In short

Conduction is the transfer of thermal energy through a material without any bulk motion of the material itself. Hot molecules jostle their cooler neighbours, handing over a little kinetic energy at each collision; the energy diffuses from the hot side to the cold side.

For a slab of material of thickness L and cross-sectional area A, with temperatures T_1 and T_2 on its two faces (T_1 > T_2) held in steady state, the rate of heat flow is

\boxed{\;\frac{Q}{t} = kA\,\frac{T_1 - T_2}{L} = kA\,\frac{\Delta T}{L}\;}

where k is the thermal conductivity of the material (in W/(m·K)). This is Fourier's law of heat conduction. Metals have large k (copper \approx 400, aluminium \approx 235), insulators have small k (air \approx 0.025, wood \approx 0.15, brick \approx 0.8, water \approx 0.6, all in W/(m·K)).

Rewriting Fourier's law as

\frac{Q}{t} = \frac{\Delta T}{R}, \qquad R = \frac{L}{kA},

exposes a clean electrical analogy: \Delta T acts like voltage, Q/t like current, and R = L/(kA) is the thermal resistance. Resistances in series add; resistances in parallel combine reciprocally. Composite walls — a brick wall backed by insulation, a double-glazed window, a Sintex water tank's two-layer plastic — are all series problems solved with this analogy.

Pour boiling chai into a brass lota and pick it up three seconds later. Your fingers jump back before your brain has finished registering what happened — the metal is already too hot to hold. Pour the same chai into an unglazed clay kulhar from a railway-station stall and you can cradle it in your palms for the full sipping ceremony without discomfort, even though the chai inside both vessels is at roughly the same temperature. Why the difference? The brass is not hotter than the clay; both vessels have come to thermal equilibrium with the tea. The difference is not in how much thermal energy each wall holds, but in how fast each wall ships it through.

This is the phenomenon physicists call conduction — the flow of heat through a stationary medium, driven by a temperature difference. It is the quietest and most ubiquitous kind of heat transfer. It is what cooks the bottom of a chapati when it touches a hot tawa, what melts the frost on a winter Shimla windowpane from the warm room-side outward, what fails when a thermos flask loses its vacuum and lets your tea go cold. Every time you wrap a hot vessel with a cloth, swaddle pipes to stop them freezing, or paint a Sintex tank's walls white so the sun warms it less, you are making a conduction decision.

This article builds conduction from the molecular picture up, derives Fourier's law as a limit of Newton's heat-flow intuition made precise, introduces the idea of thermal resistance and its electrical analogy (so powerful it lets you solve composite-wall problems almost by inspection), and closes with how insulation, good conductors, and everyday materials sit on the conductivity spectrum.

What is actually moving

A brass mug sitting on a hot tawa is not moving as a whole. Its molecules are not migrating from the hot bottom to the cooler top. Yet heat is flowing upward through the brass at a furious rate. What physically is flowing?

Not matter. What's flowing is kinetic energy, handed from one molecule to the next by local collisions. Here is the picture. In the region of the brass touching the tawa, the metal atoms vibrate about their lattice positions with large amplitudes — that is what makes this region hot. In the region near the top, atoms vibrate less — that region is cooler. At the interface between hot and cold regions, energetic atoms collide (via the springs of atomic bonds, which is what binds the lattice) with less energetic atoms, transferring some of their motion. The collision propagates: the newly-energized atoms now have more motion to share with their still-cooler neighbours. Wait long enough and this chain-reaction of jostling carries thermal energy from bottom to top through the metal, without any atom actually having to traverse the mug.

Metals have a second, faster channel: free electrons. In a metal, a sea of electrons roams independently of the lattice. These electrons pick up energy from hot atoms (fast vibrations deflect them) and deliver it to cooler atoms (they bump and dump energy). Because electrons are light and extremely fast, they can shuttle energy across many lattice spacings between collisions, making metals much better conductors than non-metallic solids. The brass lota burns you because the free electrons inside have already carried the heat up through the wall while you were still thinking about picking it up. The clay kulhar has no free electrons; energy moves only through the slower lattice vibrations, and much slower still because the clay has air-filled pores that interrupt the path entirely. By the time a noticeable amount of heat has reached your fingers through the kulhar's wall, you have already drunk the chai.

