Touch a metal spoon left in a hot pan and you’ll feel it almost instantly — touch the wooden handle next to it and you won’t. Conduction is why solids transfer heat so differently depending on what they’re made of.
📘 What you need to know
Thermal energy transfers between a hotter and cooler region through conduction, convection or radiation
Conduction is the main way thermal energy moves through solids
It happens through atomic vibrations (in all solids) and free electron collisions (mainly in metals)
A material’s ability to conduct heat is measured by its thermal conductivity, k
The rate of conductive heat flow is given by ΔQ⁄Δt = kAΔT⁄Δx
Three Ways Thermal Energy Moves
Thermal energy can transfer from a hotter region to a cooler one through three different mechanisms: conduction, convection and radiation. This page focuses on conduction — the dominant mechanism in solids — with convection and radiation covered in their own pages.
How Conduction Works
When two solids of different temperature touch, thermal energy transfers from the hotter one to the cooler one until they reach thermal equilibrium. This happens through two mechanisms working together:
Atomic vibrations — particles at the hotter end vibrate more vigorously and collide with their neighbours, passing energy along, atom by atom. This occurs in every solid, metal or not.
Free electron collisions — metals contain large numbers of delocalised electrons that move freely through the structure, colliding with atoms and speeding up the transfer of vibrations dramatically. This is why metals reach thermal equilibrium so much faster than non-metals.
Larger vibrations near the hot end pass energy along through collisions, while free electrons speed the process up further
Thermal Conductivity
Thermal conductivity, k, quantifies how effectively a material transfers heat by conduction. It’s measured in W m⁻¹ K⁻¹. Materials with a high thermal conductivity — usually those with plenty of free electrons — are excellent conductors; materials with a low thermal conductivity are good thermal insulators.
Metals like copper conduct heat far more effectively than insulators like wood or trapped air — bar widths here aren’t fully to scale, since copper’s conductivity is thousands of times greater than air’s
The Temperature Gradient Equation
Whenever a temperature difference exists across a material, thermal energy flows from the hotter side to the cooler side. This is called a temperature gradient, and the rate of that heat flow can be calculated directly.
Rate of heat transfer by conduction
ΔQ⁄Δt = kAΔT⁄Δx
Where ΔQ⁄Δt is the flow of thermal energy per second in watts, k is thermal conductivity, A is the cross-sectional area in m², ΔT is the temperature difference, and Δx is the thickness of the material in metres. Provided the cross-sectional area stays constant, this flow of energy per second is uniform throughout — much like current staying constant around a series circuit, even though different components have different resistances.
🧭 Recipe: Finding the Junction Temperature in a Composite Bar
Recognise that the rate of heat flow is the same through every section of the bar in steady state
Write kΔT = constant for each section, since A and Δx are shared across sections of equal length and cross-section
Substitute the known conductivities and end temperatures to form an equation in the unknown junction temperature
Solve for the junction temperature
Quick recap: ΔQ/Δt = kAΔT/Δx. Higher k means faster heat flow for the same area, temperature difference and thickness.
WE 1
A copper heat-exchanger plate (k = 400 W m⁻¹ K⁻¹) has a cross-sectional area of 0.020 m² and a thickness of 5.0 mm. One face is held at 85 °C and the other at 20 °C. Calculate the rate of heat transfer through the plate.
Step 1 — List the known quantities
k = 400 W m⁻¹ K⁻¹, A = 0.020 m², ΔT = 85 − 20 = 65 K, Δx = 5.0 × 10⁻³ m
Step 2 — Substitute into the equationΔQ/Δt = kAΔT/Δx = (400 × 0.020 × 65) ÷ (5.0 × 10⁻³)= 104 000 W ≈ 104 kW
WE 2
A composite bar is made of a 20 cm length of brass (k = 110 W m⁻¹ K⁻¹) joined to a 20 cm length of glass (k = 1.0 W m⁻¹ K⁻¹), both with identical cross-section, insulated along their sides. The brass end is held at 80 °C and the glass end at 10 °C. Determine the temperature at the brass–glass junction.
Step 1 — Set the rate of heat flow equal in both sectionsk_brass(80 − T) = k_glass(T − 10)Step 2 — Substitute the known values110(80 − T) = 1.0(T − 10)Step 3 — Expand and solve for T8800 − 110T = T − 10 → 8810 = 111TT ≈ 79.4 °CBecause glass is such a poor conductor compared to brass, almost the entire temperature drop happens across the glass section.
💡 Top tips
Always convert thickness into metres before substituting into the temperature gradient equation
In a composite bar, the poorer conductor always carries the larger share of the temperature drop
If a question mentions thermal energy transfer through metals, conduction is almost always the mechanism being tested
Rate of heat flow stays uniform along a bar of constant cross-section in steady state, even across different materials
⚠ Common mistakes
Forgetting to convert thickness or length into metres before substituting into the equation
Assuming the temperature drop is split evenly between sections of a composite bar, regardless of their conductivities
Mixing up thermal conductivity, k, with the spring constant or the Boltzmann constant, which use the same letter
Applying the conduction equation to a fluid that isn’t trapped, where convection is actually the dominant mechanism
Up next: Thermal Convection — where we look at how heat moves through liquids and gases instead.
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