The big idea: Thermal energy (heat) always flows from a hotter place to a colder one.
It can get there 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).
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.
Worked example — heat loss through a window
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.
Solution
- Start with the given formula:
- Find ΔT first — the temperature difference (21 − 5 = 16):
- Put in the numbers (k = 0.80, A = 2.0, Δx = 4.0 × 10⁻³):
- Work it out — the unit of a rate of energy is the watt (W):
Final answer
rate ≈ 6.4 × 10³ W (6400 J of heat per second).
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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).
Classic ask: explain how the rate changes as a layer gets thicker (Δx is on the bottom, so the rate falls).
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.
IB-style question — conduction through a growing ice layer
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.
Solution
- Start with the given formula (per unit area, so A = 1.0 m²):
- Find ΔT — the temperature difference across the ice:
- Put in the numbers (k = 2.2, A = 1.0, Δx = 5.0 × 10⁻²):
- Work it out — the unit is the watt (here W per m²):
- As the ice thickens, Δx (on the bottom) grows, so the rate falls — thicker ice insulates better:
Final answer
≈ 4.4 × 10² W m⁻² now; the rate decreases as the ice gets thicker (rate ∝ 1 ÷ thickness), which is why a frozen lake freezes more slowly the deeper its ice gets.