That is the mechanism of conduction. The question now is: how much heat, how fast, given a temperature difference and a material?

Molecular mechanism of heat conduction through a solidA horizontal rectangular slab represents a solid. On the left end, a red label reads hot, T1, with dense wavy marks indicating large molecular vibration. On the right end, a blue label reads cold, T2, with small wavy marks indicating small vibration. Inside the slab, rows of dots represent atoms arranged in a lattice, with vibration amplitudes decreasing gradually from left to right. Arrows between neighbouring atoms indicate energy being handed from atom to atom. A caption arrow above shows heat Q flowing from left to right.hot, T₁cold, T₂solid slab of length LQ / t = heat flowatoms near the hot end vibrate more; they pass energy to their neighbours
A slab of solid with a temperature gradient. Atoms on the hot side (left) vibrate vigorously; those on the cold side (right) vibrate gently. At the boundary between any two neighbouring atoms, the more energetic one gives up a little energy to the less energetic one. The cumulative effect is a steady flow of thermal energy from hot to cold.

Fourier's law

Take a slab of material — a brick wall of a Rajasthan summer house, say — of thickness L and cross-sectional area A. Hold one face at temperature T_1 (the hot outside, 45 °C in May) and the other face at T_2 < T_1 (the cool inside, 25 °C because of the thick walls and the trickle of evaporative cooling from a ghada). Wait long enough for the interior of the wall to reach a steady state — the temperature profile inside the wall no longer changes in time, even though energy is flowing continuously through it from hot to cold. In the steady state, the rate of heat flow through every cross-section of the wall is the same. (If it weren't, heat would be accumulating somewhere inside the wall and the temperature there would still be rising — not steady state.)

Empirically, the rate of heat flow Q/t (in joules per second, i.e. watts) obeys four simple rules:

  1. It is proportional to the area A. Doubling the wall area doubles the heat flow, other things equal.
  2. It is proportional to the temperature difference \Delta T = T_1 - T_2. Doubling \Delta T doubles the flow.
  3. It is inversely proportional to the thickness L. A wall twice as thick transmits half as much heat.
  4. It depends on the material. Copper transmits energy orders of magnitude faster than brick of the same dimensions.

Combining the three geometric dependences and packaging the material dependence into a single constant k gives Fourier's law:

\boxed{\;\frac{Q}{t} = kA\,\frac{T_1 - T_2}{L} = kA\,\frac{\Delta T}{L}\;}

The constant k is the thermal conductivity of the material. Its SI units are watts per metre per kelvin, W/(m·K), because (Q/t) is in watts, A in m², \Delta T in kelvins, and L in metres:

[k] = \frac{\text{W}}{\text{m}^2} \cdot \frac{\text{m}}{\text{K}} = \frac{\text{W}}{\text{m} \cdot \text{K}}.

Where Fourier's law comes from

Fourier's law is one of those miracles of nature where a simple empirical proportionality hides real microscopic complexity. Here is a qualitative chain of reasoning that produces it from the molecular picture.

Think of a gas in a box with a temperature gradient. At any cross-section, hotter molecules are drifting from the warm side to the cool side, and cooler molecules are drifting from the cool side to the warm side. Both drifts happen at the molecular speed (hundreds of m/s for air at room temperature), but the hotter molecules carry more kinetic energy than the cooler ones. So there is a net flux of kinetic energy across the cross-section — from hot to cold — even though there is no net flux of molecules. The magnitude of this energy flux turns out to be proportional to the temperature gradient \Delta T / L, with a coefficient that depends on molecular speed, mean free path, and how much energy each molecule carries — all encoded in the single number k.

