Conduction, Convection and Radiation

Beginner

Step 1. Heat always moves from warmer places to cooler ones, and it can travel in three ways. A pot of water on a gas flame shows all three at once.

Step 2. Conduction: heat passing through a material by contact. The particles in the hot underside of the pot jiggle hard and knock into their neighbours, which start jiggling harder too, and so on up through the metal. Nothing travels except the jiggling. Metals pass heat on quickly; wood, plastic, wool and still air pass it slowly and are called insulators. A fluffy coat keeps you warm because it traps air.

Step 3. Convection: heat carried by a moving liquid or gas. Water near the hot bottom warms, expands a little and becomes lighter than the cooler water above, so it rises; cooler water sinks to take its place, and a loop forms. People say that heat rises, but it is really the warm fluid that rises.

Step 4. Radiation: heat carried by invisible infrared light, which needs no material at all. That is how sunshine warms your face across empty space. Dark, dull surfaces absorb and give off this radiation well; shiny ones reflect it, which is why emergency blankets are silvery.

In the animation, gold arrows show heat crossing the metal base, where the particles jiggle. The water turns in two loops: red blobs (warm) rise in the middle and blue blobs (cool) sink along the sides. Purple waves carry radiation away from the flame.

Check yourself: why does a saucepan have a metal body but often a plastic or wooden handle? Answer: metal passes heat quickly to the food; the handle must be a poor conductor so that you can hold it.

Intermediate

Heat flow and what drives it

At room temperature, steel and wood are equally warm, yet the steel feels colder. Heat is energy that moves because of a temperature difference, on its own always from warmer to cooler. The amount, $Q$, is measured in joules, the rate $Q/t$ in watts, and the rate per square metre, the heat flux $q$, in W/m². A difference $\Delta T$ has the same value in kelvins and in degrees Celsius; only radiation depends on the temperatures themselves, which must then be in kelvins.

Heat travels in three ways: by conduction, from particle to particle through a material that stays put; by convection, carried by a moving liquid or gas; and by radiation, electromagnetic waves, mostly infrared, that need no material. The pot in the animation shows all three at once.

Conduction: Fourier's law

Through a flat layer of area $A$ and thickness $d$ whose faces differ by $\Delta T$, once the temperatures inside have settled, the heat flow is
$\frac{Q}{t}=\frac{kA\Delta T}{d}$

Twice the area offers twice as many paths, twice the difference pushes twice as hard, and twice the thickness halves the flow. The material enters through its thermal conductivity $k$, in W/(m·K): about 400 for copper, 16 for stainless steel, 0.7 for brick, 0.15 for wood, 0.04 for mineral wool and 0.025 for still air. Metals lead thanks to their free electrons; wool, foam and feathers insulate by trapping air in pockets too small for it to circulate.

Layers in series combine like resistors. Each square metre of a layer has a thermal resistance $R=d/k$, in m²·K/W, and builders sum them into a U-value:
$\frac{Q}{t}=UA\Delta T,U=\frac{1}{\sum (d/k)}$
A low $U$ means good insulation. For one layer $U=k/d$, so doubling the thickness halves the loss.

Worked example: the pot in the animation

The steel base has $k=16W/(\mathrm{m}\cdot \mathrm{K})$, $A=0.03{\mathrm{m}}^2$ and $d=4\mathrm{mm}=0.004\mathrm{m}$, and the flame keeps its underside 15 K hotter than its top, so
$\frac{Q}{t}=\frac{16\cdot 0.03\cdot 15}{0.004}=1800\mathrm{W}$

Warming 2 kg of water, with $c=4180J/(\mathrm{kg}\cdot \mathrm{K})$, from 20 °C to 80 °C needs
$Q=mc\Delta T=2\cdot 4180\cdot 60=501600\mathrm{J}\approx 502\mathrm{kJ}$
so, with no losses, it takes $t=501600/1800\approx 279\mathrm{s}$, about 4.6 minutes.

