Thermal conduction

Pick up a metal spoon left in a hot cup of tea and within seconds your fingers protest — yet a wooden spoon in the same cup stays comfortably cool. Both spoons sit in identical hot liquid; the difference is how fast heat travels through the solid itself, molecule to molecule, with no bulk flow of matter. That process is thermal conduction, and it is the reason metals feel cold in winter and warm in summer while wood never does, the reason a frying pan needs a wooden handle, and the reason double-glazed windows trap heat. This article builds the steady-state law of conduction, $\dfrac{dQ}{dt} = kA\dfrac{(T_1-T_2)}{L}$ , from the idea of a temperature gradient, applies it to a single rod and to two rods joined end to end, and closes with Searle's experiment — the standard laboratory method for measuring a good conductor's thermal conductivity.

The core insight

In steady state, the rate of heat conduction along a rod is proportional to its cross-sectional area and the temperature gradient along it, and inversely proportional to nothing else — $\dfrac{dQ}{dt} = kA\dfrac{(T_1-T_2)}{L}$ , with $k$ a fixed material property called thermal conductivity.

The core idea

What is actually moving. No matter travels from the hot end of a rod to the cold end during conduction — only energy does. At the hot end, molecules (or, in metals, free electrons) vibrate or move more vigorously; they jostle their slightly cooler neighbours and hand on some of that extra kinetic energy, which in turn passes it further along, like a relay of collisions. In metals, mobile free electrons carry energy far more efficiently than lattice vibrations alone, which is exactly why metals are, as a class, much better conductors than non-metals.

Steady state and the temperature gradient. Imagine a rod of cross-sectional area $A$ and length $L$ with its two ends clamped at fixed temperatures $T_1$ and $T_2$ , with $T_1 > T_2$ . Immediately after heating begins, every part of the rod is also warming up itself — some of the incoming heat is "spent" raising the rod's own temperature. Only once every cross-section has settled to a constant temperature that no longer changes with time do we say the rod has reached steady state. In steady state, the heat entering the hot end per second equals the heat leaving the cold end per second (assuming no heat escapes from the sides), and the rate of heat flow is

\frac{dQ}{dt} = kA\,\frac{(T_1-T_2)}{L}

where: $\dfrac{dQ}{dt}$ is the rate of heat flow in watts, $A$ is the cross-sectional area in $\text{m}^2$ , $T_1-T_2$ is the temperature difference between the ends, $L$ is the length of the rod, and $k$ is the coefficient of thermal conductivity, a property of the material with SI unit $\text{W m}^{-1}\text{K}^{-1}$ . The ratio $\dfrac{T_1-T_2}{L}$ is the temperature gradient — the drop in temperature per unit length along the direction of heat flow — so the law simply says heat current is proportional to area times gradient.

Good versus poor conductors. Metals such as silver ( $k \approx 410\ \text{W m}^{-1}\text{K}^{-1}$ ) and copper ( $k \approx 400\ \text{W m}^{-1}\text{K}^{-1}$ ) have large $k$ ; wood, glass and air have small $k$ and are used precisely because they resist heat flow. A larger $k$ at fixed $A$ , $L$ and $\Delta T$ means a larger heat current — this is why a metal spoon in hot tea heats up at its far end quickly while a wooden one barely warms.

Figure

A rod between a hot end and a cold end showing heat current flowing along the temperature gradient

Derivation AWhy heat current depends on area, gradient and k

The steady-state formula for conduction is grounded, not something a student is meant to derive from scratch on the exam; the goal here is to see clearly why heat current must depend on area, gradient and the material constant $k$ in exactly this combination, using only the physical picture of molecular collisions and the requirement of energy conservation in steady state.

Step 1. Heat current scales with area. Picture the rod's cross-section as made of many thin parallel filaments, each carrying its own share of the collision-relay described above. Doubling the cross-sectional area $A$ doubles the number of such filaments working in parallel, so it doubles the total rate of heat flow for the same temperature difference and length. This is why $\dfrac{dQ}{dt}$ is directly proportional to $A$ .

Step 2. Heat current scales with the temperature gradient, not with $\Delta T$ or $L$ alone. A steeper temperature drop per unit length means each pair of neighbouring molecular layers along the rod differs in "vigour" by more, so more energy is handed across every internal boundary per second. Doubling $\Delta T$ while keeping $L$ fixed doubles the gradient and doubles the current; equally, stretching the same $\Delta T$ over twice the length halves the gradient and halves the current. Both effects combine into the single ratio $\dfrac{T_1-T_2}{L}$ .

Step 3. The material constant $k$ absorbs everything about the substance itself — how densely packed its molecules are, how strongly they interact, and, in metals, how many free electrons are available to ferry energy. Two rods of identical shape and identical $\Delta T$ but different materials will carry different heat currents purely because their $k$ differs; this is precisely why $k$ is defined as the constant of proportionality: $\dfrac{dQ}{dt} = kA\dfrac{(T_1-T_2)}{L}$ , and a good conductor is simply one with large $k$ .