For a solid, the carriers are lattice vibrations (and electrons in metals), but the structure of the argument is the same: the flux of energy is proportional to the gradient of temperature, with a proportionality constant that depends on how fast the energy carriers move, how far they travel between collisions, and how much energy each one carries. This is the qualitative content of Fourier's law.

For the real derivation — in differential form, \mathbf{q} = -k \nabla T — and the kinetic-theory microscopic expressions for k, see the going-deeper section below.

The minus sign and the gradient form

Fourier's law is often written more precisely in one-dimensional differential form:

\frac{Q}{t} = -kA\,\frac{dT}{dx},

where T(x) is the temperature profile along the direction of heat flow. The minus sign encodes the direction: heat flows in the direction of decreasing temperature, so if dT/dx > 0 (temperature increasing with x), then Q/t is in the -x direction.

For the slab with T_1 at x = 0 and T_2 at x = L, the steady-state temperature profile is a straight line: T(x) = T_1 - (T_1 - T_2)(x/L), so dT/dx = -(T_1 - T_2)/L. Plugging into the differential form:

\frac{Q}{t} = -kA \cdot \left(-\frac{T_1 - T_2}{L}\right) = kA\,\frac{T_1 - T_2}{L},

recovering the box formula. The minus sign disappears because "temperature drops across the slab" exactly cancels the minus in -k\, dT/dx.

Thermal conductivities — who sits where

The conductivity k spans about five orders of magnitude across common materials. A feel for the numbers is more useful than any single value.

Material k (W/(m·K)) Category
Silver 430 best known conductor
Copper 400 standard excellent conductor; cooking-vessel bases
Aluminium 235 light, cheap conductor; pressure-cooker bodies
Brass 120 lota, puja bell, lassi tumbler
Steel (mild) 50 tawa, kadhai, pressure-cooker lid
Stainless steel 16 steel tumbler, double-walled vacuum flasks
Glass 0.8 window pane
Brick 0.8 Jaipur haveli wall
Water 0.6 filling a vessel
Cotton cloth 0.04 insulating a pipe
Clay (earthenware) 0.5 kulhar, matka
Wood (teak) 0.17 door, handle, spoon
Fibreglass / rockwool 0.04 industrial insulation
Air (still) 0.025 between window panes, inside double-walled tank
Vacuum 0 (no conduction) between thermos walls

A few takeaways worth internalising:

The electrical analogy: thermal resistance

Fourier's law wants to be rewritten. Move the geometric factors L and A into a single quantity:

\boxed{\;R = \frac{L}{kA},\qquad \frac{Q}{t} = \frac{\Delta T}{R}.\;}

The quantity R is the thermal resistance of the slab, with units K/W (kelvins per watt). Its role is exactly the electrical role of R in Ohm's law I = V/R, with the following dictionary:

Electrical Thermal
Voltage V Temperature difference \Delta T
Current I Heat flow rate Q/t
Resistance R = \rho L / A Thermal resistance R = L/(kA)
Resistivity \rho 1/thermal-conductivity (1/k)

The analogy is not a loose metaphor. The mathematics of steady-state heat flow in a network of slabs is identical to the mathematics of steady-state current in a network of resistors. Every trick you know from circuit analysis carries over.

Resistances in series

Two slabs stacked face-to-face, with the same heat flowing through both (because nothing else can enter or leave at their junction), add their resistances:

R_\text{series} = R_1 + R_2.

Why: in steady state, the heat flux Q/t is the same through both slabs (conservation of energy — nothing accumulates at the interface). The temperature drops across the two slabs sum to the total drop: \Delta T = \Delta T_1 + \Delta T_2. Each drop is \Delta T_i = (Q/t)\,R_i. Summing gives \Delta T = (Q/t)(R_1 + R_2), so the effective resistance is R_1 + R_2. This is identical to series resistors in a circuit.

Resistances in parallel

Two slabs side-by-side, each with the same temperature difference across its thickness, share the total heat flow between them in proportion to their conductance:

\frac{1}{R_\text{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}.

