Fourier's Law of Heat Conduction

Equivalent forms

\dot{Q} = -kA\frac{dT}{dx}

\vec{q} = -k \nabla T

\dot{Q} = \frac{kA(T_H - T_C)}{L}

The negative sign encodes nature's preference: heat always flows down the temperature gradient, never up.

Symbol	Name	SI unit	Dimension	Range
$Q_dot$	Heat flow rateoutput Rate of heat energy transfer through the material	W	M·L²·T⁻³	0 – 10000
$k$	Thermal conductivity Material's ability to conduct heat	W/(m·K)	M·L·T⁻³·Θ⁻¹	0.01 – 500
$A$	Cross-sectional area Area perpendicular to heat flow direction	m²	L²	0.0001 – 0.01
$Delta_T$	Temperature difference Temperature difference across the conductor (T_H - T_C)	K	Θ	1 – 500
$L$	Length Length of the conducting path	m	L	0.01 – 5

Unit systems

Symbol	SI	CGS	Imperial
$Q_dot$	W	erg/s	BTU/h
$k$	W/(m·K)	cal/(s·cm·°C)	BTU/(h·ft·°F)
$A$	m²	cm²	ft²
$Delta_T$	K	K	°F
$L$	m	cm	ft

Where it holds

Valid for steady-state conduction in isotropic, homogeneous solids. Breaks down for radiative or convective heat transfer, and at nanoscale where ballistic phonon transport dominates.

Dimensional analysis

[Q̇ $] = [k][$ A $][\Delta T]/[L] \to M\cdot L^{2}\cdot T^{-3} = (M\cdot L\cdot T^{-3}\cdot \Theta ^{-1})(L^{2})(\Theta )/(L) = M\cdot L^{2}\cdot T^{-3} \checkmark$

Discovery

Joseph Fourier · 1822

Published in 'Théorie analytique de la chaleur.' Fourier invented Fourier series specifically to solve the heat equation — a mathematical revolution born from a physics problem.

Try this

Why can you touch a wooden spoon in a hot pot but not a metal one?

A copper rod (k = 385 W/m·K) is 0.5 m long with cross-section 0.001 m². One end is at 100°C, the other at 20°C. Find the heat flow rate.

Research status: stable

Real-world applications

Building insulation design (R-value $= L/kA)$
CPU heat sink engineering
Cooking — why copper pans heat more evenly than steel
Spacecraft thermal protection systems

Common misconceptions

The negative sign does not mean heat flows backward — it ensures Q̇ is positive when heat flows from hot to cold (down the gradient)
Thermal conductivity k is not constant — it varies with temperature, especially in metals and semiconductors
Fourier's law applies to conduction only, not convection or radiation — real heat transfer often involves all three simultaneously

Experimental verification

Originally verified by Fourier using heated metal bars with thermometers along their length. Modern methods use laser-flash analysis and guarded hot-plate apparatus (ASTM C177) to measure

k to \pm 1

% accuracy for materials from aerogel

(k \approx 0.015)

to diamond

(k \approx 2200)

.

Derivation

Fourier postulated that heat flux is proportional to the negative temperature gradient:

q = -k(dT/dx)

.

For a slab of area A, thickness L, with faces at T_H and T_C: Q̇

= kA(T_H - T_C)/L

.

This follows from integrating q over area A with a linear temperature profile in steady state.

Limiting cases

k \to 0

(perfect insulator)⟶

Q_dot \to 0

No conduction through a perfect insulator.

\Delta _T \to 0

⟶

Q_dot \to 0

No temperature gradient means no heat flow.

L \to 0

⟶

Q_dot \to \infty

Infinitely thin conductor implies infinite flux (unphysical; contact resistance intervenes).

What if…

What if you replace copper

(k = 385)

with wood

(k = 0.15)

?

Heat flow drops by a factor $of \sim 2567$ . Wood conducts $0.15/385 \approx 0.04$ % as well as copper — this is why wooden spoon handles stay cool while copper ones burn.

What if you double the rod length?

Heat flow halves. Q̇ $\propto 1/L$ — longer conduction paths reduce heat transfer. This is why thick walls insulate better than thin ones.

What

if \Delta T = 0

but heat is being generated inside?

Fourier's law gives zero flux at that point, but the heat equation $(\partial T/\partial t = \alpha \nabla ^{2}T + q$ ̇/ $\rho c)$ handles internal generation. Steady state requires the generated heat to flow out through boundaries.

1

Copper rod heat flow

Given ·

k:: 385
A:: 0.001
T H:: 373
T C:: 293
L:: 0.5

Find · Q_dot

Steps

Temperature difference: $\Delta T = 373 - 293 = 80\,\mathrm{K}$
Apply Q̇ $= kA\Delta T/L$
Q̇ $= 385 \times 0.001 \times 80 / 0.5 = 61.6\,\mathrm{W}$

Result · Q̇

= kA(T_H - T_C)/L = 385 \times 0.001 \times 80 / 0.5 = 61.6\,\mathrm{W}

2

Styrofoam wall insulation

Given ·

k:: 0.033
A:: 10
T H:: 293
T C:: 263
L:: 0.05

Find · Q_dot

Steps

$\Delta T = 293 - 263 = 30\,\mathrm{K}$ (indoor $20^{\circ}C$ , outdoor $-10^{\circ}C)$
Q̇ $= 0.033 \times 10 \times 30 / 0.05$
Q̇ $= 198\,\mathrm{W}$ lost through $10\,\mathrm{m}^{2} of 5\,\mathrm{cm}$ styrofoam

Result · Q̇

= kA\Delta T/L = 0.033 \times 10 \times 30 / 0.05 = 198\,\mathrm{W}

Stefan-Boltzmann Law→Linear Thermal Expansion→First Law of Thermodynamics→