Conduction & Thermal Conductivity

Conduction is the physics you are quietly paying for on every electricity bill. The R-values printed on loft insulation, cavity-wall foam and double-glazing units are Fourier's law in commercial dress: a builder's R-value is $L/k$, the resistance of one square metre of the material, and the whole trade of thermal engineering is stacking enough of them between you and the winter. Vacuum flasks and cryogenic tanks go one better — remove the matter and there is nothing left to conduct, the $k \to 0$ limit made real. At the other end of the ladder, the pursuit of high conductivity is a multi-billion-dollar problem. A processor chip dissipates tens or hundreds of watts through a few square centimetres, and that heat can only leave by conduction — which is why chips are bonded to copper spreaders through a thermal paste whose sole job is to banish the air gaps ($k = 0.024$) that would otherwise throttle the path. A turbine blade, an engine block, an LED's aluminium substrate and the base of a good frying pan are the same decision: make $R = L/(kA)$ as small as possible. Diamond, oddly, is among the finest thermal conductors known despite being an electrical insulator, and spreads heat in high-power lasers. Nature engineers with the same law. Fur, feathers and a polar bear's hollow hairs trap still air; a whale's blubber is thick, low-$k$ fat, and its thickness is the $L$ in the formula. And run the equation backwards and it measures a planet: drill a borehole, read the temperature gradient in the rock, multiply by the rock's conductivity, and equation (2) hands you the heat still leaking out of the Earth's interior.

Heat a bar at one end and you are not injecting a substance — you are injecting agitation. The atoms at the hot end swing more violently about their fixed sites, and since every atom is bonded to its neighbours, a violent swing knocks into a calm one and shares its energy. That collision-by-collision handover creeps along the bar as a wave of increased jiggling, while the atoms themselves go nowhere. This is the defining feature of conduction, and the cleanest way to separate it from its two siblings: convection needs a fluid that actually moves, radiation needs no medium at all, and conduction needs matter that stays exactly where it is. In a metal there is a second, far faster courier. Metals are held together by a sea of free electrons, tethered to no particular atom, streaming through the lattice at enormous speeds and ferrying energy from the hot end to the cold end long before the lattice vibrations could. That is why good electrical conductors are almost always good thermal conductors — copper and silver top both lists, for the same reason. Wood, glass and brick have no electron sea, so they must pass energy bond by bond, and they do it feebly. Air is worse still: its molecules are far apart and collide only occasionally, so a still layer of air is one of the finest insulators known. Now put a bar between a hot reservoir and a cold one, lag the sides so nothing escapes sideways, and wait. At first the temperature everywhere is changing; eventually it stops, every point settles, and the bar has reached the steady state. Whatever heat enters the hot face each second now leaves the cold face each second — nothing accumulates anywhere. And once that is true, something beautiful happens to the temperature profile: it becomes a perfectly straight line.

That straight line is the punchline of the whole topic. Its steepness — the degrees lost per metre — is the temperature gradient, and it is the gradient, not the temperature, that pushes heat along. A bar sitting uniformly at $500\ \mathrm{^\circ C}$ conducts nothing at all, because no slice is hotter than its neighbour and none has any reason to give energy away. A bar with a steep fall over a short distance conducts furiously. Heat, like water and like electric charge, responds to a slope.

Everything the syllabus asks of you follows from one experimental law, established by Joseph Fourier and now carrying his name. Take a uniform bar or slab of a single material, hold its two faces at steady temperatures $T_1$ (hot) and $T_2$ (cold), and measure the energy crossing it per second. The flow doubles if you double the face area, doubles if you double the temperature difference, and halves if you double the thickness. Package those three proportionalities with a material constant and you have the law:

where: $H$ — the heat current, the rate of heat flow in watts ($\mathrm{J\,s^{-1}}$); $k$ — the thermal conductivity of the material, in $\mathrm{W\,m^{-1}K^{-1}}$; $A$ — the cross-sectional area perpendicular to the flow, in $\mathrm{m^2}$; $\Delta T = T_1 - T_2$ — the temperature difference between the two faces, in kelvin (identical to the number in $\mathrm{^\circ C}$, because this is a difference); $L$ — the thickness, or the length of the path the heat must travel, in metres. Read the equation as a sentence and it tells you how to keep a house warm. Heat leaks faster through a bigger wall ($A$ up), a colder night ($\Delta T$ up) and a better-conducting material ($k$ up); it leaks slower through a thicker wall ($L$ up). Every insulation decision ever made — thicker loft lagging, smaller windows, foam instead of brick — is one of those four knobs being turned. And note what $H$ is not: it is not an amount of energy, it is joules per second. Writing "$H = 8000\ \mathrm{J}$" is one of the surest ways to lose a mark here. Strip the area out and you get the flow per square metre of wall — the heat flux, which depends only on the material and on how steeply the temperature falls:

where: $H/A$ — the heat flux, in $\mathrm{W\,m^{-2}}$; $\Delta T/L$ — the temperature gradient, in $\mathrm{K\,m^{-1}}$, i.e. the slope of the straight line in the figure above. (In calculus language the exact statement is that the flux equals $-k$ times the slope $\mathrm{d}T/\mathrm{d}x$, the minus sign recording that heat flows down the slope. For a uniform bar the slope is constant and equal to $\Delta T/L$, so nothing is lost by using the school form — no calculus is needed anywhere in this article.) Equation (2) is the sharpest statement of the physics: flux is conductivity times gradient. It also defines $k$ honestly — $k$ is the flux you would get through a material carrying a gradient of one kelvin per metre. That is why $k$ belongs to the substance and nothing else: it neither knows nor cares how long your rod is. Copper is $385\ \mathrm{W\,m^{-1}K^{-1}}$ whether you shape it into a wire or a saucepan. Exam questions often ask not for the rate but for the energy delivered in a given time. In the steady state $H$ is constant, so energy is simply rate times time:

where: $Q$ — the heat energy conducted, in joules; $t$ — the time in seconds. The physical reading is trivial but the mark scheme is not: "how much heat flows in 10 minutes" wants joules, and therefore wants $t = 600\ \mathrm{s}$, not $t = 10$. Watts times seconds give joules; watts times minutes give nonsense.

Look hard at equation (1) and you may feel a flicker of recognition. A flow is proportional to a difference, divided by something that depends on geometry and material. That is precisely the shape of Ohm's law, $I = V/R$ — and the resemblance is not a mnemonic but a historical fact: Georg Ohm, who announced his experimental law in 1826, cast its theory the following year in a monograph explicitly modelled on Fourier's mathematics of heat, published five years earlier. Electricity borrowed the shape of its equation from heat, not the other way round. So take the analogy seriously and give conduction its own resistance, by grouping the geometric and material factors of equation (1) into one quantity:

where: $R$ — the thermal resistance of the slab or rod, in kelvin per watt ($\mathrm{K\,W^{-1}}$); $L$ — thickness along the flow; $k$ — conductivity; $A$ — face area. The physical reading is the whole point: $R$ measures how many kelvin of temperature difference you must spend to push one watt of heat through the object. A double-glazed window has a big $R$ (few watts leak even for a large $\Delta T$); a copper saucepan base has a tiny $R$ (a huge heat current crosses it for a drop of a degree or two, which is exactly what you want from a pan). The dictionary between the two subjects is exact, and worth keeping as a table in your head: $\Delta T$ plays the part of the potential difference $V$; the heat current $H$ (watts) plays the part of the electric current $I$ (amperes); thermal resistance $L/(kA)$ plays the part of electrical resistance $\rho L/A$. Even the geometry matches — long, thin and poorly-conducting resists in both worlds; short, fat and well-conducting does not.

Once you think in resistances, insulation stops being a mystery and becomes arithmetic. Lagging a pipe, double-glazing a window, wearing a fleece — every one of them simply adds thermal resistance to the path of the heat. You cannot change the $\Delta T$ between your body and a January night; what you can do is raise $R$, and $H = \Delta T/R$ collapses accordingly. Layers stacked one behind another have their resistances add, which is why a thin sheet of foam can beat a thick wall of brick; the full arithmetic of stacked layers and junction temperatures is the business of the companion archetype on conduction through slabs, rods and composites, and everything it needs is already in equation (4).

Thermal conductivities span five orders of magnitude between copper and still air, and the spread is not random. It tracks exactly what the material has available to carry energy.

At the top sit the metals. Copper ($k = 385$) and aluminium ($k = 205$) have the sea of free electrons, and electrons are light, fast and mobile, so they ferry energy far more effectively than sluggish lattice vibrations ever could. Steel is a metal too, but a much poorer one ($k = 50$): its irregular alloy lattice scatters the electrons, so copper out-conducts steel by $385/50 = 7.7$. Below the metals come the non-metals — glass ($0.8$), brick ($0.6$) — where atoms can only nudge their neighbours, and $k$ collapses by a factor of nearly five hundred ($385/0.8 = 481$ for copper against glass). At the bottom lies gas: air has $k = 0.024$, its molecules far apart and passing energy only when they happen to collide. Copper beats still air by $385/0.024 \approx 1.6\times10^{4}$. That last number founds an entire industry. Wool, fur, feathers, polystyrene, cavity walls, fibreglass loft rolls and a diver's neoprene suit are not made of magically insulating stuff — they are ordinary fibres and plastics whose real job is to divide air into pockets too small to circulate. Kill the circulation and you kill convection; what remains is still air, doing almost nothing. The jumper is a scaffold. The insulator is the air.

