Heat transfer by conduction

Pick up a metal spoon left resting in a hot pan and within seconds the handle is too hot to hold. Leave a wooden spoon in the very same pan and it stays cool enough to touch far longer. Both spoons are sitting in the same fire, both are made of solid matter, and both are transferring heat the same way — by conduction, molecule jostling molecule without any large-scale flow of material. Yet one carries that heat almost instantly and the other barely at all. The difference is a single number attached to each material, its thermal conductivity, and untangling exactly what that number means, how it enters a formula, and how to use that formula on rods, slabs and layered walls is the whole of this article. Conduction turns out to behave in a way that should feel oddly familiar if you have ever studied electric current. A temperature difference drives heat through a material the way a potential difference drives current through a wire; a poor conductor resists that flow the way a resistor resists current; and when several layers of different materials are stacked one after another, their resistances to heat simply add, exactly as resistors do in series. That analogy is not a coincidence — it is the same "flow down a gradient" physics — and once you see it, composite-wall problems that look intimidating collapse into simple series-and-parallel resistance arithmetic. By the end of this article you should be able to state the steady-state conduction law, read off the units and physical meaning of thermal conductivity $k$ , work a single-slab problem confidently, and — the part CBSE loves to test — combine two or more layers of different materials using thermal resistance in series. We will keep every number and every result from the original worked examples on this topic and build outward from them, adding the underlying molecular picture, a full derivation of the steady-state formula from first principles of heat flow, a second, harder worked example, and the common slip-ups that cost marks in board and competitive exams alike.

Key takeaways

Conduction transfers heat through a material via molecular collisions (and free electrons in metals) without any bulk motion of the material itself — it dominates in solids.
The steady-state rate of heat flow through a slab of area $A$ , thickness $L$ , and face temperatures $T_1 > T_2$ is $\dfrac{dQ}{dt} = \dfrac{kA\,\Delta T}{L}$ , where $k$ is the material's thermal conductivity, unit $\text{W m}^{-1}\text{K}^{-1}$ .
$k$ is a material property only — it does not depend on the rod's length, area, or shape; metals (copper, silver) have large $k$ , insulators (wood, glass wool, air) have small $k$ .
The quantity $R = \dfrac{L}{kA}$ behaves as a thermal resistance; for layers placed one after another (in series), resistances add and the steady-state heat current through every layer is the same.
"Steady state" is the essential qualifier — it means the temperature at every point inside the slab has settled and stopped changing with time; the formula does not apply while the material is still warming up.
A copper rod ( $k = 385\ \text{W m}^{-1}\text{K}^{-1}$ , length $0.50\ \text{m}$ , area $2.0\times10^{-4}\ \text{m}^2$ , ends at $100^\circ\text{C}$ and $0^\circ\text{C}$ ) conducts heat at $15.4\ \text{W}$ — the anchor number to remember for this topic.
Key idea: in the steady state, heat current through a slab is $\frac{dQ}{dt} = \frac{kA\,\Delta T}{L}$ , exactly analogous to Ohm's law, with thermal resistance $L/kA$ adding in series across composite layers.

The core insight

Heat conducted per second through a slab of area $A$ and thickness $L$ is $\frac{dQ}{dt} = \frac{kA\,\Delta T}{L}$ in the steady state, with the material's thermal conductivity $k$ playing the role of a conductance and the geometric factor $L/(kA)$ behaving exactly like an electrical resistance that adds in series across layered materials.

Where it fits in the syllabus

Conduction is the first of the three heat-transfer mechanisms in the CBSE Class 11 Thermal Properties of Matter chapter, studied alongside convection and radiation. The board syllabus for 2025-26 restricts conduction to the steady-state formula $\frac{dQ}{dt} = \frac{kA\,\Delta T}{L}$ and the meaning of thermal conductivity — you are expected to apply it to single rods/slabs and to series combinations of different materials (composite walls or rods). Newton's law of cooling, treated as a full exponential decay, and the greenhouse effect are kept out of the theory syllabus at this level (Newton's cooling appears only as a qualitative practical idea), so this article stays entirely within the conduction formula and its series-resistance extension.