These are the final values on the readouts: $t$ reaches 4.6 min, the water 80 °C and $Q$ 502 kJ, time-lapsed into eight seconds. The gold arrows mark the steady flow up through the base, drawn far thicker than 4 mm so that its jiggling atoms show. Steel is a mediocre conductor, but 4 mm is so thin that 15 K drives 1800 W through it. A wooden base, with $k=0.15W/(\mathrm{m}\cdot \mathrm{K})$, would pass about a hundred times less.

Convection: the loops in the pot

Water touching the hot base expands slightly and becomes less dense than the cooler water around it, as the density label in the scene states, so buoyancy lifts it and denser water sinks to take its place. The dashed green streamlines close into two loops that rise together above the flame and sink along the walls; each parcel turns red on its way up and blue on its way down. This is natural convection; with a fan or a pump it becomes forced convection.

No simple formula covers a whole flow, so engineers write the heat exchanged between a surface and a fluid as
$\frac{Q}{t}=hA\Delta T$
with $\Delta T$ between the surface and the distant fluid, and a measured coefficient $h$: about 10 W/(m²·K) for still air, around 100 for air from a fan and tens of thousands for boiling water. This is Newton's law of cooling; wind chill is the same law, with the wind raising $h$.

Radiation: the fourth power of temperature

Every surface radiates, near room temperature mostly infrared around 10 µm. The Stefan–Boltzmann law gives the power from an area $A$ at absolute temperature $T$:
$P=\epsilon \sigma A{T}^4$
with $\sigma =5.67\times 10^{-8}W/({\mathrm{m}}^2\cdot {\mathrm{K}}^4)$ and an emissivity $\epsilon$ from 0 (a perfect mirror) to 1 (a black body). Heating a plate from 27 °C to 327 °C (300 K to 600 K) doubles $T$ and multiplies $P$ by $2^4=16$.

A body also absorbs radiation from its surroundings, and a good emitter is an equally good absorber. With surroundings at $T_0$ the net loss is
$P_{\mathrm{net}}=\epsilon \sigma A(T^4-T_0^4)$
Bare skin at 306 K (33 °C), with $\epsilon \approx 0.97$ and about 1.5 m² radiating, in a room with walls at 293 K loses $0.97\cdot 5.67\times 10^{-8}\cdot 1.5\cdot (306^4-293^4)\approx 115\mathrm{W}$, more than the 100 W or so a resting body produces. The purple waves carry energy sideways from the flame, where conduction and convection send almost none.

Worked example: a single-glazed window

A 1.5 m² window has one pane of glass 4 mm thick, with $k=1.0W/(\mathrm{m}\cdot \mathrm{K})$, between a room at 20 °C and the outside at 0 °C. For the glass alone, Fourier's law gives $1.0\cdot 1.5\cdot 20/0.004=7500\mathrm{W}$, the output of several heaters. Real windows lose far less: a film of slow air clings to each face and adds a resistance $1/h$, with $h\approx 8W/({\mathrm{m}}^2\cdot \mathrm{K})$ indoors (half convection, half radiation) and $h\approx 25W/({\mathrm{m}}^2\cdot \mathrm{K})$ outdoors in a light wind.

Per square metre, the three resistances add up:
$R=\frac{1}{8}+\frac{0.004}{1.0}+\frac{1}{25}=0.125+0.004+0.040=0.169{\mathrm{m}}^2\cdot K/W$
$\frac{Q}{t}=\frac{A\Delta T}{R}=\frac{1.5\cdot 20}{0.169}\approx 178\mathrm{W},U=\frac{1}{R}\approx 5.9W/({\mathrm{m}}^2\cdot \mathrm{K})$

The glass supplies only about 2 % of the resistance; the air films do the insulating. The flux is $20/0.169\approx 118W/{\mathrm{m}}^2$, so the indoor film takes $118\cdot 0.125\approx 15\mathrm{K}$ and the inner face of the glass sits near 5 °C, below the dew point of room air at 50 % humidity, about 9 °C: single panes run with condensation. Double glazing with $U\approx 1.2W/({\mathrm{m}}^2\cdot \mathrm{K})$ would lose about 36 W here, and the brick-and-wool wall of the expert view only about 11 W.