Board level — the single-glazed window that bleeds eight kilowatts (CBSE/ICSE/IGCSE pattern)

A single glass pane has area $A = 2\ \mathrm{m^2}$ and thickness $L = 4\ \mathrm{mm}$. The inside surface is at $20\ \mathrm{^\circ C}$ and the outside surface is at $0\ \mathrm{^\circ C}$. Take $k_{\text{glass}} = 0.8\ \mathrm{W\,m^{-1}K^{-1}}$. Find the rate of heat loss through the pane, and the energy lost in one hour.

Step 1. Convert to SI before anything else: $L = 4\ \mathrm{mm} = 0.004\ \mathrm{m}$. This one conversion is where most of the marks are lost. The temperature difference is $\Delta T = 20 - 0 = 20\ \mathrm{K}$; a difference needs no $273$ added, because the offset cancels.

Step 2. Apply equation (1): $H = kA\,\Delta T/L = (0.8 \times 2 \times 20)/0.004 = 32/0.004 = \boxed{8000\ \mathrm{W} = 8\ \mathrm{kW}}$ — the pane is haemorrhaging heat at the rate of eighty $100\ \mathrm{W}$ light bulbs.

Step 3. For the energy, use equation (3) with the time in seconds: $t = 1\ \mathrm{hour} = 3600\ \mathrm{s}$, so $Q = Ht = 8000 \times 3600 = 2.88\times10^{7}\ \mathrm{J} = 28.8\ \mathrm{MJ}$, which in domestic units is $\boxed{8\ \mathrm{kWh}}$ — a whole unit of electricity every seven and a half minutes.

Step 4. Sense-check by the circuit route: $R = L/(kA) = 0.004/(0.8 \times 2) = 0.0025\ \mathrm{K\,W^{-1}}$, a pitiful resistance, and $H = \Delta T/R = 20/0.0025 = 8000\ \mathrm{W}$ — the same answer. Why so bad? Because $L$ is tiny and it sits on top in $R$. One honest caveat: a real window loses less than $8\ \mathrm{kW}$, because the glass faces never quite sit at the room and outdoor air temperatures — thin stagnant air films cling to both sides and add resistance of their own. The figure is the upper limit, and it is exactly why double glazing (a sealed layer of still air, $k = 0.024$) exists.

Board level — heat through a brick wall (CBSE/IGCSE pattern)

A brick wall is $10\ \mathrm{m^2}$ in area and $10\ \mathrm{cm}$ thick, with $k_{\text{brick}} = 0.6\ \mathrm{W\,m^{-1}K^{-1}}$. The inside face is at $20\ \mathrm{^\circ C}$, the outside face at $5\ \mathrm{^\circ C}$. Find the rate of heat flow through it and the heat flux.

Step 1. SI first: $L = 10\ \mathrm{cm} = 0.1\ \mathrm{m}$; $\Delta T = 20 - 5 = 15\ \mathrm{K}$.

Step 2. Equation (1): $H = kA\,\Delta T/L = (0.6 \times 10 \times 15)/0.1 = 90/0.1 = \boxed{900\ \mathrm{W}}$.

Step 3. The flux is the flow per square metre, equation (2): $H/A = 900/10 = 90\ \mathrm{W\,m^{-2}}$. Check it the other way: $k \times \text{gradient} = 0.6 \times (15/0.1) = 0.6 \times 150 = 90\ \mathrm{W\,m^{-2}}$. The two routes agree, as they must.

Step 4. Read the comparison. The wall is $25$ times thicker than the window pane and has $5$ times the area, yet it leaks only $900\ \mathrm{W}$ against the window's $8000\ \mathrm{W}$. Its resistance $R = 0.1/(0.6 \times 10) = 0.0167\ \mathrm{K\,W^{-1}}$ is nearly seven times the pane's $0.0025\ \mathrm{K\,W^{-1}}$ — and that is after spreading over five times the area. In any real building it is the windows, not the walls, where the heat escapes.

