Fourier’s law of heat conduction
Conductive heat flux points opposite the temperature gradient; a plane layer gives a familiar steady one-dimensional heat rate.
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Formula
q in W/m2; Q̇ in W; k in W/(m·K); T in K; x in m
Variables & units
| Symbol | Meaning | Unit |
|---|---|---|
| q⃗ | Heat-flux vector | W/m2 |
| Thermal conductivity | W/(m·K) | |
| Temperature gradient | K/m | |
| Q̇ | Heat-transfer rate through a section | W |
| Area normal to heat flow | m2 |
LaTeX
\vec q=-k\nabla T, \qquad \dot Q=-kA\frac{dT}{dx}
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Derivation & background
The minus sign encodes heat flow from hot toward cold. For a homogeneous plane layer with constant conductivity, constant area and steady one-dimensional conduction, Q̇=kA(Thot−Tcold)/L. Temperature-dependent conductivity, contact resistance, internal heat sources or transient temperatures require an adjusted model. The differential vector law is local; the wall equation is its integrated special case.
Exam blueprint
Validity range
The plane-wall form assumes steady one-dimensional conduction, constant k and area.
Derivation steps
Integrate local flux opposite the temperature gradient through the layer.
- 1Locally qx=−k dT/dx.
- 2With constant k and steady profile, the gradient is ΔT/L.
- 3Multiply flux density by area A.
Rearrangements
Heat rate
Units: W, W/(m·K), m2, K and m.
Task variant
Given k=0.8, A=10, L=0.2 and ΔT=15, find Q̇.
Q̇=0.8×10×15/0.2=600 W.
Common mistakes
Treating the minus sign as magnitude rather than direction.
It shows heat flow opposite the temperature gradient.
Exam context
- Heat loss through walls and insulation.
Formula cluster
Heat transfer
Conduction is distinct from convection and radiation.
Worked example
A wall with k=0.8 W/(m·K), area 10 m2, thickness 0.20 m and a 15 K temperature difference conducts about 600 W at steady state.
Applications
Estimate heat loss through walls and insulation and model temperature gradients in conducting materials.
How do you calculate with Fourier’s law of heat conduction?
Here is how to work through a typical Fourier’s law of heat conduction (q⃗ = −k∇T; Q̇ = −kA·dT/dx) task step by step:
- 1
Task
Given k=0.8, A=10, L=0.2 and ΔT=15, find Q̇.
Solution path
Q̇=0.8×10×15/0.2=600 W.
Frequently asked questions about Fourier’s law of heat conduction
What does the minus sign in Fourier’s law mean?+
The minus sign says heat flux points opposite the temperature gradient. If T increases with x, dT/dx is positive and qx negative: heat flows toward lower temperature. It encodes direction under the chosen coordinate convention, not negative conductivity. k is positive for passive materials. Thermal conductivity k depends on material and often temperature. The simple wall form is a steady special case; transient temperature requires heat storage and boundary conditions. Distinguish flux density q in W/m2 from heat rate Q̇ in W. Contact resistance, layers and heat sources change the calculation.
How do I calculate heat transfer through a wall?+
For a homogeneous plane wall in steady one-dimensional conditions, Q̇=kAΔT/L. Multiply conductivity and area by temperature difference, then divide by thickness. For k=0.8, A=10 m2, ΔT=15 K and L=0.20 m, the result is 600 W. Use consistent SI units and hot minus cold temperature. Thermal conductivity k depends on material and often temperature. The simple wall form is a steady special case; transient temperature requires heat storage and boundary conditions. Distinguish flux density q in W/m2 from heat rate Q̇ in W. Contact resistance, layers and heat sources change the calculation.
What is the difference between q and Q̇?+
q is heat-flux density, or heat rate per area in W/m2. Q̇ is total heat transferred per time through a surface, in W. For uniform flux, Q̇=qA. If flux varies across the surface, integrate q over the area instead of multiplying a single value. Thermal conductivity k depends on material and often temperature. The simple wall form is a steady special case; transient temperature requires heat storage and boundary conditions. Distinguish flux density q in W/m2 from heat rate Q̇ in W. Contact resistance, layers and heat sources change the calculation.
When is the simple wall formula invalid?+
It assumes steady one-dimensional conduction, constant material behavior and constant area. Time changes require thermal storage; multidimensional cases have spatially varying gradients. Temperature-dependent k, internal heat sources and contact resistance also require an extended model. Local Fourier law remains a starting point, but boundary conditions determine the solution. Thermal conductivity k depends on material and often temperature. The simple wall form is a steady special case; transient temperature requires heat storage and boundary conditions. Distinguish flux density q in W/m2 from heat rate Q̇ in W. Contact resistance, layers and heat sources change the calculation.
How do multiple wall layers affect heat flow?+
For steady conduction through layers in series, the same heat rate passes through each layer. Temperature drops add, and each layer contributes thermal resistance L/(kA). Total resistance is their sum and Q̇=ΔT/Rth. Interface resistance can add further terms and should not be ignored in precise component calculations. Thermal conductivity k depends on material and often temperature. The simple wall form is a steady special case; transient temperature requires heat storage and boundary conditions. Distinguish flux density q in W/m2 from heat rate Q̇ in W. Contact resistance, layers and heat sources change the calculation.
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Which formula describes Fourier’s law of heat conduction?
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Subject references and further reading
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Related formulas
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Create a curated FSRS exam set for q⃗ = −k∇T; Q̇ = −kA·dT/dx: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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