Common mistakes

• "Heat rises" is half true: warm fluid rises, but heat itself spreads in every direction by conduction and radiation, so a ceiling heater warms people 2 m below it.
• Insulation does not stop heat; it slows it. Even 10 cm of mineral wool, with $U=0.04/0.1=0.4W/({\mathrm{m}}^2\cdot \mathrm{K})$, passes 8 W per square metre across 20 K.
• Metal and wood at room temperature are equally warm. Steel feels colder because it conducts about a hundred times better (16 against 0.15 W/(m·K)), so it draws heat from your hand faster.
• Never put degrees Celsius into $T^4$. From 27 °C to 327 °C the radiated power rises 16 times (300 K to 600 K), not 12 times, let alone 21 000.
• The difference $T^4-T_0^4$ is not $(T-T_0{)}^4$. For 306 K and 293 K the first gives about 79 W per square metre of black surface, the second less than 0.002 W.

Where you meet it

Every heating bill is a sum of $UA\Delta T$ terms over walls, roof and windows. A vacuum flask defeats all three mechanisms: the vacuum stops conduction and convection, and silvering stops most radiation. Weather is convection on a grand scale, and the Earth's temperature balances absorbed sunlight against infrared radiated to space. Next, Kinetic Theory of Gases shows conduction molecule by molecule, Blackbody Radiation and the Quantum explains where $\sigma T^4$ comes from, and The Second Law of Thermodynamics and Entropy explains why heat flows on its own only from hot to cold.

Advanced

Heat flux and the temperature gradient

The slab law $Q/t=kA\Delta T/d$ covers only a flat layer whose temperatures have settled. To treat bodies of any shape, and temperatures that change in time, we need a law that holds at every point. This section sets it up.

At each point we use two vectors. The heat flux density $\vec{q}$ points the way energy flows; its size is the energy crossing unit area, set square to the flow, per unit time, in W/m². The temperature gradient $\nabla T$ points the way temperature rises fastest; its size is that rise per metre, in K/m. Fourier's law in local form says that heat flows straight down the gradient, at a rate set by the conductivity $k$:
$\vec{q}=-k\nabla T$
The minus sign makes heat run from hot to cold, against the gradient.

For a flat layer the gradient is simply $\Delta T/d$, and the local law gives back the slab law. In the animation's pot the steel base is 4 mm thick, with $k=16W/(\mathrm{m}\cdot \mathrm{K})$, and its underside is 15 K hotter than its top. So $|\nabla T|=15/0.004=3750K/m$ and $q=16\cdot 3750=60000W/{\mathrm{m}}^2$; over the base's 0.03 m² that is the 1800 W of the slab law.

The law has the form of Ohm's law written at a point, $\vec{J}=-{\sigma }_{\mathrm{el}}\nabla V$: temperature plays the potential, $\vec{q}$ the current density and $k$ the electrical conductivity ${\sigma }_{\mathrm{el}}$. Like Ohm's law, it is a material law that holds for gentle gradients, not a fundamental principle. In fibrous or layered materials $k$ depends on direction: wood conducts heat about twice as well along the grain as across it, so there $\vec{q}$ need not point exactly down the gradient.

Deriving the heat equation

Fourier's law gives the flow at one instant. To predict how a temperature profile changes in time, we combine it with conservation of energy. The result, the heat equation, governs every problem in the rest of this lesson.