The physical picture

At the molecular level, a solid is a lattice of atoms vibrating about fixed positions, plus, in metals, a sea of relatively free electrons. Heat one end of a rod and the atoms there vibrate more violently; those atoms jostle their slightly-cooler neighbours through the bonds holding the lattice together, handing on some of that extra vibrational energy, which jostle the next neighbours in turn, and so on down the rod. In metals this molecular hand-off is helped enormously by the free electrons, which can carry energy quickly across large distances before colliding again — this is exactly why metals are both excellent electrical conductors and excellent thermal conductors, and why a metal spoon heats up in your hand far faster than a wooden one whose heat can travel only atom-to-atom through a much stiffer, more disordered structure. Now picture a slab of this material, of cross-sectional area $A$ and thickness $L$ , with its two faces held at fixed temperatures $T_1$ (hot) and $T_2$ (cold) by whatever is touching them. Heat flows in at the hot face and out at the cold face. Once enough time has passed for every point inside the slab to settle into a fixed temperature that no longer drifts — the steady state — the rate at which heat crosses any cross-section of the slab is the same everywhere along its length, and is given by

\frac{dQ}{dt} = \frac{kA\,\Delta T}{L}

where: $\dfrac{dQ}{dt}$ is the heat current, the heat energy crossing the slab per second, measured in watts; $A$ is the cross-sectional area perpendicular to the direction of heat flow; $\Delta T = T_1 - T_2$ is the temperature difference between the two faces; $L$ is the thickness of the slab measured along the direction of heat flow; and $k$ is the thermal conductivity of the material, with SI unit $\text{W m}^{-1}\text{K}^{-1}$ . The physical reading is straightforward on each side of the fraction. Heat flow grows in direct proportion to $A$ and $\Delta T$ — a wider slab offers more parallel "channels" for heat to travel through, and a bigger temperature gap pushes heat through any one channel harder. Heat flow shrinks in inverse proportion to $L$ — a thicker slab makes each unit of heat travel farther through resistive material before it gets out, so less gets through per second. And $k$ sits out front as a pure multiplier fixed by the material alone: copper's $k = 385\ \text{W m}^{-1}\text{K}^{-1}$ is around ten thousand times glass wool's $k \approx 0.04\ \text{W m}^{-1}\text{K}^{-1}$ , which is precisely why one is used for cooking vessels and the other for insulating walls.

Figure

A rod with a hot left face and a cold right face, arrows showing heat current flowing from hot to cold, and a temperature axis below showing temperature falling in a straight line from the hot end to the cold end across the thickness L

For composite slabs — several different materials stacked so that heat must pass through one layer and then the next, such as a brick wall lined with insulation — it helps to define the thermal resistance of a layer as $R = \dfrac{L}{kA}$ , so that the conduction law can be rewritten in the Ohm's-law-like form $\dfrac{dQ}{dt} = \dfrac{\Delta T}{R}$ . In steady state, the same heat current must pass through every layer in turn (none of it can pile up anywhere), which is exactly the condition for resistors in series: the total resistance is the simple sum $R_{\text{total}} = R_1 + R_2 + R_3 + \dots$ , and the overall temperature drop splits across the layers in proportion to each one's own resistance. A thin layer of a very poor conductor can therefore contribute far more resistance — and do far more of the actual insulating work — than a much thicker layer of a good conductor, which is the entire logic behind lining a thick brick wall with a thin layer of foam insulation.

Derivation ASteady-state conduction current through a slab

The goal of this derivation is to obtain the formula $\dfrac{dQ}{dt} = \dfrac{kA\,\Delta T}{L}$ for the rate at which heat crosses a slab of material once the slab has settled into a steady state, starting only from the basic empirical fact that heat conducted through a thin layer is proportional to the area available, to the temperature drop across that layer, to the time elapsed, and is inversely proportional to the layer's thickness — the same proportionality that defines thermal conductivity $k$ as the constant that ties these quantities together, and showing why, once every layer of the slab has reached a fixed temperature profile, the heat current becomes one single constant number rather than something that varies from point to point or moment to moment.

Step 1. Start from the defining proportionality for a thin slice of material of area $A$ and small thickness $dx$ , with a small temperature difference $dT$ across it. Experiment shows the heat $dQ$ conducted across this slice in a time $dt$ is proportional to $A$ , to the magnitude of the temperature gradient $dT/dx$ , and to $dt$ itself: $\dfrac{dQ}{dt} = -kA\dfrac{dT}{dx}$ , where $k$ is the constant of proportionality for that material — its thermal conductivity — and the minus sign simply records that heat flows from hot toward cold, i.e. in the direction in which $T$ decreases.

Step 2. Impose the steady-state condition. In the steady state, no point inside the slab is gaining or losing energy over time — if it were, its temperature would be changing, contradicting "steady." Since none of the heat entering any thin slice is retained there, the same heat current $\dfrac{dQ}{dt}$ must be flowing across every cross-section of the slab, front to back — it cannot be larger just inside the hot face than just inside the cold face, or energy would be piling up somewhere in between. So $\dfrac{dQ}{dt}$ is one single constant number, call it $I$ , independent of position $x$ along the slab.

Step 3. Since $I = -kA\,dT/dx$ is constant and $k$ , $A$ are constants of the (uniform) slab, the temperature gradient $dT/dx$ must itself be constant along the slab — meaning $T$ falls off linearly with position from the hot face to the cold face, exactly the straight-line temperature profile drawn in the figure above. This lets us replace the differential $dT/dx$ with the simple ratio of the total drop over the total length: $\dfrac{dT}{dx} \to -\dfrac{\Delta T}{L}$ , where $\Delta T = T_1 - T_2$ is the (positive) temperature difference between the hot and cold faces and $L$ is the full slab thickness.

