B.1 · IB Physics
Thermal conduction
Energy flows by conduction through a slab whose faces are held at fixed temperatures. Change the material, thickness, area and temperatures, add a second layer in series as in an insulated wall, and compare the temperature profile or the rate for different materials.
Predict first
With the default settings (brick, Δx₁ = 100 mm, A = 1.0 m², 20 °C and 0 °C), double Δx₁ to 200 mm. What happens to the rate of transfer P?
P = kAΔT ÷ Δx, so P ∝ 1/Δx. Doubling the thickness halves the temperature gradient, and P falls from 120 W to 60 W.
Check it in the model: Move the Thickness Δx₁ slider from 100 to 200 mm and compare Rate of transfer P in the readout.
profile / selected material other materials · colour of the slab shows temperature, blue cold to gold hot
Layers in series: P = (T₁ − T₂) ÷ Σ[Δx ÷ (kA)]
ΔT across a layer = P Δx ÷ (kA)
Q = P t
Try this
- Press 'Brick wall', then double the thickness Δx₁ from 100 mm to 200 mm. What happens to P and to the temperature gradient?
- Press 'Insulated wall' and look at the T against x graph. Which layer takes most of the temperature drop, and why?
- Press 'Compare materials' and find how many orders of magnitude separate copper from still air for the same slab.
Common error: adding or averaging the conductivities of two layers. In series it is Δx ÷ (kA) for each layer that adds, like resistors in series, and the same power passes through every layer, so use ΔT = P Δx ÷ (kA) for the drop across each. Convert thickness from mm to m first.