IB DP Physics: Conduction, convection and thermal radiation: Stefan-Boltzmann and Wien laws (DP Physics)

This study guide follows the IB Physics guide, first assessment 2025, and covers the second half of topic B.1 (Thermal energy transfers): conduction, convection and thermal radiation. It applies to Standard Level and Higher Level alike, because B.1 contains no material reserved for Higher Level. The first half (molecular model, temperature, internal energy, heating calculations) is taught in a separate unit.

Pair this page with the revision notes and the practice set. The course page is IB DP Physics and the printable list is the IB DP Physics checklist. For the heating and molecular side of B.1 see the molecular model study guide. To find weak spots first, try the free 10-minute diagnostics.

What this unit covers

B.1 understanding	What you must be able to do	Level
Mechanisms	Name conduction, convection and thermal radiation as the primary ways thermal energy is transferred	SL and HL
Conduction, particle model	Explain conduction using differences in the kinetic energy of particles	SL and HL
Conduction, rate	Use the rate of transfer ΔQ/Δt = kA ΔT/Δx	SL and HL
Convection	Describe it qualitatively using fluid density differences	SL and HL
Radiation	Describe emission of electromagnetic waves from a surface; use L = σAT⁴ for a black body	SL and HL
Apparent brightness	Use the concept of b and b = L/(4πd²)	SL and HL
Black-body spectrum	Describe the emission spectrum; find temperature with λmax T = 2.9 × 10⁻³ m K	SL and HL

Luminosity may be quoted in watts or as a multiple of the Sun’s output, written L⊙. Where a constant is needed, this page states it in the example. The values used are σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and L⊙ = 3.83 × 10²⁶ W.

The three mechanisms

Thermal energy always flows from the hotter place to the colder one. It does so in three main ways.

• Conduction needs matter and passes energy from particle to particle without bulk movement of the material.
• Convection needs a fluid (a liquid or a gas) and moves energy by the movement of the fluid itself.
• Thermal radiation needs no medium. A surface emits electromagnetic waves, and these carry energy across a vacuum.

Many real situations use all three at once. A hot drink loses energy through the cup wall by conduction, through the air above it by convection and from every surface by radiation.

Conduction as a difference in kinetic energy

Particles in a hot region have more kinetic energy on average than particles in a cold region. In a solid they vibrate about fixed positions. A vibrating particle in the hot region pushes on its neighbours and gives some of its energy to them. Those neighbours pass it on in turn. The result is a net flow of energy towards the cold region, with no net movement of the material.

In a metal the free electrons add to this. Fast electrons in the hot region drift into the cold region and collide with ions and other electrons, passing on energy. This is why metals are better conductors than most non-metals. Gases conduct poorly because the particles are far apart and collide rarely.

Conduction stops when there is no temperature difference. The direction of the transfer is always set by the temperature difference, never by which body holds more total energy.

Rate of conduction

The rate of thermal energy transfer through a uniform slab or rod is

ΔQ/Δt = kA ΔT/Δx

• ΔQ/Δt is the rate of transfer, in watts (W).
• k is the thermal conductivity, a property of the material measured in W m⁻¹ K⁻¹. A larger k means a better conductor.
• A is the cross-sectional area at right angles to the flow, in m².
• ΔT/Δx is the temperature gradient, in K m⁻¹: the temperature difference divided by the distance over which it occurs.

Because only a difference in temperature enters, you can use kelvin or degrees Celsius for ΔT. Convert lengths to metres and areas to m² before you substitute.

The rate is proportional to k, to A and to ΔT, and inversely proportional to the thickness Δx. Doubling the thickness halves the rate.

Worked example 1. The lid of a polystyrene cool box has area 0.60 m² and thickness 4.0 cm. Its inner face is at 2 °C and its outer face is at 22 °C. The thermal conductivity of polystyrene is 0.035 W m⁻¹ K⁻¹. Find the rate of transfer and the energy gained by the box in 10 minutes.

ΔT = 22 − 2 = 20 K, and Δx = 0.040 m

ΔQ/Δt = kA ΔT/Δx = 0.035 × 0.60 × 20 / 0.040 = 10.5 W

Energy in 600 s = 10.5 × 600 = 6.3 × 10³ J. The gradient is 20/0.040 = 500 K m⁻¹.

Worked example 2. An aluminium rod has radius 1.0 cm and length 0.40 m. Its ends are held at 100 °C and 20 °C. Take k = 205 W m⁻¹ K⁻¹. The sides are insulated. Find the rate of transfer and the energy transferred in 5.0 minutes.

A = πr² = π × (0.010)² = 3.14 × 10⁻⁴ m²

ΔT/Δx = 80 / 0.40 = 200 K m⁻¹

ΔQ/Δt = 205 × 3.14 × 10⁻⁴ × 200 = 12.9 W

Energy in 300 s = 12.9 × 300 = 3.9 × 10³ J (3.86 × 10³ J before rounding).

The insulated sides matter. They mean all the energy that enters at the hot end leaves at the cold end, so the temperature falls at a steady rate along the rod and the gradient is the same everywhere. Where that holds, the temperature 0.10 m from the hot end is 100 − 200 × 0.10 = 80 °C.

Convection

When part of a fluid is heated, its particles move faster and spread further apart. The same mass takes up more volume, so that part of the fluid has a lower density. Denser, cooler fluid around it pushes it upwards, and the cooler fluid sinks to take its place. This sets up a circulating flow, called a convection current, which carries thermal energy through the fluid.