Competitive level — the lagged copper rod, end to end (JEE-Main/NEET/AP-2 pattern)

A copper rod of length $0.5\ \mathrm{m}$ and cross-sectional area $1\ \mathrm{cm^2}$ is lagged along its sides. One end is held in steam at $100\ \mathrm{^\circ C}$, the other in melting ice at $0\ \mathrm{^\circ C}$. Take $k_{\text{Cu}} = 385\ \mathrm{W\,m^{-1}K^{-1}}$, $k_{\text{steel}} = 50\ \mathrm{W\,m^{-1}K^{-1}}$. Find (a) the thermal resistance, (b) the steady heat current, (c) the temperature at the midpoint, (d) the heat conducted in $10$ minutes, and (e) the heat current if the rod were steel instead.

Step 1. Convert the area — the trap is hiding here. $1\ \mathrm{cm^2} = (10^{-2}\ \mathrm{m})^2 = 10^{-4}\ \mathrm{m^2}$, not $10^{-2}\ \mathrm{m^2}$. Squaring the centimetre is where a third of candidates lose the question before they start.

Step 2. (a) Thermal resistance, equation (4): $R = L/(kA) = 0.5/(385 \times 10^{-4}) = 0.5/0.0385 = \boxed{13.0\ \mathrm{K\,W^{-1}}}$.

Step 3. (b) Heat current, the circuit way: $H = \Delta T/R = 100/13.0 = \boxed{7.7\ \mathrm{W}}$. Cross-check directly: $H = kA\Delta T/L = 385 \times 10^{-4} \times 100/0.5 = 3.85/0.5 = 7.7\ \mathrm{W}$. Agreed.

Step 4. (c) The midpoint. The steady-state profile is a straight line (second derivation), so the temperature falls at a uniform $\Delta T/L = 100/0.5 = 200\ \mathrm{K\,m^{-1}}$. At $x = 0.25\ \mathrm{m}$ it has fallen $200 \times 0.25 = 50\ \mathrm{K}$, giving $\boxed{T = 50\ \mathrm{^\circ C}}$. (For a uniform rod the midpoint is the plain average of the ends — a shortcut that dies the instant two different materials are joined, since then the poorer conductor swallows most of the drop.)

Step 5. (d) Energy in ten minutes, equation (3): $t = 10 \times 60 = 600\ \mathrm{s}$, so $Q = Ht = 7.7 \times 600 = \boxed{4620\ \mathrm{J}}$.

Step 6. (e) Swap the material. Only $k$ changes and $H \propto k$, so $H_{\text{steel}} = 7.7 \times (50/385) = \boxed{1.0\ \mathrm{W}}$ — down by the factor $385/50 = 7.7$. Check directly: $50 \times 10^{-4} \times 100/0.5 = 1.0\ \mathrm{W}$. Same steam, same ice, and the copper carries nearly eight times the heat; aluminium would give $205 \times 10^{-4} \times 100/0.5 = 4.1\ \mathrm{W}$. Timing note: three minutes for all five parts. Parts (a) and (b) are one substitution; part (c) needs no arithmetic once you know the profile is linear; part (d) is a multiplication; part (e) is a ratio — never recompute from scratch, scale the answer you already have. In an MCQ the whole question is really testing whether you turned $1\ \mathrm{cm^2}$ into $10^{-4}\ \mathrm{m^2}$.

Competitive level — the scaling question (JEE-Advanced pattern)

A cylindrical rod conducting heat in the steady state between fixed temperatures has its radius doubled and its length doubled. By what factor does the heat current change? What if, instead, the same rod were replaced by one of the same shape but made of a material with a lower conductivity, so that the whole rod's resistance doubled?

Step 1. Write $H$ in terms of what actually changed. From equation (1) with $A = \pi r^2$ for a cylinder, $H \propto r^2/L$ at fixed material and fixed $\Delta T$. That proportionality is the question — never substitute numbers into a scaling problem.

Step 2. Double the radius: area goes as $r^2$, so $A$ multiplies by $2^2 = 4$, and $H$ with it. Double the length: $H \propto 1/L$, so that divides $H$ by $2$.

Step 3. Combine: $H_{\text{new}}/H_{\text{old}} = 4/2 = \boxed{2}$ — the heat current doubles. The classic wrong answers are $4$ (forgetting the length) and $1$ ("double and double must cancel" — forgetting that area goes as the square of the radius).

Step 4. The second part is a one-liner in circuit language, which is why the language is worth having: $H = \Delta T/R$ at fixed $\Delta T$, so doubling $R$ halves $H$, whatever combination of $k$, $L$ and $A$ did the doubling. Factor $= \boxed{1/2}$. Timing note: under sixty seconds. Scaling questions are pure proportionality — set up $H \propto r^2/L$, multiply the factors, done. Reaching for a calculator here is itself the mistake, and JEE-Advanced sets these to reward the student who sees $r^2$ instead of $r$.

