🌡️ Heat Conduction Calculator

Calculate the rate of heat transfer through a material using Fourier's Law of heat conduction.

🌡️ Heat Conduction Calculator

Calculate the rate of heat transfer through a material using Fourier's Law of heat conduction.

Surface Area (A)

Material Thickness (d)

Thermal Conductivity (k)

Hot Temperature (T₁)

Cold Temperature (T₂)

✅ Calculation Result

Heat Transfer Rate (Q)-

Heat Flux (q = Q/A)-

Temperature Gradient

T₁: 25°C

k = 0.04 W/(m·K)

Q ➔

d = 0.15 m

T₂: 5°C

Calculator Description

Heat conduction is the transfer of thermal energy from a hotter region to a colder one through molecular vibration and free-electron motion inside a material. It is the fundamental mechanism governing heat flow through stationary solids such as walls, insulation, windows, and pipe walls.

What this calculator finds

This calculator determines the steady-state heat transfer rate (Q) and the heat flux (Q/A) through a flat slab from the material's thermal conductivity, cross-sectional area, thickness, and the temperature difference across the two faces.

Why it matters

Evaluating insulation performance of walls/roofs and estimating heat loss (heating/cooling load)
Comparing how changes in insulation thickness or material affect heat loss
Designing insulation for process equipment and pipes to limit surface heat loss

Formula

Fourier's Law of Conduction

For one-dimensional steady-state conduction through a flat slab, the heat transfer rate is given by Fourier's law. It is proportional to the temperature difference, area, and conductivity, and inversely proportional to thickness.

Q = \dfrac{k A \Delta T}{d}

q = \dfrac{Q}{A}

Q — Heat transfer rate [W]
q — Heat flux (rate per unit area) [W/m²]
k — Thermal conductivity of the material [W/(m·K)]
A — Cross-sectional area for heat flow [m²]
d — Material thickness [m]
ΔT — Temperature difference across the faces [°C]

How the formula works

A larger temperature difference ΔT or a larger area A increases the heat transfer rate Q.
A thicker material d lengthens the heat path and lowers Q — the insulating effect.
A material with lower conductivity k (an insulator) loses less heat under the same conditions.

Worked example

For a fiberglass-insulated wall (k ≈ 0.04 W/(m·K)) that is 0.15 m thick and 10 m² in area with a 20 °C (= 20 K) difference across it, Q = (0.04 × 10 × 20) / 0.15 ≈ 53.3 W. The heat flux is 53.3 / 10 ≈ 5.3 W/m².

Useful Tips

Practical tips

For multilayer walls, sum each layer's thermal resistance R = d/(k·A) and then compute Q = ΔT / R_total.
The temperature difference ΔT is the same whether in Celsius or Kelvin, so you can use the difference directly.

Limitations & cautions

This formula covers only steady-state (time-invariant temperatures) and 1-D conduction; it does not apply to transient situations.
Real surfaces also involve convective/radiative and contact resistances, so a pure-conduction estimate can overpredict heat loss.
Thermal conductivity k can vary with temperature, so use the value appropriate for the operating range.