IB Physics HL Topic 2 — Matter, Heat & Electricity Paper 1 & 2 Thermal Conduction ~10 min read

Thermal Conduction

Grab a metal spoon left in a hot pan and you’ll drop it fast; a wooden spoon you can hold all day. Same heat, same pan — the difference is how well each material lets energy travel through it. That’s thermal conduction: energy passed particle-to-particle through a material that itself stays put. And for HL there’s an equation for exactly how fast it flows.

📚 What you need to know

How conduction actually works

Heat one end of a solid and its particles there gain kinetic energy — they vibrate harder. Jostling against their neighbours, they pass some of that energy along, which sets those particles vibrating more, and so on down the material. The energy travels from hot to cold, but the particles themselves stay roughly in place: the material doesn’t flow, only the energy does.

ALL SOLIDS: vibration HOT COLD vibrations pass energy along · material stays put METALS: + free electrons HOT COLD + + + + + + + + + + free electrons carry energy fast → best conductors
In all solids, vibrations pass energy along by collisions (left). In metals, free electrons also shuttle energy from hot to cold — which is why metals conduct so much better (right).
The free-electron mechanism is the whole story of why metals feel cold and conduct heat so fast. Metals have a “sea” of delocalised electrons free to move through the whole structure. Heat one end and those electrons pick up energy and zip to the cold end almost immediately — far quicker than waiting for vibrations to ripple across. Non-metals have no such electrons, so they’re stuck with the slow vibration route: good insulators.

The rate of conduction

How fast does energy conduct through something? For HL, four things set the rate, and they all sit in one equation. The rate of thermal energy transfer (in watts) is:

Rate of thermal conduction ΔQ / Δt = k A ΔT / Δx
heat flow hot cold A Δx (thickness) ΔQ / Δt = k A ΔT / Δx
Heat conducts through a slab of area A and thickness Δx, driven by the temperature difference ΔT across it. A higher conductivity k means faster transfer.

where ΔQ/Δt is the rate of energy transfer (W), k the thermal conductivity of the material (W m−1 K−1), A the cross-sectional area (m2), ΔT the temperature difference across the material, and Δx its thickness. The ΔT/Δx part is the temperature gradient — how sharply the temperature drops across the material.

Conduction is faster when… k ↑ higher conductivity A ↑ larger area ΔT ↑ larger temp difference Δx ↓ smaller thickness … all follow straight from ΔQ/Δt = k A ΔT / Δx
Everything speeds conduction up except thickness: a thicker material slows it down, because the same temperature difference is spread over a longer distance.
This equation quietly explains a lot of everyday design. Radiators are made of metal (high k) with large surface area (big A). Loft insulation and double glazing trap air (tiny k) in a thick layer (large Δx) to make the rate as small as possible. Same physics, opposite goals.

Worked examples

WE 1

A single-glazed window measures 0.80 m × 1.2 m and is 4.0 mm thick. Inside is 21 °C, outside 4 °C. Taking kglass = 0.80 W m−1 K−1, find the rate of heat loss through it.

Area: A = 0.80 × 1.2 = 0.96 m² Temperature difference: ΔT = 21 − 4 = 17 K · thickness Δx = 4.0 mm = 0.0040 m Rate = kAΔT / Δx = 0.80 × 0.96 × 17 ÷ 0.0040 ΔQ/Δt ≈ 3300 W (3.3 kW) A huge rate — which is exactly why double glazing (a thicker, air-filled gap) is worth it.
WE 2

A metal rod 0.50 m long with cross-sectional area 2.0 × 10−4 m² conducts energy at 12 W when its ends differ by 60 K. Find the material’s thermal conductivity.

Rearrange ΔQ/Δt = kAΔT/Δx for k: k = (ΔQ/Δt) × Δx / (AΔT) = 12 × 0.50 ÷ (2.0 × 10−4 × 60) = 6 ÷ 0.012 k = 500 W m−1 K−1 A high value — this is a good metallic conductor (copper and silver are around 400 and higher).
WE 3

A wall loses heat by conduction at 200 W. Using the rate equation, find the new rate if (a) its thickness is doubled, or (b) instead the temperature difference across it is doubled (everything else unchanged).

(a) Rate ∝ 1/Δx, so doubling the thickness halves the rate. (a) 100 W (b) Rate ∝ ΔT, so doubling the temperature difference doubles the rate. (b) 400 W Reading the proportionalities straight off the equation is often faster than recomputing from scratch.

🔧 Using the conduction rate equation

  1. Identify k, A (m2), ΔT and Δx (m) — convert mm and cm to metres.
  2. Rate: ΔQ/Δt = kAΔT/Δx, in watts.
  3. Rearrange for whichever quantity is unknown (k, A, Δx…).
  4. Ratios? Read them off: rate ∝ k, ∝ A, ∝ ΔT, ∝ 1/Δx.
  5. Sense check: metals give big rates; insulators and thick layers give small ones.
good conductor
high k (metals)
vs
good insulator
low k (air, wood)
rate =
kAΔT/Δx
Quick recap: Conduction passes thermal energy through a material without the material moving — by particle vibrations in all solids, and by free electrons in metals (which is why metals conduct best). The HL rate is ΔQ/Δt = kAΔT/Δx: faster for a higher conductivity k, larger area A and bigger temperature difference ΔT, and slower for a greater thickness Δx.

💡 Top tips

⚠ Common mistakes

Conduction is energy creeping through a material that stays still. But energy can also move by the material itself flowing — warm fluid rising, cool fluid sinking, carrying heat bodily with it. That’s the next method of thermal energy transfer. Coming up: thermal convection.

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