B.1.12—Conduction rate

Conduction rate

The rate of thermal energy transfer through a uniform slab is

$$\frac{\Delta Q}{\Delta t}=\frac{kA\Delta T}{\Delta x}$$

where k is the material’s thermal conductivity, A is cross-sectional area, ΔT is the temperature difference and Δx is the transfer distance.

Read the proportionalities

The rate increases with larger k, larger area and larger temperature difference. It decreases when the material is thicker, because Δx is in the denominator.

Calculation checks

Use consistent SI units: area in m², distance in m, temperature difference in K or °C, and k in W m⁻¹ K⁻¹. The rate is measured in watts, because 1 W = 1 J s⁻¹.

Worked example from local Question Bank row 127628

Ice has $k=2.3\,\mathrm{W\,m^{-1}\,K^{-1}}$, thickness $0.019\,\mathrm{m}$ and temperature difference $6\,\mathrm{K}$. Per unit area,

$$\frac{1}{A}\frac{\Delta Q}{\Delta t}=\frac{k\Delta T}{\Delta x}=\frac{(2.3)(6)}{0.019}=7.3\times10^2\,\mathrm{W\,m^{-2}}$$

The result is a heat flux; multiply by area to obtain total power.

Common trap

Use the temperature difference across the slab, not an absolute temperature. A temperature gradient is a change per distance, so do not omit Δx.