Legend

$x$	Position along the rod.	$\mathrm{m}$
$T$	Temperature profile along the rod at one moment; its tangents are dashed.	$\mathrm{K}$
$dx$	Thickness of the thin slice.	$\mathrm{m}$
$q(x)$	Heat flux density entering the slice through its left face.	$W/{\mathrm{m}}^2$
$q(x+dx)$	Heat flux density leaving through its right face; smaller than $q(x)$.	$W/{\mathrm{m}}^2$
$\rho c$	Heat capacity per unit volume: density times specific heat capacity.	$J/({\mathrm{m}}^3\cdot \mathrm{K})$
$\partial T/\partial t$	Rate at which the slice warms.	$K/s$

Figure 1 shows a rod in which heat flows along $x$ only, and a thin slice of it, of thickness $dx$. Above the rod is the temperature profile $T(x)$ at one moment, with dashed tangents at the two faces of the slice. Look at the tangents: the left one is steeper than the right one, because the profile is convex (it curves upwards). The flux at each face is set by the local slope, $q=-k\partial T/\partial x$, so the arrow $q(x)$ entering on the left is longer than the arrow $q(x+dx)$ leaving on the right. More heat enters than leaves, and the surplus warms the slice at the rate $\partial T/\partial t$ of the green arrow.

Now count the energy, per unit area of the rod's cross-section. Each second the slice gains $q(x)-q(x+dx)$. For a thin slice this difference is the rate of change of $q$ along the rod times the thickness, $-(\partial q/\partial x)dx$. The slice has volume $dx$ per unit area, and $\rho c$ is the heat capacity per unit volume (density times specific heat capacity), so warming it at the rate $\partial T/\partial t$ takes $\rho cdx\partial T/\partial t$. Setting gain equal to storage and cancelling $dx$:
$\rho c\frac{\partial T}{\partial t}=-\frac{\partial q}{\partial x}$

Now substitute Fourier's law, $q=-k\partial T/\partial x$. With $k$ constant, $\partial q/\partial x=-k{\partial }^{2}T/\partial x^2$, and dividing by $\rho c$ gives the heat equation:
$\frac{\partial T}{\partial t}=\alpha \frac{{\partial }^{2}T}{\partial x^2},\alpha =\frac{k}{\rho c}$

The same count works for a small box in three dimensions. The net heat flowing out of it per unit volume is the divergence $\nabla \cdot \vec{q}$, so $\rho c\partial T/\partial t=-\nabla \cdot \vec{q}$, and ${\partial }^{2}T/\partial x^2$ becomes the Laplacian ${\nabla }^{2}T$. The constant $\alpha$ is the thermal diffusivity, in m²/s. It grows with $k$, which moves heat faster, and falls with $\rho c$, which demands more heat for each kelvin.

The equation has a simple reading. The second derivative ${\partial }^{2}T/\partial x^2$ is the curvature of the profile: the change of slope between the two faces of the slice in Figure 1. Where the profile curves upwards, a point is colder than the average of its neighbours, and it warms; where it curves downwards, the point is warmer than its neighbours, and it cools. So every bump in a temperature profile flattens out in time.

How far and how fast heat spreads

Before solving anything, the heat equation already tells us the time scales of a problem. This is often all an engineer needs.

The equation contains a single constant, $\alpha$, in m²/s. The only length that can be built from $\alpha$ and a time $t$ is $\sqrt{\alpha t}$, so after a time $t$ a disturbance has spread a distance of order $\sqrt{\alpha t}$. Turned round, a layer of thickness $d$ settles to its steady state in a time of order $d^2/\alpha$. The time grows with the square of the thickness: twice as thick, four times as slow.

Three examples. The pot's steel base, with density 7900 kg/m³ and specific heat capacity 500 J/(kg·K), has $\alpha =16/(7900\cdot 500)\approx 4.1\times 10^{-6}{\mathrm{m}}^2/s$ and settles in about $0.004^2/(4.1\times 10^{-6})\approx 4\mathrm{s}$. That is tiny next to the 279 s warm-up, so using the steady slab law for the pot was justified. Twenty centimetres of brick, with $\alpha \approx 4.6\times 10^{-7}{\mathrm{m}}^2/s$, take about 24 h. The 6.7 cm of water in the pot would need some 9 hours to warm through by conduction alone; convection does it in minutes.

The wall in the graph

The Advanced view of the animation shows a wall warming up. Following it shows the heat equation at work: heat creeping in, energy being stored, and a final steady state.