Step 4. In steady state, energy conservation forces the heat current to be the same at every cross-section along the rod (nothing is left over to warm the rod further), which is exactly why the single formula above, evaluated with the overall $\Delta T$ and $L$ , correctly gives the heat flow at any point — and why, when two rods are joined end to end, the same heat current must pass through both.

Worked example

A copper rod of length $0.50\ \text{m}$ and cross-sectional area $2.0\times10^{-4}\ \text{m}^2$ has its ends maintained at $100^\circ\text{C}$ and $20^\circ\text{C}$ , with $k_{Cu} = 400\ \text{W m}^{-1}\text{K}^{-1}$ .

Step 1. Identify the quantities: $A = 2.0\times10^{-4}\ \text{m}^2$ , $L = 0.50\ \text{m}$ , $T_1 - T_2 = 100 - 20 = 80^\circ\text{C}$ , $k = 400\ \text{W m}^{-1}\text{K}^{-1}$ .

Step 2. Substitute into $\dfrac{dQ}{dt} = kA\dfrac{(T_1-T_2)}{L}$ :

\frac{dQ}{dt} = (400)(2.0\times10^{-4})\frac{(80)}{0.50} = (400)(2.0\times10^{-4})(160)

Step 3. Multiply through: $400 \times 2.0\times10^{-4} = 0.080$ , and $0.080 \times 160 = 12.8$ .

\boxed{\frac{dQ}{dt} = 12.8\ \text{W}}

Heat flows through the rod at a steady rate of 12.8 joules every second.

Competitive stretch

Now join two rods A and B end to end (same length $L$ and area $A$ ), with $k_A = 200\ \text{W m}^{-1}\text{K}^{-1}$ and $k_B = 400\ \text{W m}^{-1}\text{K}^{-1}$ , between a hot reservoir at $90^\circ\text{C}$ and a cold reservoir at $0^\circ\text{C}$ . In steady state the same heat current flows through both rods, so at the junction temperature $\theta$ : $k_A(90-\theta) = k_B\theta$ , giving $200(90-\theta) = 400\theta$ , so $18000 = 600\theta$ and $\boxed{\theta = 30^\circ\text{C}}$ . The junction sits closer to the cold end because the better conductor, B, offers less resistance to heat flow, so most of the temperature drop occurs across the poorer conductor, A.

Common traps

Applying $\dfrac{dQ}{dt} = kA\dfrac{\Delta T}{L}$ before steady state is reached — during the transient warming phase, some heat goes into raising the rod's own temperature and the simple formula does not yet hold.
Confusing thermal conductivity $k$ , a fixed material property, with the rate of heat flow itself, which also depends on the rod's geometry ( $A$ , $L$ ) and the temperature difference applied.
In rods joined end to end, forgetting that it is the heat current (rate of heat flow) that must match at the junction in steady state, not the temperature or the gradient.
Mixing up conduction with convection — conduction has no bulk motion of the medium, while convection (which occurs in fluids) does; this distinction is a favourite conceptual trap.

Ask the tutor: if the series-rod junction algebra or the logic of Searle's experiment still feels shaky, ask the AI tutor to walk through it step by step with a labelled diagram.

Why do metals feel cold to touch even at room temperature, while wood at the same temperature feels normal?

Both objects are at the same room temperature, but your skin is usually a little warmer than that, so heat flows out of your finger into whatever you touch. Metal has a large thermal conductivity $k$ , so it conducts that heat away from your fingertip very quickly, and your nerve endings register the fast heat loss as "cold." Wood has a small $k$ , so heat leaves your finger much more slowly, and the surface barely feels cooler than your skin. It is a difference in the rate of heat conduction, not in actual temperature.

Why does the steady-state formula only apply after some time has passed, and not the instant you start heating?

When you first heat one end of a rod, not all the incoming heat travels straight through to the cold end — a portion is used to raise the temperature of the rod's own material at each point along the way, since the rod itself was initially cooler. Only once every cross-section has settled to a fixed temperature, so no more heat is being absorbed to warm the rod further, does the heat entering equal the heat leaving at every instant. That settled condition is steady state, and it is only then that $\dfrac{dQ}{dt} = kA\dfrac{\Delta T}{L}$ correctly describes the flow.

In Searle's experiment, why is water circulated through the far end instead of just measuring the rod's temperature directly?

Searle's method measures $k$ by comparing the heat conducted along the rod to the heat actually carried away, and the easiest way to measure "heat carried away per second" reliably is to let it warm a known, steadily flowing mass of water and record its temperature rise. Once steady state is reached, the heat conducted through the rod, $kA\dfrac{(\theta_1-\theta_2)}{L}$ , must equal the heat absorbed by the water, $m_w c_w\dfrac{(\theta_4-\theta_3)}{t}$ , since none is lost from the sides; equating the two lets $k$ be calculated from four measured temperatures and the water's mass flow rate, without needing to measure heat directly.

Thermal conduction

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads