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Fourier's Law of Heat Conduction

Relates the rate of heat conduction through a material to its thermal conductivity and the temperature gradient across it.

Materials ScienceThermal PropertiesProcess Design

Fourier's Law CalculatorHeat Conduction

q = −k · (dT/dx)
q = heat flux (W/m²)  ·  k = thermal conductivity (W/m·K)  ·  dT/dx = temperature gradient (K/m)
⟹ Solveq, k, dT/dx
W/m·K
K/m
W/m²
Please fix the errors above.
Solve for:
Materials:
Heat Flux
k: dT/dx: q:
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q = −k · (dT/dx)  ·  The negative sign indicates heat flows from hot to cold

Interpretation

q = −k(dT/dx). Heat flux due to temperature gradient. k is thermal conductivity. Used in steady‑state heat transfer analysis. Foundation of thermal design.

q = -k * (dT/dx)
Fourier's Law of Heat Conduction

Variables

SymbolQuantityUnit
qHeat fluxW/m2
kThermal conductivityW/m.K
dT/dxTemperature gradientK/m

What it means

This is identical to id=56. Fourier’s law is the basic equation for conduction heat transfer. It states that the heat flux (q) is proportional to the negative temperature gradient. The thermal conductivity k is a material property that measures the ability to conduct heat. This law is used to compute heat loss through walls, insulate pipes, cool electronics, and design heat exchangers. In solids, heat conduction is due to lattice vibrations and free electrons. Understanding Fourier’s law is crucial for engineers in building, chemical, mechanical, and aerospace disciplines to manage energy and maintain safe operating temperatures.

Worked example

Fourier's Law – Two Examples

Real‑World
Scenario: A copper heat sink (k = 401 W/m·K) has a temperature gradient of −100 K/m. The thermal engineer calculates the heat flux to verify the heat sink can dissipate 40,000 W/m² from a CPU.
ParameterValue
k401 W/m·K
dT/dx−100 K/m
1q = −401 × (−100) = 40,100 W/m²
Result 40,100 W/m² ✓ High
Scenario: A building wall with insulation (k = 0.04 W/m·K) has a temperature gradient of −20 K/m. The energy engineer calculates the heat loss to size the heating system for a passive house.
ParameterValue
k0.04 W/m·K
dT/dx−20 K/m
1q = −0.04 × (−20) = 0.8 W/m²
Result 0.8 W/m² ✓ Low loss
Materials insight: Fourier's law is the fundamental equation for heat conduction. The negative sign indicates heat flows from high to low temperature.

Common mistakes

  • Same as ID 56 – see that entry.
  • Note: For multi‑dimensional heat flow, use the full heat equation.

Applications

Fourier's law (q = −k·dT/dx) is the fundamental equation for conductive heat transfer. It is used in design of thermal insulation, heat exchangers, and thermal management systems. Engineers determine heat flux, temperature gradients, and thermal conductivity. The law is critical for sizing cooling systems in electronics, for designing furnaces and ovens, and for evaluating building energy efficiency. By applying Fourier's law, engineers can predict temperature profiles, ensure safe operating temperatures, and optimise the thermal performance of devices and systems. It is also the basis for thermal conductivity measurement techniques.

  • Design of heat sinks, heat spreaders, and thermal interface materials
  • Insulation design for buildings, pipes, and cold chains
  • Thermal modelling of power electronics and LED lighting
  • Furnace and reactor thermal design
  • Thermal property characterisation of materials

Frequently Asked Questions

Q01What is Fourier's law of heat conduction?
A01

Fourier's law states that the heat flux is proportional to the temperature gradient: q = –k · ∇T. For one dimension, q = –k · dT/dx. It is the fundamental equation for heat conduction.

Q02What is the physical meaning of thermal conductivity (k)?
A02

k is a measure of a material's ability to conduct heat. High k (e.g., copper ~400 W/m·K) indicates good conduction; low k (e.g., air ~0.025) indicates good insulation.

Q03What is the difference between steady‑state and transient heat conduction?
A03

Steady‑state: temperature does not change with time (dT/dt=0). Fourier's law directly gives the heat flux. Transient: temperature changes with time, requiring the heat equation: ρ·c_p·∂T/∂t = ∇·(k∇T) + Q_gen.

Q04How do you apply Fourier's law to a composite wall (multiple layers)?
A04

For layers in series, the heat flux is constant through all layers. The total thermal resistance is the sum of individual resistances (L/kA). The heat flux is q = (T₁ – T₂) / Σ(L_i/k_i).

Q05What are the typical thermal conductivities of different materials?
A05

  • Metals: 50‑400 W/m·K.
  • Ceramics: 2‑10 W/m·K.
  • Polymers: 0.1‑1 W/m·K.
  • Insulators: 0.01‑0.05 W/m·K.

Q06How does thermal conductivity change with temperature?
A06

For metals, k generally decreases with increasing temperature (electron scattering). For insulators, k often increases with temperature (phonon transport). At low temperatures, k can peak.

Q07What is the heat equation and how is it derived?
A07

The heat equation combines Fourier's law with energy conservation: ρ·c_p·∂T/∂t = k·∇²T + Q_gen. It is the governing equation for transient conduction.

Q08What are the common mistakes when using Fourier's law?
A08

  • Using a constant k when it varies with temperature.
  • Ignoring the negative sign.
  • Applying steady‑state to transient problems.
  • Not including heat generation.

Q09How do you measure thermal conductivity?
A09

Using steady‑state methods (guarded hot plate) or transient methods (laser flash). The method depends on the material and temperature range.

Q10What are some applications of Fourier's law in engineering?
A10

  • Designing thermal insulation.
  • Heat exchanger design.
  • Electronic cooling.
  • Thermal stress analysis.