Legend

$x$	Distance from the inner (room-side) face.	$\mathrm{m}$
$T$	Temperature in the wall six hours after the start.	$\mathrm{K}$
$k_1$	Thermal conductivity of the brick, the inner layer.	$W/(\mathrm{m}\cdot \mathrm{K})$
$k_2$	Thermal conductivity of the mineral wool, the outer layer.	$W/(\mathrm{m}\cdot \mathrm{K})$
$\sqrt{\alpha t}$	How far the warming has crept into the brick after time $t$; there the profile stands near half its surface value.	$\mathrm{m}$
$E$	Energy stored in the brick per unit area of wall: $\rho c$ times the shaded area.	$J/{\mathrm{m}}^2$
$T_{\infty }$	Steady-state profile that the wall approaches, straight within each layer.	$\mathrm{K}$

Figure 2 shows the wall six hours after the start. It is 20 cm of brick ($k_1=0.7W/(\mathrm{m}\cdot \mathrm{K})$) on the room side and 10 cm of mineral wool ($k_2=0.04W/(\mathrm{m}\cdot \mathrm{K})$) outside, drawn as bars under the axis $x$, to the same scale. Everything was at 0 °C until the inner face, at $x=0$, was held at 20 °C; the outer face stays at 0 °C.

Read the red profile $T(x)$ from left to right. It falls steeply near the inner face and has barely reached the wool: the warming has crept only about $\sqrt{\alpha t}$ into the brick, the distance marked under the axis. After $t=6\mathrm{h}=21600\mathrm{s}$ this is $\sqrt{4.6\times 10^{-7}\cdot 21600}\approx 0.10\mathrm{m}$. At that depth, marked by the dot, the profile stands at about 10 °C, near half its value at the face.

The shaded area between the profile and the initial 0 °C line measures stored energy: multiplied by $\rho c$, it is the heat $E$ taken up by the brick per square metre of wall. The wool's share is tiny, because its $\rho c$ is 60 times smaller than the brick's. All this energy came in through the inner face and has not left through the outer one, so $E$ is, very nearly, the time integral of $q(0)-q(d)$: the flux in at the inner face minus the flux out at the outer face, where $x=d$.

The animation's readouts show these two fluxes. The inner flux $q(0)$ starts huge and falls roughly as $1/\sqrt{t}$, like the flux into a very thick block of brick, $\Delta T\sqrt{k\rho c/(\pi t)}$, which is about 190 W/m² after one hour. The outer flux $q(d)$ stays at zero until the warming reaches it. At the six hours of Figure 2 they are about 76 and 2.3 W/m²: most of what enters is still charging the brick. Both approach 7.2 W/m², and they are within 10 % of each other after 48 hours and within 1 % after 72.

The dashed grey line $T_{\infty }$ is the steady state the wall ends up in. There $\partial T/\partial t=0$, so the heat equation says ${\partial }^{2}T/\partial x^2=0$: the profile is straight within each layer. In the steady state no energy piles up anywhere, so the same flux $q$ crosses both layers. Each layer's slope is therefore $q/k$, so the slopes are in the inverse ratio of the conductivities, and the line through the wool is $k_1/k_2=17.5$ times steeper.

The size of that common flux follows from adding the two layers' resistances $d/k$ per square metre, with $d_1=0.20\mathrm{m}$ of brick and $d_2=0.10\mathrm{m}$ of wool:
$q=\frac{\Delta T}{d_1/k_1+d_2/k_2}=\frac{20}{0.286+2.5}\approx 7.2W/{\mathrm{m}}^2,U\approx 0.36W/({\mathrm{m}}^2\cdot \mathrm{K})$
The drop across each layer is $q$ times its resistance. The wool's resistance is about nine times the brick's, so it takes 17.9 K of the 20 K and the brick only 2.1 K.

Modes of decay

How quickly does the wall forget its start and settle on $T_{\infty }$? The answer comes from splitting the profile into simple shapes that each fade at their own rate.

