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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover the second half of topic B.1 (Thermal energy transfers) in the IB Physics guide, first assessment 2025, for Standard Level and Higher Level; B.1 has no additional higher level content. Use σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and L⊙ = 3.83 × 10²⁶ W where needed, together with the Wien constant 2.9 × 10⁻³ m K. See the study guide and revision notes. Course page: IB DP Physics. Checklist: IB DP Physics checklist.

Questions

1. Explain, in terms of the kinetic energy of particles, how thermal energy is conducted along a solid rod when one end is heated. [3]

2. A glass window pane has thermal conductivity 1.0 W m⁻¹ K⁻¹, area 1.5 m² and thickness 6.0 mm. One face is at 18 °C and the other at 10 °C.

(a) Calculate the rate of thermal energy transfer through the pane. [3] (b) The pane is replaced by one of the same glass and area but twice as thick, and the temperature difference across it is halved. State the new rate. [2]

3. A small electric heater is to warm a room. Describe, with reference to density, why the heater warms the room more effectively when it is placed near the floor than near the ceiling. [4]

4. A copper rod has cross-sectional area 2.0 × 10⁻⁴ m² and length 0.50 m. Its sides are insulated. One end is held at 100 °C and the other at 20 °C. The thermal conductivity of copper is 385 W m⁻¹ K⁻¹.

(a) Calculate the temperature gradient along the rod. [1] (b) Calculate the rate of thermal energy transfer along the rod. [2] (c) Calculate the energy transferred in 10 minutes. [2] (d) Assuming the gradient is uniform, determine the temperature 0.20 m from the hot end. [2]

5. A lamp filament radiates as a black body with a luminosity of 60 W at a temperature of 2700 K.

(a) Calculate the surface area of the filament. [3] (b) The temperature is raised to 3000 K with the area unchanged. Calculate the new luminosity. [2]

6. Star X has three times the radius of star Y and half its surface temperature. Both behave as black bodies. Determine L(X)/L(Y). [3]

7. The surface of human skin is at 306 K.

(a) Calculate the peak wavelength of its emission, treating it as a black body. [2] (b) A hot star has a peak wavelength of 290 nm. Calculate its surface temperature. [2] (c) State the region of the electromagnetic spectrum in which each peak lies. [2]

8. A star has luminosity 2.5 × 10²⁷ W and lies at a distance of 4.0 × 10¹⁷ m.

(a) Calculate its apparent brightness. [3] (b) A second star with four times this luminosity has the same apparent brightness. Determine its distance. [2] (c) Express the luminosity of the first star in terms of L⊙. [1]

9. A star has a spectrum with a peak at 580 nm. Its apparent brightness at Earth is 3.0 × 10⁻⁹ W m⁻² and it is 2.0 × 10¹⁷ m away.

(a) Determine its surface temperature. [2] (b) Calculate its luminosity. [3] (c) Calculate its radius. [3] (d) Express its luminosity in terms of L⊙. [1]

10. Star P has surface temperature 9000 K and radius 5.6 × 10⁸ m. Star Q has surface temperature 3500 K and radius 2.8 × 10¹⁰ m. Treat both as black bodies.

(a) Calculate the luminosity of P. [3] (b) Calculate the luminosity of Q. [2] (c) Show, using proportionality, that L(Q)/L(P) is about 57. [2] (d) Calculate the peak wavelength of each star. [2]

11. The temperature of a black body is raised from 3000 K to 4500 K.

(a) Describe three changes to its emission spectrum. [3] (b) Calculate the new peak wavelength. [2] (c) Determine the factor by which the power radiated per unit area changes. [2]

12. The insulated panels of a cold room have total area 30 m², thickness 0.12 m and thermal conductivity 0.035 W m⁻¹ K⁻¹. The inner surface is at −18 °C and the outer surface at 24 °C.

