Heat Transfer: Conduction, Convection & Radiation

Conduction: Fourier's law and thermal resistance

Jean-Baptiste Joseph Fourier showed in 1822 that heat flux through a material is proportional to the temperature gradient and the material's thermal conductivity k. Written in resistance form, Q̇ = (T₁−T₂)/R_th with R_th = L/(kA), the equation has exactly the structure of Ohm's law — temperature difference plays the role of voltage, heat flow the role of current, and layered materials simply add resistances in series. Copper conducts at k ≈ 385 W/(m·K), still air at only 0.026 — a factor of nearly 15,000 that explains why a stagnant air gap is such an effective insulator.

q = −k · dT/dx                      (Fourier's law, W/m²)
Q̇ = k·A·(T₁ − T₂)/L                 R_th = L/(k·A)   [K/W]

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Convection and the heat diffusion PDE

When a moving fluid carries heat away from a surface, Newton's law of cooling parameterises it with a convective coefficient h: Q̇ = h·A·(T_s − T_∞). Natural (buoyancy-driven) convection in air gives h ≈ 5-25 W/(m²·K); forced convection from a fan raises that to 25-250; liquid cooling can reach 100-20,000 W/(m²·K). Combining Fourier's law with energy conservation across a continuum gives the heat diffusion equation — one of the most important PDEs in all of physics — which a finite-difference scheme integrates forward in time step by step.

∂T/∂t = α · ∇²T              α = k/(ρ·Cₚ)   (thermal diffusivity, m²/s)
Explicit finite-difference stability condition: α·Δt/Δx² ≤ 0.5

Radiation and the Stefan-Boltzmann law

Every body above absolute zero emits electromagnetic radiation. Josef Stefan (1879) and Ludwig Boltzmann (1884) showed a perfect black body radiates Q̇ = σ·A·T⁴, with real surfaces scaling that by emissivity ε. Because this term grows with the fourth power of absolute temperature, radiation's share of total heat loss rises sharply with temperature: at 300 K a blackened surface in still air loses roughly 60% by convection and 40% by radiation, but by 1000 K radiation accounts for about 85% of the total. In the vacuum of space there is no convection at all, which is exactly why spacecraft thermal control depends entirely on surface emissivity and multi-layer insulation blankets.

Q̇_bb = σ · A · T⁴          σ = 5.670×10⁻⁸ W/(m²·K⁴)
Wien's law: λ_peak · T = 2.898×10⁻³ m·K
Sun (5778 K): λ_peak ≈ 500 nm    Room temp (300 K): λ_peak ≈ 9700 nm (IR)

Frequently asked questions

Why does thermal resistance behave like electrical resistance?

Fourier's law in resistance form, Q-dot = (T1-T2)/R_th with R_th = L/(kA), has exactly the same structure as Ohm's law with temperature difference playing the role of voltage and heat flow playing the role of current. Layered materials, such as a wall plus insulation, simply add their resistances in series just like resistors.

Why can radiation be ignored at room temperature but not at high temperature?

Radiated power scales with the fourth power of absolute temperature through the Stefan-Boltzmann law, while convective power scales only linearly with temperature difference. At 300 K a blackened surface in still air loses roughly 60% of its heat by convection and 40% by radiation, but by 1000 K radiation accounts for around 85% of total heat loss.

How does a spacecraft lose heat with no air to convect into?

In the vacuum of space there is no convection at all, so radiation is the only heat-transfer mode available. Spacecraft thermal control is therefore dominated by surface emissivity and multi-layer insulation blankets that manage how much of the Stefan-Boltzmann radiated power actually reaches or leaves the spacecraft body.

Try it live

Everything above runs in your browser — open Heat Conduction and paint hot and cold sources on a grid to watch thermal energy diffuse in real time under the 2-D heat equation. Nothing is installed, nothing is uploaded.

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