The heat equation is linear: the sum of two solutions is again a solution. So any departure from the steady state can be written as a sum of sine-shaped modes, and each one decays on its own, keeping its shape. Put a mode $\sin (2\pi x/\lambda )$ of wavelength $\lambda$ into the heat equation: two $x$-derivatives multiply it by $-(2\pi /\lambda {)}^2$, so its amplitude must fall as $e^{-t/\tau }$, with
$\tau =\frac{{\lambda }^2}{4{\pi }^2\alpha }$
Because the decay time goes as ${\lambda }^2$, fine wiggles vanish almost at once and only the broadest mode lingers. Joseph Fourier developed his series, sums of sines, for exactly this problem, in his theory of heat of 1822.

In the wall, the slowest mode has to fit the brick's two faces. At the inner face the temperature is held, so the departure from the steady state is zero there. At the outer face the wool nearly stops the flow, so the slope of the departure is nearly zero. The broadest sine with a zero at one end and a flat top at the other is a quarter wave, $\lambda =4d$, which gives $\tau =4d^2/({\pi }^2\alpha )\approx 9.7\mathrm{h}$. The wool does leak a little, so the real decay is slightly quicker: the gap between the two readouts shrinks by a factor $e$ about every 9 hours, from under 10 % at 48 h to under 1 % at 72 h.

Temperature waves

When the surface temperature rises and falls every day, or every year, the heat equation turns the swing into a wave that fades as it travels inward. This is why thick walls and deep cellars stay at nearly constant temperature.

Legend

$t$	Time.	$\mathrm{s}$
$T$	Temperature, measured from its mean.	$\mathrm{K}$
$T_0$	Surface temperature, swinging with angular frequency $\omega$.	$\mathrm{K}$
$T_x$	Temperature at depth $x$ below the surface (here 20 cm).	$\mathrm{K}$
$A$	Swing of the surface temperature about its mean.	$\mathrm{K}$
$A_x$	Swing at depth $x$: $A{e}^{-x/\delta }$, with $\delta =\sqrt{2\alpha /\omega }$.	$\mathrm{K}$
$\Delta t$	How much later the peak arrives at depth $x$ than at the surface: $x/(\delta \omega )$.	$\mathrm{s}$

Figure 3 plots two temperatures against time $t$, each measured from its mean. The large curve $T_0$ is the surface, swinging with angular frequency $\omega$ and amplitude $A$. The small curve $T_x$ is the temperature at a depth $x$ inside. Notice two things: $T_x$ swings much less, by only $A_x$, and its peaks come later, by $\Delta t$.

Both effects come from one solution of the heat equation, a damped wave travelling inward:
$T(x,t)-T_{\mathrm{mean}}=A{e}^{-x/\delta }\cos (\omega t-x/\delta ),\delta =\sqrt{2\alpha /\omega }$
To see why $\delta$ has this value, write the wave as the real part of $A{e}^{i\omega t}{e}^{-(1+i)x/\delta }$. A time derivative multiplies it by $i\omega$; two space derivatives multiply it by $(1+i{)}^2/{\delta }^2=2i/{\delta }^2$. The heat equation then demands $i\omega =2i\alpha /{\delta }^2$, which is $\delta =\sqrt{2\alpha /\omega }$.

Read the solution term by term. The amplitude falls by a factor $e$ every distance $\delta$, so at depth $x$ the swing is $A_x=A{e}^{-x/\delta }$. The phase $x/\delta$ subtracted from $\omega t$ means that each peak arrives later than at the surface by $\Delta t=x/(\delta \omega )$. Slow cycles reach deeper, because $\delta$ grows as $1/\sqrt{\omega }$.

For brick and the daily cycle, $\omega =2\pi /(24\mathrm{h})$, the depth $\delta$ is about 11 cm. The figure is drawn for $x=20\mathrm{cm}$, where $x/\delta \approx 1.8$: 20 cm into thick brickwork the swing is cut to about a sixth, $e^{-1.8}\approx 0.17$, and arrives some 7 hours late. In soil, with $\alpha \approx 5\times 10^{-7}{\mathrm{m}}^2/s$, the yearly wave has $\delta \approx 2.2\mathrm{m}$: a cellar 3 m down sees a quarter of the surface swing, about two and a half months late.

