The big idea: Wrap your hands round a hot mug and they warm up; steam rises off the top; and you feel a fire's glow from across the room. That's thermal energy on the move — always from a hotter place to a colder one.
It can travel in three ways: conduction, convection and radiation.
They differ in what actually moves — and whether any material is needed at all.
| Mechanism | What moves | Needs a material? | Everyday example |
|---|---|---|---|
| Conduction | energy passes along, particles stay put | Yes — best in solids (esp. metals) | a metal spoon's handle getting hot |
| Convection | the hot fluid itself rises and circulates | Yes — only in fluids (liquids/gases) | warm air rising off a radiator |
| Radiation | infrared waves (no particles needed) | No — even works through a vacuum | the Sun's heat reaching Earth |
Spot the difference: Conduction = energy passed particle-to-particle (the particles stay put).
Convection = the hot fluid itself moves and carries the energy.
Radiation = infrared waves — the only one that crosses empty space (a vacuum).
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In a solid, faster-vibrating hot particles jostle their cooler neighbours, passing energy along. In a metal, free-moving electrons carry it too — which is why metals conduct so well.
The data booklet gives an equation for how fast heat flows by conduction through a flat slab:
- rate of heat flow — energy transferred each second (W, i.e. J s⁻¹)
- thermal conductivity of the material (W m⁻¹ K⁻¹)
- cross-sectional area the heat flows through (m²)
- temperature difference across the slab (K, or °C — a difference is the same in both)
- thickness of the slab (m)
What makes conduction faster?: Bigger k, bigger area A, or a bigger temperature difference ΔT → faster heat flow.
A thicker slab (bigger Δx) → slower heat flow. Δx is on the bottom, so rate ∝ 1 ÷ thickness.
A glass window has area 2.0 m², thickness 4.0 × 10⁻³ m, and thermal conductivity k = 0.80 W m⁻¹ K⁻¹. Inside is 21 °C, outside is 5 °C. Find the rate at which heat conducts through it.
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How this is tested — heat transfer shows up on both papers:
Paper 1A / short answer
- Describe the conduction mechanism.
- Or outline why a cooling graph's gradient flattens.
Paper 2
- Estimate the rate of heat flow through a wall, ice layer or window with ΔQ/Δt = kA·ΔT/Δx — and state the unit (W).
The classic trap: Explain how the rate changes as a layer gets thicker — Δx is on the bottom, so the rate falls (rate ∝ 1 ÷ thickness).
Thicker layer ⇒ slower conduction: In ΔQ/Δt = kA·ΔT/Δx, the thickness Δx is in the denominator.
So if everything else stays the same, doubling the thickness halves the rate of heat flow: rate ∝ 1 ÷ Δx.
A frozen lake loses heat by conduction up through its ice. The ice has conductivity k = 2.2 W m⁻¹ K⁻¹. The water below is at 0 °C and the top of the ice is at −10 °C. When the ice is 5.0 × 10⁻² m thick, find the rate of heat loss per square metre (take A = 1.0 m²), then state what happens to this rate as the ice gets thicker.
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