Step 4. Substitute this constant gradient back into the Step 1 relation and drop the sign, keeping only the physically meaningful positive heat current flowing from hot to cold: $I = \dfrac{dQ}{dt} = \dfrac{kA\,\Delta T}{L}$ . This is the steady-state conduction law used throughout this article — it holds once, and only once, the linear temperature profile has been established throughout the slab.

Worked examples

Board level - single copper rod

A copper rod of length $0.50\ \text{m}$ and cross-sectional area $2.0\times10^{-4}\ \text{m}^2$ has its two ends maintained at $100^\circ\text{C}$ and $0^\circ\text{C}$ . Find the rate of heat flow through the rod in the steady state. Take $k_{\text{copper}} = 385\ \text{W m}^{-1}\text{K}^{-1}$ .

Step 1. Identify the given quantities and find $\Delta T$ : $k = 385\ \text{W m}^{-1}\text{K}^{-1}$ , $A = 2.0\times10^{-4}\ \text{m}^2$ , $L = 0.50\ \text{m}$ , and $\Delta T = 100 - 0 = 100\ \text{K}$ (a $1^\circ\text{C}$ difference is the same size as a $1\ \text{K}$ difference, so no conversion is needed for $\Delta T$ itself).

Step 2. Substitute directly into the steady-state conduction formula: $\dfrac{dQ}{dt} = \dfrac{kA\,\Delta T}{L} = \dfrac{(385)(2.0\times10^{-4})(100)}{0.50}$ .

Step 3. Simplify the numerator first, then divide: numerator $= 385 \times 2.0\times10^{-4} \times 100 = 7.7$ , so $\dfrac{dQ}{dt} = \dfrac{7.7}{0.50} = 15.4\ \text{W}$ .

The heat flows through the copper rod at a rate of $\mathbf{15.4\ W}$ in the steady state — that is, $15.4$ joules of heat cross any section of the rod every second.

Competitive level - composite wall in series

A composite wall is made of a layer of brick ( $k_1 = 0.6\ \text{W m}^{-1}\text{K}^{-1}$ , thickness $0.20\ \text{m}$ ) and a layer of insulation ( $k_2 = 0.04\ \text{W m}^{-1}\text{K}^{-1}$ , thickness $0.05\ \text{m}$ ), placed one after the other, each with cross-sectional area $2.0\ \text{m}^2$ . The outer brick surface is at $30^\circ\text{C}$ and the inner insulation surface is at $10^\circ\text{C}$ . Find the steady-state rate of heat flow through the wall, and the temperature at the brick-insulation interface.

Step 1. Recognise this as a series-resistance problem and compute each layer's thermal resistance $R = L/(kA)$ : $R_1 = \dfrac{0.20}{(0.6)(2.0)} = \dfrac{0.20}{1.2} = 0.1667\ \text{K/W}$ for the brick, and $R_2 = \dfrac{0.05}{(0.04)(2.0)} = \dfrac{0.05}{0.08} = 0.625\ \text{K/W}$ for the insulation.

Step 2. Add the resistances in series (the same heat current must pass through both layers, so resistances simply sum): $R_{\text{total}} = R_1 + R_2 = 0.1667 + 0.625 = 0.7917\ \text{K/W}$ .

Step 3. Apply the total temperature drop $\Delta T = 30 - 10 = 20\ \text{K}$ across the total resistance: $\dfrac{dQ}{dt} = \dfrac{\Delta T}{R_{\text{total}}} = \dfrac{20}{0.7917} \approx 25.3\ \text{W}$ .

Step 4. Find the interface temperature $T_i$ by applying the same heat current to the brick layer alone: $\dfrac{dQ}{dt} = \dfrac{30 - T_i}{R_1}$ , so $30 - T_i = (25.3)(0.1667) \approx 4.2\ \text{K}$ , giving $T_i \approx 25.8^\circ\text{C}$ .

The steady-state heat flow through the composite wall is about $\mathbf{25.3\ W}$ , with the brick-insulation interface sitting at about $\mathbf{25.8^\circ C}$ — most of the $20\ \text{K}$ drop (about $15.8\ \text{K}$ of it) happens across the thin insulation layer, not the thick brick, exactly because the insulation's resistance $R_2$ is nearly four times $R_1$ despite being a quarter the thickness. Timing note: on a timed paper, do not recompute $\dfrac{dQ}{dt}$ from scratch for the interface temperature — reuse the value from Step 3 and apply it to just one layer, as in Step 4; recomputing everything from a fresh energy balance wastes a minute you don't have.