Steady states and Laplace's equation

Once nothing changes in time, heat flow becomes the same mathematical problem as electrostatics, so every solution found there can be reused.

In a steady state $\partial T/\partial t=0$, and the heat equation reduces to ${\nabla }^{2}T=0$. This is Laplace's equation, which the electrostatic potential obeys in empty space. Every electrostatic solution therefore describes a heat flow too, with temperature in place of the potential and the flux $\vec{q}$ in place of the field.

Take a small source of power $W$ deep in a large block. In the steady state all of $W$ crosses every sphere of radius $r$ around it, so $q=W/(4\pi r^2)$, just like the field of a point charge. With $q=-kdT/dr$, integrating inward from far away, where the block is undisturbed, gives a temperature rise of $W/(4\pi kr)$. For 10 W released deep in soil with $k=1W/(\mathrm{m}\cdot \mathrm{K})$, the soil 0.5 m away is warmer by $10/(4\pi \cdot 1\cdot 0.5)\approx 1.6\mathrm{K}$.

Worked problem: insulation inside or outside?

Does it matter which side of a wall the insulation goes on? The steady heat loss says no; the stored heat and the warm-up time say yes, very much.

Legend

$x$	Distance from the room-side face.	$\mathrm{m}$
$T$	Temperature in the wall in the steady state.	$\mathrm{K}$
$k_1$	Thermal conductivity of the brick.	$W/(\mathrm{m}\cdot \mathrm{K})$
$k_2$	Thermal conductivity of the mineral wool.	$W/(\mathrm{m}\cdot \mathrm{K})$
$T_m$	Temperature at the interface between the two layers.	$\mathrm{K}$
$\Delta T$	Total temperature difference across the wall; the same in both graphs.	$\mathrm{K}$
$E$	Energy stored in the brick per unit area of wall: $\rho c$ times the shaded area.	$J/{\mathrm{m}}^2$

Figure 4 compares the two arrangements in the steady state, with the room on the left at $x=0$ and the same total difference $\Delta T=20\mathrm{K}$ across the wall. The upper graph is the animation's wall: brick ($k_1$) on the room side, wool ($k_2$) outside. The lower graph swaps the layers: wool on the room side, brick outside. Compare where the interface temperature $T_m$ sits, and how large the shaded area under the brick's profile is.

Step 1, the steady loss. Series resistances add in any order, so the total is still $0.286+2.5=2.786{\mathrm{m}}^2\cdot K/W$ and the loss stays $q=7.2W/{\mathrm{m}}^2$.

Step 2, the profile. The drop across each layer is $qd/k$, so it travels with the layer. With the wool outside (upper graph), the brick runs from 20 °C down to $T_m\approx 17.9{}^{\circ }\mathrm{C}$. With the wool inside (lower graph), the wool takes its 17.9 K first, so $T_m=20-7.18\cdot 2.5\approx 2.1{}^{\circ }\mathrm{C}$, and the brick spans only 2.1 °C to 0 °C.

Step 3, the stored heat. Per square metre the brick stores $\rho cd$ times its mean temperature rise above 0 °C, with $\rho c=1.51\times 10^{6}J/({\mathrm{m}}^3\cdot \mathrm{K})$; this is the energy $E$, $\rho c$ times the shaded area in Figure 4. With the wool outside the brick averages 19.0 °C and holds $1.51\times 10^6\cdot 0.20\cdot 19.0\approx 5.7\mathrm{MJ}$; with the wool inside it averages about 1 °C and holds only 0.31 MJ. Adding the wool's few hundredths of a megajoule, the totals are 5.8 and 0.34 MJ per square metre, a factor of 17. With the wool outside, the room must supply those 5.8 MJ while the wall warms, on top of what leaves through the outer face: as much as 9 days of the steady loss.