(a) Calculate the rate at which thermal energy is conducted into the room. [3] (b) The cooling unit can remove only 250 W in steady state. Determine the largest temperature difference across the panels for which the unit can cope. [3] (c) Calculate the energy conducted in through the panels in 8.0 hours at the rate found in (a). [2]

Answers

1. Particles at the hot end have greater average kinetic energy and vibrate more [1]. They interact with neighbouring particles and pass on some of their energy [1]. Each neighbour passes it on in turn, so there is a net flow of energy towards the cold end without any bulk movement of the solid [1]. [3] Examiner insight: “Heat moves from hot to cold” gives the direction but not the mechanism, so state the particle energy and the neighbour interaction.

2. (a) ΔT = 18 − 10 = 8 K [1]. ΔQ/Δt = kAΔT/Δx = 1.0 × 1.5 × 8 / 6.0 × 10⁻³ [1]. = 2.0 × 10³ W [1]. (b) The rate is proportional to ΔT/Δx. Half the ΔT and double the thickness gives one quarter [1]. New rate = 500 W [1]. [5] Examiner insight: Convert 6.0 mm to 6.0 × 10⁻³ m before substituting; a thickness left in mm gives an answer a thousand times too small, and writing the proportionality in (b) shows the reasoning even if the arithmetic slips.

3. Air next to the heater is warmed and its particles spread out, so its density falls [1]. The less dense warm air rises [1]. Cooler, denser air sinks or flows in to take its place, setting up a circulating convection current that carries energy through the room [1]. A heater near the ceiling warms air that is already at the top, so little circulation forms and the lower air stays cool [1]. [4] Examiner insight: Credit tends to follow the chain heating, lower density, rise, replacement; writing “hot air rises” with no mention of density usually leaves out a link.

4. (a) Gradient = ΔT/Δx = 80 / 0.50 = 160 K m⁻¹ [1]. (b) Rate = kA × gradient = 385 × 2.0 × 10⁻⁴ × 160 [1] = 12.3 W [1]. (c) E = 12.32 × 600 [1] = 7.39 × 10³ J [1]. (d) Fall in temperature = 160 × 0.20 = 32 K [1]. Temperature = 100 − 32 = 68 °C [1]. [7] Examiner insight: Carry the unrounded rate into part (c) so that rounding does not drift the third significant figure.

5. (a) Rearrange L = σAT⁴ to A = L/σT⁴ [1]. T⁴ = 2700⁴ = 5.31 × 10¹³ [1]. A = 60 / (5.67 × 10⁻⁸ × 5.31 × 10¹³) = 1.99 × 10⁻⁵ m² [1]. (b) L ∝ T⁴, so L = 60 × (3000/2700)⁴ [1] = 91.4 W [1]. [5] Examiner insight: In (b) a ratio method avoids repeating the area calculation and reduces the chance of an arithmetic error.

6. L ∝ r²T⁴ [1]. Ratio = 3² × (1/2)⁴ [1] = 9/16 = 0.563 [1]. [3] Examiner insight: State the proportionality before inserting numbers; a squared radius and a fourth-power temperature are the two details most often mismatched.

7. (a) λmax = 2.9 × 10⁻³ / 306 [1] = 9.48 × 10⁻⁶ m [1]. (b) T = 2.9 × 10⁻³ / 290 × 10⁻⁹ [1] = 1.0 × 10⁴ K [1]. (c) Skin: infrared [1]. Star: ultraviolet [1]. [6] Examiner insight: Convert nanometres to metres before dividing; the region should be named for each peak separately, since a bare “infrared and ultraviolet” does not show which belongs to which.

8. (a) b = L/4πd² [1] = 2.5 × 10²⁷ / (4π × (4.0 × 10¹⁷)²) [1] = 1.24 × 10⁻⁹ W m⁻² [1]. (b) For equal b, d² ∝ L [1], so d = 4.0 × 10¹⁷ × √4 = 8.0 × 10¹⁷ m [1]. (c) 2.5 × 10²⁷ / 3.83 × 10²⁶ = 6.53 L⊙ [1]. [6] Examiner insight: In (b), recognise that four times the luminosity needs only twice the distance.

