Transport: Heat Conduction & Diffusion Scaling (IPhO Thermo/Stat)
IPhO thermo/stat lesson on transport estimates: Fourier and Fick laws, thermal and mass diffusion time scales, and how to spot L^2 scaling quickly.
On this page
Transport problems are “physics of time scales”. If you can see why diffusion time grows like L², you can estimate heating times, mixing times, and signal delays without solving full PDEs. IPhO questions often reward the right scaling argument plus one clean steady-state calculation.
- Fourier law (1D): qₓ = -k dT/dx, heat current Q dot = qₓA.
- Fick law (1D): jₓ = -D dc/dx, particle current N dot = jₓA.
- Thermal diffusion time: t∼ L²/α where α = k/(ρ c).
- Mass diffusion time: t∼ L²/D.
1. Definitions (Must Know)
- Thermal conductivity k: material property linking heat flux to temperature gradient (units W m⁻¹ K⁻¹).
- Thermal diffusivity α: α = k/(ρ c) (units m² s⁻¹), sets how fast temperature disturbances spread.
- Diffusion coefficient D: sets how fast concentration spreads (units m² s⁻¹).
- Heat flux q: power per area (W m⁻²). In 1D steady conduction, it is constant through the cross-section.
- Thermal resistance (slab): Rₜₕ = L/kA so Q dot = (Δ T)/Rₜₕ.
- Random walk scaling: after time t, typical displacement x_typ∼ square root of Dt.
2. Key Ideas (What Earns Marks)
- See the L² time scale: diffusion-like processes are slow on large length scales. Doubling a length scale multiplies diffusion time by four.
- Separate steady-state from transient:
- steady-state conduction is algebra (thermal resistances)
- transients are usually scaling estimates (thermal penetration depth)
- Map heat to diffusion: temperature evolves by a diffusion equation with diffusivity α.
- Order-one constants rarely matter: t∼ L²/α is often enough unless the question explicitly asks for a factor like π².
3. Detailed Explanations
A. Why heat conduction has a diffusion equation
Energy conservation in a slab plus Fourier law produces (∂ T)/(∂ t) = α (∂² T)/(∂ x²), α = k/(ρ c).
This is structurally identical to the diffusion equation for concentration: (∂ c)/(∂ t) = D (∂² c)/(∂ x²).
So any intuition you have for diffusion carries over to heat.
B. Thermal resistance networks (steady state)
For a plane slab of thickness L: Q dot = (kA/L)(Tₕₒₜ-T_cold).
For multiple layers in series, add thermal resistances: R_(th,tot) = ∑ᵢ Lᵢ/kᵢA, Q dot = (Δ T)/R_(th,tot).
C. Diffusion length and time scale
From x_typ∼ square root of Dt (random walk), t∼ L²/D.
For heat, replace D by α: t∼ L²/α.
Micro-pattern: penetration depth for a short heating pulse
If you heat one surface for a time t, only a layer of thickness ℓ∼ square root of (α t) has time to respond. This is the fastest way to estimate how much mass is “thermally active” during a transient.
D. Kinetic-theory scaling for D and k (when a gas is involved)
For a dilute gas with typical speed v and mean free path λ: D∼ (1/3)vλ.
Thermal conductivity scaling is similar: k∼ (1/3)cᵥ^((vol)) vλ, where cᵥ^((vol)) is heat capacity per volume. These are not exact, but they explain trends with density and temperature.
4. Common Mistakes
- Confusing conductivity k (steady-state flux) with diffusivity α (time-dependent spreading).
- Forgetting area A in Rₜₕ = L/(kA) and therefore getting heat currents wrong by large factors.
- Treating diffusion as a constant-speed front. Diffusion spreads as square root of t, not as t.
- Mixing units (cm and m) inside L² scaling, which blows up errors.
- Solving a full PDE when the question only needs a scaling estimate.
5. Exam Tips
- Ask first: is this steady-state or transient?
- steady: use Q dot = Δ T/Rₜₕ
- transient: use t∼ L²/α or L²/D
- For transient heating of a thick object from one side, use the penetration depth idea to find the effective heated mass.
- If you see “composite wall” or “layers”, draw the thermal-resistance circuit.
- If the problem is about gases, expect λ and v to enter through D∼ vλ.
6. Worked Examples
Example 1: How long does it take to equilibrate across a slab?
A material has thermal diffusivity α = 1.0 × 10⁻⁵ m²/s. Estimate the time for a temperature change to spread across thickness L = 1 cm.
Worked solution (t ~ L^2/alpha)
Use the diffusion time scale t∼ L²/α.
Here L = 1 cm = 1.0 × 10⁻² m, so t∼ ((1.0 × 10⁻²)²)/(1.0 × 10⁻⁵) = (1.0 × 10⁻⁴)/(1.0 × 10⁻⁵) ≈ 10 s.
Order-one geometric factors can shift this, but the scaling and magnitude are the key.
Example 2: Steady heat flow through two layers (thermal resistances)
Two layers of equal area A are in series between temperatures Tₕ and T_c:
- Layer 1: thickness L₁, conductivity k₁
- Layer 2: thickness L₂, conductivity k₂
Find the steady heat current Q dot and the interface temperature Tᵢ.
Worked solution (series resistances)
Thermal resistances: R₁ = L₁/k₁A, R₂ = L₂/k₂A, Rₜₒₜ = R₁ + R₂.
Heat current: Q dot = (Tₕ-T_c)/Rₜₒₜ.
Temperature drop across layer 1 is Δ T₁ = Q dot R₁, so the interface temperature is
Equivalently, Tᵢ is a weighted average: Tᵢ = (Tₕ R₂ + T_c R₁)/(R₁ + R₂).
Example 3: Diffusive spread of a concentration blob
A dye initially occupies a narrow region. After time t, what is the typical width of the dye distribution in 1D in terms of D?
Worked solution (random walk scaling)
Diffusion gives a typical displacement scaling x_typ∼ square root of Dt .
So the width of the distribution grows like square root of Dt (up to a factor of order 1 that depends on the exact definition of width).
7. Mind Stretchers
- A surface temperature oscillates sinusoidally with angular frequency ω. How deep does the temperature oscillation penetrate into the material?
- Two identical rods are joined end-to-end, one with conductivity k and one with 2k. In steady state with fixed end temperatures, where is the midpoint temperature? Generalize to unequal lengths.
- For a gas, explain why decreasing pressure can increase the mean free path but does not necessarily increase thermal conductivity in the way naive k∼ vλ reasoning suggests (hint: density also changes).
Mind-stretcher (solution idea): oscillatory heating penetration depth
The diffusion equation sets the characteristic balance (∂ T)/(∂ t)∼ ω T, (∂² T)/(∂ x²)∼ T/ℓ².
With ∂ T/∂ t = α ∂² T/∂ x², this gives ω T∼ αT/ℓ² ⇒ ℓ∼ square root of (α/ω) .
A more careful solution gives an order-one factor (often square root of 2), but the key IPhO insight is the square-root dependence on both α and 1/ω.
8. Practice
- Do five quick estimates: pick L and compute t∼ L²/α for common materials (metals, glass, water).
- Drill steady-state conduction as a resistance circuit (series layers, parallel paths).
Syllabus and review details
No official syllabus alignment is listed for this lesson.