Step 4, the timing. With the wool inside, the layer facing the room is light, with $\alpha \approx 1.6\times 10^{-6}{\mathrm{m}}^2/s$, and settles in about $d^2/\alpha =0.10^2/(1.6\times 10^{-6})\approx 1.75\mathrm{h}$. After two hours the room side loses 7.8 W/m², within 10 % of its final value. With the wool outside, $q(0)$ needs two days to come that close.

The verdict. Inside insulation suits rooms heated now and then, which then warm up within hours. Outside insulation turns the brick into a warm store that smooths out daily swings. It also keeps the brick warm: behind inside insulation the brick sits near 0 to 2 °C, where moisture from the room can condense in the wall unless a vapour barrier stops it.

Where the coefficients come from

So far the conductivity $k$, the convective coefficient and the radiative exchange have been taken as given. Simple models explain their sizes.

Gases. Kinetic theory gives $k\approx \frac{1}{3}n\bar{v}\ell c_1$: there are $n$ molecules per unit volume, moving at a mean speed $\bar{v}$, and each carries a heat capacity $c_1$ over one mean free path $\ell$ before it collides. Since $\ell \propto 1/n$, the density cancels: a gas conducts as well at half the pressure. This holds until $\ell$ becomes comparable with the container. In air at 1 atm $\ell \approx 70\mathrm{nm}$, so a vacuum flask needs a pressure below about 1 Pa before $\ell$ exceeds its gap of a few millimetres and conduction collapses.

Metals. Free electrons carry the heat as well as the charge, which is why good electrical conductors also conduct heat well.

Convection. A fluid layer of depth $d$ heated from below starts to circulate when its Rayleigh number, $g\beta \Delta T{d}^3/(\nu \alpha )$, exceeds a critical value: about 1700 between two solid plates, and about 1100 when the top is a free surface, as in the pot. Here $g$ is the gravitational acceleration, $\beta$ the fluid's expansion coefficient and $\nu$ its kinematic viscosity. The number compares buoyancy, which drives the flow, with viscosity and conduction, which damp it. For the 6.7 cm of water in the pot, less than a thousandth of a kelvin across the layer is enough.

Radiation. For small differences $T^4-T_0^4\approx 4T_0^3\Delta T$, so radiation also behaves like a coefficient: a nearly black surface near room temperature exchanges about $4\sigma T_0^3\approx 6W/({\mathrm{m}}^2\cdot \mathrm{K})$, as much as natural convection in air.

Limits of validity

Fourier's law assumes that the carriers of heat (molecules, lattice vibrations or electrons) collide many times over the distances and times of interest. It fails in structures comparable with their mean free path, which is a few hundred nanometres for lattice vibrations in silicon at room temperature. In transistors tens of nanometres across, heat flows partly ballistically, without collisions, and the effective $k$ falls well below the bulk value. The heat equation also lets a sudden heating be felt at once everywhere, however faintly; this fails on time scales of picoseconds, before the carriers have collided.

Conductivity varies with temperature, which makes the equation non-linear over wide ranges. The Stefan–Boltzmann law assumes grey surfaces much larger than the thermal wavelength, about 10 µm at room temperature. Across much narrower gaps, near-field radiation can exceed the black-body limit many times over.

Where this leads

Molecular diffusion obeys the same equation as heat, Electric Current and Resistance is the electrical twin of conduction, and steady heat flow repeats Electric Field and Potential. Kinetic Theory of Gases supplies the mean free path behind $k$, and Blackbody Radiation and the Quantum derives $\sigma$ from Planck's law. The Second Law of Thermodynamics and Entropy shows that heat flowing down a temperature difference produces entropy, about $7.2(1/273-1/293)\approx 1.8\times 10^{-3}W/({\mathrm{m}}^2\cdot \mathrm{K})$ in the wall, and Schrödinger's Equation and Tunnelling meets a diffusion equation with an imaginary coefficient.