Common traps

Forgetting that the conduction formula applies only in the steady state, where every point's temperature has settled — before that, the material itself is still absorbing heat and warming up, and the simple formula does not apply.
Treating thermal conductivity $k$ as though it depends on the rod's length or area — $k$ is a property of the material only; changing the geometry changes the heat current, not $k$ itself.
Adding thermal conductivities directly for a composite slab instead of adding thermal resistances $L/(kA)$ — this single error is the most common way marks are lost on composite-conduction questions.
Confusing thermal conductivity $k$ (how easily heat flows through a material, unit $\text{W m}^{-1}\text{K}^{-1}$ ) with specific heat capacity $c$ (how much heat a given mass needs to change temperature by one degree, unit $\text{J kg}^{-1}\text{K}^{-1}$ ) — the two describe entirely different physical properties and are never interchangeable in a formula.
Skipping the interface-temperature step in composite problems and assuming the temperature drop is shared equally between layers — it is shared in proportion to each layer's resistance, not equally, unless the resistances happen to be equal.

Common trapAdding conductivities instead of resistances for layers in series is the single most common exam error on this topic

For a composite wall or rod made of several materials placed one after another, it is tempting to simply average or add the $k$ values, the way you might for two resistors you mistake for being in parallel. That is wrong. The correct rule mirrors Ohm's law for series resistors exactly: thermal resistances $R = L/(kA)$ add, $R_{\text{total}} = R_1 + R_2 + \dots$ , and only then is $\dfrac{dQ}{dt} = \Delta T / R_{\text{total}}$ computed. A thin, low- $k$ insulating layer can dominate the total resistance even though its $k$ value looks unremarkable next to a thick, high- $k$ layer — check resistance, not conductivity, when judging which layer matters most.

You should be able to…

State the steady-state conduction formula $\dfrac{dQ}{dt} = \dfrac{kA\,\Delta T}{L}$ and identify the SI unit of thermal conductivity.
Explain conduction at the molecular level, and why metals conduct heat better than most non-metals.
Calculate the steady-state rate of heat flow through a single uniform slab or rod.
Define thermal resistance $R = L/(kA)$ and combine resistances in series for a composite wall or rod made of multiple materials.
Find the temperature at the interface between two layers of a composite slab once the overall heat current is known.
Explain, using $k$ , why particular materials are chosen as thermal conductors versus thermal insulators.

Ask the tutor: need a hand setting up a three-layer composite wall or comparing the conduction formula side by side with Ohm's law — ask the AI tutor to walk through it step by step.

Why does a metal spoon feel hotter than a wooden spoon even though both are at the same temperature in the same pan?

It is not really that the metal spoon is hotter — both spoons are at essentially the pan's temperature near the point of contact. What differs is how fast each one delivers heat into your hand, which is governed by thermal conductivity $k$ . Copper or steel has a $k$ thousands of times larger than wood, so heat conducts from the pan, through the handle, and into your skin far more quickly for the metal spoon, making your hand's temperature rise fast enough for nerves to register pain. The wooden spoon conducts heat so slowly that your hand's own blood flow can carry the small trickle of heat away almost as fast as it arrives, so it never feels dangerously hot.

Why does the composite-wall problem need thermal resistance at all — can't I just use the conduction formula on the whole wall at once?

You could, only if you already knew the effective $k$ of the whole two-layer wall treated as one material — but you don't, because the two layers have different $k$ values and different thicknesses. The trick is that in steady state, the same heat current has to pass through both layers one after another, exactly like current through resistors in series. Writing each layer's resistance as $L/(kA)$ turns the whole problem into simple series addition, $R_{\text{total}} = R_1 + R_2$ , after which one application of $\Delta T / R_{\text{total}}$ gives the answer for the entire wall in one step.

Is the temperature profile inside a conducting slab always a straight line?

Only under the conditions used in this article: a single uniform material, in the steady state, with no heat being generated or absorbed inside the slab itself. Under those conditions, Step 3 of the derivation shows the temperature gradient must be constant, which forces a straight-line profile. In a composite wall each individual layer has its own straight-line profile (with its own slope, set by its own $k$ ), but the overall profile bends at each interface between materials — it is piecewise linear, not one single straight line across the whole wall.

How is conduction different from convection and radiation, which come later in this chapter?

Conduction moves heat through direct molecular contact within a material, with no bulk movement of the material itself — think of heat crawling down a solid metal rod. Convection instead relies on the bulk motion of a fluid (liquid or gas) itself carrying hot regions to cooler ones, which is why it needs a fluid and cannot occur in a rigid solid. Radiation needs no material medium at all — it is heat carried by electromagnetic waves and can cross empty space, which is how the Sun's heat reaches Earth. All three can act together in real situations, but each has a distinct formula and a distinct physical mechanism, and this chapter treats conduction first because its formula is the most directly derivable of the three.