Why: in parallel, \Delta T is the same across both slabs. The heat flow splits — (Q/t) = (Q/t)_1 + (Q/t)_2 = \Delta T/R_1 + \Delta T/R_2 = \Delta T\,(1/R_1 + 1/R_2). The effective 1/R is therefore the sum of the individual 1/R's.

Why this is powerful

A Sintex-style water tank has a two-layer plastic wall with a thin air gap between them. A traditional haveli wall in Jodhpur is thick sandstone, often with a plaster finish on the inside. A double-glazed window (Delhi airport) has two glass panes with an air gap between them. Each of these is a series stack of thermal resistances, and the total heat loss through the wall per unit area is limited by the total resistance of the stack — which is dominated by the largest single resistance. For the double-glazed window, the air gap is by far the largest R; almost all the temperature drop happens across the air, and the glass panes are almost at the same temperature as the rooms they face. This is what makes double-glazing effective.

Similarly, a composite wall of brick (R_\text{brick}) plus an insulation layer (R_\text{insul}) with R_\text{insul} \gg R_\text{brick} has total resistance close to R_\text{insul}. The brick becomes thermally invisible. A single expensive insulation layer replaces any amount of cheap brick.

Worked examples

Example 1: How fast does a brass lota lose heat?

A brass lota has a cylindrical wall of thickness L = 0.002 m (2 mm), a cross-sectional area for heat flow A = 0.025 m² (the outer surface of the wall), and contains hot chai at T_1 = 95\,^\circC. The surrounding air is at T_2 = 25\,^\circC. Treat the lota wall alone as the resistance (i.e., assume the outside of the wall is held at the air temperature). Thermal conductivity of brass, k = 120 W/(m·K). Find the rate of heat loss through the wall.

Cross-section of a brass lota wall conducting heatA vertical cross-section of a cylindrical lota shown as a thin-walled rectangle. Inside the vessel is hot chai at 95 degrees Celsius with a wavy red fill. The wall thickness is 2 mm, labelled L. Outside the wall is ambient air at 25 degrees Celsius with a thin blue label. An arrow indicates heat flow Q over t going outward through the wall. The wall cross-sectional area is 0.025 square metres.hot chaiT₁ = 95 °Coutside airT₂ = 25 °Coutside airT₂ = 25 °CL = 2 mmQ/tQ/t(A = 0.025 m², k = 120 W/m·K)
Cross-section of the lota wall. Hot chai inside at 95 °C; outside air at 25 °C. Heat flows outward through the 2 mm brass wall. The rate is set by Fourier's law with the given conductivity $k$ and geometry $L$, $A$.

Step 1. Write Fourier's law.

\frac{Q}{t} = kA\,\frac{T_1 - T_2}{L}.

Why: the slab form of Fourier's law applies because the wall is thin relative to the lota's radius, so we can treat it as a flat slab to a very good approximation.

Step 2. Substitute the numbers. \Delta T = 95 - 25 = 70 K.

\frac{Q}{t} = 120 \times 0.025 \times \frac{70}{0.002}.

Step 3. Evaluate.

\frac{Q}{t} = 120 \times 0.025 \times 35{,}000 = 3.0 \times 35{,}000 = 105{,}000 \text{ W}.

Why: the factor (T_1 - T_2)/L = 70/0.002 = 35{,}000 K/m is the temperature gradient across the wall — 35,000 kelvin per metre, which is enormous because the wall is so thin. Fourier's law just multiplies this gradient by kA.

Step 4. Reality check.

105{,}000 W is 105 kilowatts — the power of about thirty electric kettles. That seems absurd for a small chai vessel. The resolution: the number is the instantaneous rate of heat flow if the wall really were held at 95 °C inside and 25 °C outside. In practice, the air next to the outer wall acts as an insulator (it has k = 0.025, far smaller than the brass), so the actual temperature gradient inside the brass is much smaller than we computed — most of the temperature drop happens across the thin layer of still air just outside the lota. That stagnant-air layer plus radiation are what really limit the heat loss in real life; the brass wall itself is thermally almost transparent. Example 3 (below) will redo the calculation with the air layer in series, and the answer becomes physically sensible.

