(m·K), m², K and m — or the watts will be wrong: convert centimetres to metres by dividing by 100 before you enter the thickness.