9. (a) T = 2.9 × 10⁻³ / 580 × 10⁻⁹ [1] = 5.0 × 10³ K [1]. (b) L = 4πd²b [1] = 4π × (2.0 × 10¹⁷)² × 3.0 × 10⁻⁹ [1] = 1.51 × 10²⁷ W [1]. (c) A = L/σT⁴ = 1.51 × 10²⁷ / (5.67 × 10⁻⁸ × (5000)⁴) = 4.26 × 10¹⁹ m² [1]. r = √(A/4π) [1] = 1.84 × 10⁹ m [1]. (d) 1.51 × 10²⁷ / 3.83 × 10²⁶ = 3.94 L⊙ [1]. [9] Examiner insight: In a chain of steps, show each intermediate value with its unit so that a reader can follow your method.

10. (a) A = 4πr² = 4π × (5.6 × 10⁸)² = 3.94 × 10¹⁸ m² [1]. L = σAT⁴ = 5.67 × 10⁻⁸ × 3.94 × 10¹⁸ × 9000⁴ [1] = 1.47 × 10²⁷ W [1]. (b) A = 4π × (2.8 × 10¹⁰)² = 9.85 × 10²¹ m² [1]. L = 5.67 × 10⁻⁸ × 9.85 × 10²¹ × 3500⁴ = 8.38 × 10²⁸ W [1]. (c) L ∝ r²T⁴, so ratio = (2.8 × 10¹⁰ / 5.6 × 10⁸)² × (3500/9000)⁴ [1] = 50² × 0.389⁴ = 2500 × 0.0229 = 57 [1]. (d) P: 2.9 × 10⁻³ / 9000 = 3.2 × 10⁻⁷ m [1]. Q: 2.9 × 10⁻³ / 3500 = 8.3 × 10⁻⁷ m [1]. [9] Examiner insight: In a “show that” the value is given, so display the substituted ratio and an unrounded value before the result.

11. (a) The peak moves to a shorter wavelength [1]. The curve rises at every wavelength, so more power is emitted overall [1]. The area under the curve increases, since total power per unit area grows as T⁴ [1]. (b) λmax = 2.9 × 10⁻³ / 4500 [1] = 6.4 × 10⁻⁷ m [1]. (c) Factor = (4500/3000)⁴ = 1.5⁴ [1] = 5.06 [1]. [7] Examiner insight: Make each spectrum feature a separate comparative statement, such as “peak at shorter wavelength”, not a general “it gets brighter”.

12. (a) ΔT = 24 − (−18) = 42 K [1]. ΔQ/Δt = 0.035 × 30 × 42 / 0.12 [1] = 368 W (367.5 W) [1]. (b) ΔT = (ΔQ/Δt)Δx / kA [1] = 250 × 0.12 / (0.035 × 30) [1] = 28.6 K [1]. (c) 8.0 h = 28 800 s [1]. E = 367.5 × 28 800 = 1.06 × 10⁷ J [1]. [8] Examiner insight: Rearrange before substituting, as in (b); a temperature difference is the same number in kelvin and in degrees Celsius.

Where marks are usually lost

• Putting Celsius values into σAT⁴ or into Wien’s law.
• Leaving out the 4π, or using the radius where the area is needed.
• Not converting millimetres, centimetres or nanometres to metres.
• Explaining convection as “heat rises” with no mention of density.
• Describing conduction without mentioning the kinetic energy of the particles.
• Mixing up luminosity, which belongs to the star, with apparent brightness, which depends on distance.
• Squaring the temperature, or taking the fourth power of the radius, in scaling questions.
• Giving no unit, or W where W m⁻² is wanted.

Next steps

• Review the formulas in the revision notes.
• Work back through the examples in the study guide.
• Return to the IB DP Physics course page and tick off the checklist.
• Take the free 10-minute diagnostics to find your next weak area.
• Book a free trial class.

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.