The guide asks for a qualitative description, so there is no equation. A good answer has four links: heating, lower density, rise of the warm fluid, sinking of the cool fluid.

Two uses show the idea. A radiator warms a room because the air above it rises and cooler air flows in at the bottom. A freezer cabinet with its cooling coils at the top also works well, because the chilled air becomes denser and sinks through the food. Convection cannot occur in a solid, and it cannot occur if the heating is placed at the top of a fluid, since the warm fluid is already as high as it can go.

Thermal radiation and black bodies

Every surface emits electromagnetic waves because of the thermal motion of its particles. A hotter surface emits more power, and the wavelengths it emits shift towards shorter values. You notice this when a heated metal first glows dull red and then becomes white.

A black body is an idealised object that absorbs all radiation falling on it and, at a given temperature, emits the maximum power possible from its surface. Stars and glowing filaments are often modelled as black bodies. The guide asks you to model the emission from a black body using the Stefan-Boltzmann law.

The luminosity L of a body is the total power it radiates, over all wavelengths. For a black body

L = σAT⁴

• L is the luminosity, in W.
• σ is the Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
• A is the surface area, in m². For a sphere of radius r, A = 4πr².
• T is the absolute (kelvin) temperature of the surface.

T must be in kelvin. A temperature in °C gives a wrong answer by a large factor, because of the fourth power. Dividing by L⊙ gives the luminosity in solar units: L/L⊙.

Worked example 3. A spherical lamp bulb element of radius 2.0 cm is at 1500 K. Treat it as a black body. Find its luminosity.

A = 4πr² = 4π × (0.020)² = 5.03 × 10⁻³ m²

L = σAT⁴ = 5.67 × 10⁻⁸ × 5.03 × 10⁻³ × 1500⁴ = 1.44 × 10³ W

Worked example 4 (scaling). A black-body sphere has its radius doubled and its absolute temperature halved. By what factor does L change?

L ∝ r²T⁴, so the factor is 2² × (1/2)⁴ = 4 × 1/16 = 1/4. The luminosity falls to one quarter. Scaling questions need no constants at all, and they are quick marks if you write the proportionality first.

Apparent brightness

Luminosity belongs to the source. Apparent brightness b is the power received per unit area at an observer, in W m⁻². Energy from a source spreads evenly over a sphere of surface area 4πd² at distance d, so

b = L / (4πd²)

This is an inverse square law. Double the distance and b falls to a quarter. It assumes that nothing absorbs light on the way.

Worked example 5. A star has luminosity 5.0 L⊙ and lies 3.0 × 10¹⁷ m away. Find b.

L = 5.0 × 3.83 × 10²⁶ = 1.915 × 10²⁷ W

b = 1.915 × 10²⁷ / (4π × (3.0 × 10¹⁷)²) = 1.69 × 10⁻⁹ W m⁻²

Worked example 6. A star of luminosity 1.2 × 10²⁸ W has b = 2.4 × 10⁻⁹ W m⁻². Find its distance.

Rearrange: d = √(L / 4πb) = √(1.2 × 10²⁸ / (4π × 2.4 × 10⁻⁹)) = 6.3 × 10¹⁷ m

Black-body spectrum and Wien’s law

A graph of emitted intensity against wavelength for a black body is a smooth hump with a peak. It has these features.

• The curve starts at zero at short wavelengths, rises steeply to one peak and then falls more slowly towards long wavelengths.
• Raising the temperature raises the whole curve, because more power is emitted at every wavelength.
• The peak wavelength λmax moves to a shorter wavelength.
• The area under the curve is the total power emitted per unit area. It grows as T⁴.

Wien’s displacement law links the peak wavelength to the temperature:

λmax T = 2.9 × 10⁻³ m K

Worked example 7. A star’s spectrum peaks at 725 nm. Find its surface temperature.

T = 2.9 × 10⁻³ / 725 × 10⁻⁹ = 4.0 × 10³ K

Worked example 8 (combined). The star in example 7 has a luminosity of 2.0 × 10²⁸ W. Find its radius and its luminosity in solar units.

A = L / σT⁴ = 2.0 × 10²⁸ / (5.67 × 10⁻⁸ × (4000)⁴) = 1.378 × 10²¹ m²

r = √(A / 4π) = 1.05 × 10¹⁰ m

L/L⊙ = 2.0 × 10²⁸ / 3.83 × 10²⁶ = 52

Wien’s law supplies T from the spectrum, Stefan-Boltzmann turns T and L into an area, and the sphere formula turns the area into a radius. This chain, and the same chain run with b and d, is the usual pattern for extended questions.

Common errors

• Putting T in °C into L = σAT⁴ or Wien’s law.
• Using the radius where the area belongs, or forgetting the 4π.
• Writing that heat “rises” in convection; say the warm fluid is less dense.
• Reading the peak of the spectrum as the total power; the total is the area under the curve.
• Confusing luminosity (a property of the star) with apparent brightness (depends on distance).
• Entering thickness in cm or mm without converting to metres.

Next steps

Try the practice set, then compress the formulas with the revision notes.

Official syllabus

Source: the Physics guide for the IB Diploma Programme (first assessment 2025), International Baccalaureate Organization. Topic B.1, Thermal energy transfers, second half.