Result: Treating the brass wall alone, Q/t \approx 10^5 W. In reality, the still-air layer outside (and any hand holding the lota) adds much more thermal resistance in series, and the actual heat loss is on the order of tens of watts. The 2 mm of brass has a negligible resistance compared to the 1 mm of still air next to it.

What this shows: A 2 mm brass wall by itself is a thermal near-short-circuit. That is why the brass lota burns your hand: nothing stops the chai's 95 °C from reaching the outer surface. The clay kulhar's wall, in contrast, has a much lower k (\sim 0.5 W/(m·K), 240 times smaller than brass) and a greater thickness, so the outside of the kulhar stays much cooler even when the chai inside is at 95 °C.

Example 2: Composite wall — a Sintex-style water tank

A Sintex-style domestic water tank has a two-layer plastic wall with a thin air gap in between. The wall stack (from inside to outside) is: 4 mm plastic, then 10 mm air, then 4 mm plastic. Thermal conductivities: plastic, k_p = 0.20 W/(m·K); still air, k_a = 0.025 W/(m·K). The tank has cross-sectional area A = 1 m² for heat flow. On a Delhi May afternoon, water inside is at T_1 = 30\,^\circC; outside air is at T_2 = 45\,^\circC. Find (a) the rate of heat flow into the tank per square metre of wall, (b) the temperatures at the two plastic-air interfaces.

Cross-section of a Sintex-style water tank wall: plastic-air-plasticA horizontal cross-section of a composite wall showing three layers side by side: a plastic layer on the left, 4 millimetres thick, an air gap in the middle, 10 millimetres wide, and a plastic layer on the right, 4 millimetres thick. Inside temperature on the far left is 30 degrees Celsius, labelled T1. Outside temperature on the far right is 45 degrees Celsius, labelled T2. Interface temperatures labelled T_a and T_b are shown between the layers. Arrows show heat flowing from right to left because outside is hotter. A thermal circuit diagram below shows three resistors in series labelled R1 plastic, R2 air, R3 plastic, with a voltage source labelled delta T equals 15 kelvin.plastic4 mmair gap10 mmplastic4 mmwaterT₁ = 30°CoutsideT₂ = 45°CT_aT_bQ/tR₁R₂R₃0.0200.4000.020K/WK/WK/WΔT = 15 K acrossR_total = 0.44 K/W
Above: three-layer wall cross-section. Below: the equivalent thermal circuit — three resistances in series, driven by the 15 K temperature difference. The air gap is the dominant resistance.

Step 1. Compute each layer's thermal resistance using R = L/(kA) with A = 1 m².

Plastic layer: R_p = \dfrac{0.004}{0.20 \times 1} = 0.020 K/W.

Air layer: R_a = \dfrac{0.010}{0.025 \times 1} = 0.400 K/W.

Why: the air gap is 10 mm thick (not much) and air has tiny conductivity — this ratio of L/k is large, which is why the air gap carries most of the thermal resistance.

Step 2. Total resistance (three slabs in series).

R_\text{total} = R_p + R_a + R_p = 0.020 + 0.400 + 0.020 = 0.440 \text{ K/W}.

Why: heat flows through plastic, then air, then plastic — three resistors in series — so their resistances add directly.

Step 3. Heat flow rate.

Outside is hotter than inside (Delhi May), so heat flows from outside into the water: \Delta T = T_2 - T_1 = 45 - 30 = 15 K. By Fourier's law (electrical form):

\frac{Q}{t} = \frac{\Delta T}{R_\text{total}} = \frac{15}{0.440} \approx 34.1 \text{ W}.

So 34 watts are flowing continuously into the water per square metre of wall area.

Step 4. Interface temperatures.

Heat flows through the outer plastic first (coming from outside). The temperature drop across the outer plastic:

T_2 - T_b = (Q/t) \cdot R_p = 34.1 \times 0.020 = 0.68 \text{ K}.

So T_b = 45 - 0.68 \approx 44.32\,^\circC. Similarly, the inner plastic has the same resistance, and the same heat flow, so the temperature drop across the inner plastic is also 0.68 K, giving

T_a = T_1 + 0.68 = 30.68\,^\circ\text{C}.

The air gap accounts for the remaining drop: T_b - T_a = 44.32 - 30.68 = 13.64 K — almost the full 15 K.

Why: the two plastic layers together take up only 4.5% of the total resistance; the air gap takes up the remaining 91%, so it absorbs 91% of the temperature drop. Most of the thermal defence is done by the 10 mm of still air.

Result: The tank absorbs heat at 34 W/m². The two plastic layers are nearly thermal pass-throughs (each drops less than a kelvin); the air gap does almost all the insulating work.

What this shows: The air gap is the reason a Sintex tank stays cool enough to be useful in a Delhi summer. Without it (a single 8 mm plastic wall, R = 0.04 K/W), the heat flow would be 15/0.04 = 375 W/m², more than ten times larger. The air gap — 10 mm of nothing but still air — is worth eleven times its weight in plastic.

Common confusions

If you came here to understand how conduction works, use Fourier's law, solve composite-wall problems, you have what you need. What follows is for readers who want the general form of Fourier's law, its microscopic foundation, the transient (non-steady-state) extension, and why heat flow can be described as a diffusion process.

Fourier's law in three dimensions

In full vector form, Fourier's law is

\mathbf{q} = -k \nabla T,

where \mathbf{q} is the heat flux vector (W/m², pointing in the direction of heat flow), and \nabla T is the temperature gradient (pointing in the direction of increasing temperature). The minus sign flips this so the flow goes from hot to cold. For an isotropic material (same k in every direction), this is the law; for anisotropic materials (fibres, laminates, some crystals), k becomes a tensor.

The one-dimensional slab formula comes from integrating this across the slab thickness. The 3D form allows you to handle curved geometries (cylindrical pipes, spherical tanks) and irregular shapes (insulation around an engine, thermal design of a satellite).

Microscopic expression for k

From the kinetic theory of gases, a simple estimate for thermal conductivity of a gas is

k \approx \tfrac{1}{3}\, n\, \bar{v}\, \lambda\, c_v^\text{(per molecule)},

where n is the number density of molecules, \bar{v} is the mean molecular speed (from \tfrac{1}{2} m \bar{v}^2 \sim k_B T), \lambda is the mean free path (the average distance between collisions), and c_v^\text{(per molecule)} is the heat capacity per molecule. The argument: each molecule, moving at \bar{v} for a distance \lambda before colliding, carries thermal energy proportional to its heat capacity and the temperature at its starting point. Averaging the net flux across a cross-section gives the formula.

For gases, this predicts k roughly independent of pressure (because n \propto P, but \lambda \propto 1/P, and they cancel) — a counterintuitive prediction that is experimentally confirmed down to quite low pressures. Only in a high vacuum, where \lambda becomes comparable to the container size, does conduction finally fail. This is the principle exploited in a thermos flask: evacuate the space between the walls until \lambda exceeds the gap, and conduction is suppressed.

For solids, the analog is Debye's theory of phonons (quantized lattice vibrations), with \bar{v} \to v_\text{sound} and \lambda the phonon mean free path. For metals, add an electron-gas contribution; the Wiedemann-Franz law says that the electronic contribution to k is proportional to the electrical conductivity \sigma:

\frac{k_\text{elec}}{\sigma T} = \mathcal{L} = \frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2 \approx 2.44 \times 10^{-8} \text{ W·Ω/K}^2,

known as the Lorenz number. The same electron sea that makes a metal conduct electricity well also makes it conduct heat well, in a precisely predictable ratio. This is why the list of "good electrical conductors" almost exactly matches the list of "good thermal conductors."

