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Thermal conduction

Apply Fourier’s law, thermal resistance and sign conventions to conduction, with links to convection and radiation.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Conduction. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

At a glance

  • Prerequisites: temperature gradient, energy conservation, basic calculus.
  • Outcomes: apply Fourier’s law in 1D and interpret thermal conductance/resistance.
  • Key result: Q dot = -kA dT/dx.
  • Common trap: dropping the minus sign without controlling direction conventions.

Setup

Model a homogeneous slab/rod in 1D (x direction):

  • Cross-sectional area A constant.
  • Thermal conductivity k constant (for first-pass UY1 problems).
  • T(x) smooth.

Physical interpretation:

  • k (thermal conductivity) tells you how strongly a material conducts heat for a given gradient. Large k (metals) means a small gradient can carry large heat flow.
  • Heat flow is driven by a gradient: conduction is the “diffusion of microscopic energy” through matter.

Sign convention:

  • Positive x to the right.
  • Positive Q dot taken as heat flow in +x direction.

Bridge note (O-Level to UY1):

  • O-Level: heat flows hot to cold.
  • UY1 upgrade: quantify local flux with a differential law and sign.

Core derivation / explanation

Fourier’s law (1D):

Q dot = -kAdT/dx.

The minus sign enforces hot-to-cold flow:

  • If dT/dx < 0 (temperature decreases rightward), then Q dot > 0 (flow rightward).

For steady conduction through uniform slab length L with end temperatures Tₕ > Tc:

Q dot = (kA(Tₕ-Tc))/L.

Define thermal resistance:

Rₜₕ = L/kA, Q dot = (Δ T)/Rₜₕ.

This mirrors Ohm’s-law structure.

Limit checks:

  • k → 0 gives Q dot → 0 (perfect insulator in this model).
  • L → 0 gives very large Q dot for a fixed Δ T (the “short path” limit).
Pitfalls (conduction)
  • Keep a consistent direction convention: the minus sign in Fourier’s law enforces “hot to cold”. - If k depends strongly on T, the steady slab result Q dot = kAΔ T/L must be replaced by an integral form. - Real systems can have contact resistances and multi-dimensional spreading; the 1D slab model is a controlled approximation.

Units check:

[k] = W m⁻¹K⁻¹ ⇒ [kAdT/dx] = W.

Physical sanity check:

  • Larger k or A increases heat rate.
  • Larger L lowers heat rate.

Heat rate through a copper rod

Worked example 1

Conduction through copper

Problem

A uniform copper rod has k = 400 W m⁻¹K⁻¹, A = 2.0 × 10⁻⁴ m² and L = 0.50 m. Its ends differ by 40 K. Find the steady heat-transfer rate, assuming insulated sides.
Show full solution

  1. Apply the steady slab law

    Method

    Use Q dot = kAΔ T/L for the hot-to-cold rate.

    Reason

    Constant conductivity and insulated sides give a linear steady temperature gradient along the rod.

    Working

    Answer

    Q dot = (kAΔ T)/L = ((400)(2.0 × 10⁻⁴)(40))/0.50 = 6.4 W.

  2. Interpret the rate

    Working

    Answer

    The rod transfers 6.4 J each second from hot to cold: Q dot = 6.4 W. This is a power, not a total heat energy.

Practice set

Check your understanding 1

For steady conduction through a uniform slab, what happens to Q dot when thickness doubles with conductivity, area and end temperatures fixed?

Show hint
Locate thickness in Q dot = kAΔ T/L.
Show answer
It halves: Q dot ∝ 1/L, so replacing L by 2L gives half the original heat rate.

Check your understanding 2

A uniform wall has k = 0.8 W m⁻¹K⁻¹, area 10 m², thickness 0.20 m and a temperature difference of 15 K. Find the steady heat rate, neglecting edge losses.

Show hint
Divide conductivity times area times temperature difference by thickness.
Show answer
Q dot = (0.8)(10)(15)/0.20 = 600 W, flowing from the hotter side to the colder side.

Check your understanding 3

Temperature decreases as x increases in a rod. A calculation using positive k,A gives negative Q dot, defined as positive toward increasing x. What sign error should you check?

Show hint
Apply the minus sign to the negative temperature gradient.
Show answer
Fourier’s law gives Q dot = -kA(dT/dx) > 0 when dT/dx < 0. Check for a missing minus sign or an inconsistently assigned gradient direction. The heat must flow toward increasing x in this case.

Summary + next steps

  • Conduction is quantified by local gradient-driven transport.
  • Fourier’s law plus geometry gives practical heat-rate formulas.
  • Thermal resistance form helps with multilayer systems.

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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