Forgetting to square the centimetre. $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$, not $10^{-2}\ \mathrm{m^2}$; likewise $1\ \mathrm{mm} = 10^{-3}\ \mathrm{m}$ and $4\ \mathrm{mm} = 0.004\ \mathrm{m}$. This is the single most-punished slip in JEE-Main and NEET conduction questions, and it is worth a factor of one hundred.

Quoting $H$ as an energy. $H$ is a rate, in watts. If the question asks for the heat conducted in a stated time, multiply by that time in seconds ($10$ minutes $= 600\ \mathrm{s}$, one hour $= 3600\ \mathrm{s}$). CBSE and ICSE mark schemes award a separate mark for the unit.

Saying "cold flows in". Heat flows out, from hot to cold, always. NEET and IGCSE both set assertion-reason and multiple-choice items whose wrong options are phrased in terms of cold entering — and the metal-feels-colder-than-wood question is the same trap in disguise (it is your hand losing heat quickly, not the metal being colder).

Adding $273$ to a temperature difference. $\Delta T$ in Fourier's law is a difference, so $20\ \mathrm{^\circ C}$ of difference is exactly $20\ \mathrm{K}$. You convert to kelvin for the gas laws and for Stefan's radiation law, never for a $\Delta T$ in conduction.

Thinking $k$ depends on the shape. Thermal conductivity is a property of the material only. A short fat copper rod and a long thin one have wildly different resistances $R$ but the identical $k = 385$. If a question says two rods are "of the same material", their $k$ is one number and it cancels in any ratio.

Ignoring the geometry because the material sounds good. "Copper is a great conductor, so it conducts a lot" is not an argument: $H$ depends on $A$ and $L$ as strongly as on $k$, and a long thin copper wire can easily carry less heat than a short fat brick. Write out all four quantities every time.

Confusing conduction with convection. Conduction happens in matter that stays put. If the question describes fluid actually circulating — warm water rising, a fan blowing — that is convection, a different article and a different formula.

Assuming a steady state that has not arrived. $H = kA\Delta T/L$ describes the steady state, after temperatures stop drifting. In the first moments after a rod is dipped into steam, some of the incoming heat is still warming the rod itself and the profile is not yet straight. Boards only ask the steady case — but they expect to see the word "steady".

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — the CBSE/JEE treatment of conduction and thermal conductivity · Cambridge IGCSE Physics 0625, Thermal Physics (conduction, insulation, everyday applications) · Halliday, Resnick & Walker, "Conduction, Convection and Radiation".

Provenance. Standard results. A verifier should recompute: the copper rod $R = 0.5/(385\times10^{-4}) = 13.0\ \mathrm{K\,W^{-1}}$ and $H = 100/13.0 = 385\times10^{-4}\times100/0.5 = 7.7\ \mathrm{W}$ (both routes agreeing), its linear profile $T(x) = 100 - 200x$ giving $50\ \mathrm{^\circ C}$ at the midpoint, $Q = 7.7\times600 = 4620\ \mathrm{J}$ in ten minutes, and the same rod in steel $= 1.0\ \mathrm{W}$, in aluminium $= 4.1\ \mathrm{W}$; the single glass pane $0.8\times2\times20/0.004 = 8000\ \mathrm{W}$, with $R = 0.0025\ \mathrm{K\,W^{-1}}$ and $Q = 8000\times3600 = 2.88\times10^{7}\ \mathrm{J} = 8\ \mathrm{kWh}$; the brick wall $0.6\times10\times15/0.1 = 900\ \mathrm{W}$, flux $90\ \mathrm{W\,m^{-2}}$, $R = 0.0167\ \mathrm{K\,W^{-1}}$; the scaling result that doubling both radius and length multiplies $H$ by $4/2 = 2$; and the ratios $385/50 = 7.7$, $385/0.8 = 481$, $385/0.024 \approx 1.6\times10^4$.

Prerequisites. Heat vs Temperature · Thermal Equilibrium & the Zeroth Law. Leads to. Convection · ARCHETYPE: Conduction Through Slabs, Rods & Composites (series and parallel conductors, junction temperatures). Go deeper (undergraduate). "Transport Coefficients and Green–Kubo Relations" derives the conductivity $k$ from the microscopic motion of the carriers themselves, rather than measuring it — the next tier's answer to why copper has the $k$ it does.