The heat equation — transient conduction

Fourier's law is a local statement: the flux at a point is proportional to the local gradient. Combining it with conservation of energy (rate of heat entering a small volume = rate of change of internal energy in that volume) gives the heat equation:

\frac{\partial T}{\partial t} = \alpha\,\nabla^2 T,

where \alpha = k / (\rho c_p) is the thermal diffusivity (in m²/s). This is the equation that governs how temperature profiles evolve in time — how a slug of hot material cools, how a heated end of a rod warms up the rest, how the daily temperature swing outside a house penetrates into its walls over the course of a day.

Thermal diffusivity tells you how quickly temperature changes spread. The characteristic distance a temperature wave travels in time t is \sqrt{\alpha t}. For stone (\alpha \sim 5 \times 10^{-7} m²/s), in 12 hours (half a day), \sqrt{\alpha t} \sim \sqrt{5 \times 10^{-7} \times 43200} \approx 0.15 m — 15 cm. This is why a half-metre-thick haveli wall protects the inside of the house from the day's heat: the thermal wave from the 45 °C afternoon takes longer than 12 hours to cross the wall, and by the time it gets to the inner face, it is night and already cooler outside. The wall acts as a phase-lag, not just an insulator.

Temperature profile inside a slab in steady state — why it's linear

In one dimension, with no heat sources, the heat equation in steady state reduces to

\frac{d^2 T}{dx^2} = 0,

whose solutions are straight lines: T(x) = a + bx. Applying boundary conditions T(0) = T_1 and T(L) = T_2 gives

T(x) = T_1 + (T_2 - T_1)\,\frac{x}{L}.

This is why the temperature inside a steady-state slab drops linearly from one face to the other. If you instrument a brick wall with thermocouples at several depths during a stable weather day, the thermocouples' readings line up on a straight line — this is conduction being directly visible.

The linearity is only true for a slab with no internal heat generation (no chemical reactions, no electrical heating, no absorbed radiation). A rod carrying electrical current (which heats it throughout its length via Joule heating) has a parabolic temperature profile; a nuclear-fuel rod with distributed reactions has a more complex shape. The general tool is the heat equation with a source term Q_\text{gen}:

\frac{\partial T}{\partial t} = \alpha\,\nabla^2 T + \frac{Q_\text{gen}}{\rho c_p}.

Heat conduction through a cylindrical pipe

For a pipe of inner radius r_1 and outer radius r_2, carrying hot fluid inside and surrounded by cooler air outside, the 1D radial heat equation in steady state gives a logarithmic temperature profile, not a linear one:

T(r) = T_1 - \frac{(T_1 - T_2)}{\ln(r_2/r_1)}\,\ln(r/r_1),

and the heat flow per unit length of pipe is

\frac{Q/t}{L_\text{pipe}} = \frac{2\pi k (T_1 - T_2)}{\ln(r_2/r_1)}.

The corresponding thermal resistance per unit length is R_\text{cyl} = \ln(r_2/r_1) / (2\pi k). This is why the thickness of insulation around a pipe matters in a specific way: adding insulation reduces k's effective contribution only logarithmically, and beyond a certain "critical radius" it can actually increase total heat loss by increasing the exposed outer surface. Industrial insulation design uses these formulas to pick the right thickness for steam pipes, cold-storage pipes, and LNG lines.

The analogy with electrical resistance is exact

The reason the electrical analogy works so perfectly is that the steady-state equation \nabla^2 T = 0 (Laplace's equation) is the same equation that governs the electrostatic potential in a region with no charge (again \nabla^2 \phi = 0). The flux relations (\mathbf{q} = -k\nabla T and \mathbf{j} = -\sigma \nabla \phi by Ohm's law) have the same form. The temperature in a conducting region and the potential in a dielectric obey the same mathematics, so every geometric insight from circuit analysis carries over. Feynman put it succinctly: the same equations have the same solutions.

For a transient problem (including time evolution), the thermal analog becomes a diffusion equation rather than Laplace's — parallel to the diffusion of charge in a resistor-capacitor network rather than a pure resistor. A thermocouple rod with heat capacity plus conductivity is an RC thermal network.

Where this leads next