IB DP Physics: The greenhouse effect and planetary energy balance (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 all of topic B.2 (Greenhouse effect) in the IB Physics guide, first assessment 2025, for Standard Level and Higher Level; the guide sets no extra content for Higher Level alone in B.2. Use σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and Wien’s constant 2.9 × 10⁻³ m K where needed. See the study guide and revision notes first if needed. Course page: IB DP Physics. Checklist: IB DP Physics checklist.

Questions

1. (a) State what is meant by the albedo of a planet. [1] (b) State two reasons why the albedo of Earth is not constant. [2]

2. A planet of radius 3.4 × 10⁶ m is exposed to a solar constant of 590 W m⁻² and has an albedo of 0.25. (a) Calculate the total power incident on the planet. [2] (b) Calculate the power scattered by the planet. [1] (c) Calculate the mean absorbed intensity over the whole surface of the planet. [2]

3. A flat plate of area 0.50 m² is at a temperature of 400 K and radiates a total of 620 W. (a) Calculate the emissivity of the plate. [3] (b) The plate is changed to a surface of emissivity 0.85. Calculate the temperature at which it radiates 1000 W m⁻². [3]

4. Explain why the mean intensity of sunlight arriving at a planet is S/4, where S is the solar constant. [3]

5. A planet has no atmosphere, a solar constant of 1900 W m⁻² and an albedo of 0.40. (a) Calculate its equilibrium temperature if it radiates as an ideal black surface. [3] (b) Calculate the equilibrium temperature if its emissivity is 0.90. [2]

6. Name the four main greenhouse gases. For each, state one natural source or one source created by human activity. [4]

7. Explain, using molecular energy levels, how a carbon dioxide molecule absorbs infrared radiation from the Earth’s surface and what happens afterwards. [5]

8. Describe how the greenhouse effect can be explained using a resonance model. [3]

9. The table gives energy flows for a model planet, averaged over its whole surface, in W m⁻².

Flow	Value
Solar radiation incident at the top of the atmosphere	340
Solar radiation reflected to space	95
Solar radiation absorbed by the surface	170
Solar radiation absorbed by the atmosphere	75
Infrared emitted by the surface	380
Infrared from the surface absorbed by the atmosphere	350
Infrared from the surface passing straight to space	30
Energy moved from surface to atmosphere by evaporation and convection	120
Infrared emitted by the atmosphere towards the surface	330

(a) Calculate the albedo of the planet. [1] (b) Show that the surface is in energy balance. [2] (c) Calculate the infrared emitted by the atmosphere towards space. [2] (d) Show that the planet as a whole is in balance. [1] (e) Calculate the surface temperature, assuming it radiates as an ideal black surface. [1]

10. A planet has a solar constant of 3000 W m⁻² and an albedo of 0.20. Treat it as radiating as an ideal black surface. (a) Show that the planet would have an equilibrium temperature of 321 K with no atmosphere. [3] (b) A model atmosphere is added as a single layer. It lets sunlight through, absorbs all the infrared from the surface, and radiates equally upward and downward. Show that the surface intensity σTs⁴ is 1200 W m⁻² and find Ts. [2] (c) In a second model the layer absorbs only 0.60 of the surface infrared and radiates with emissivity 0.60, equally up and down. Determine Ts. [3] (d) Explain why the answer to (c) is lower than the answer to (b). [1]

11. An Earth-like planet absorbs a mean intensity of 238 W m⁻². Its atmosphere is modelled as a single layer that absorbs a fraction a of the surface infrared and radiates with emissivity a, equally up and down. In this model σTs⁴(1 − a/2) is equal to the absorbed intensity. (a) Calculate the surface temperature when a = 0.78. [2] (b) Burning fossil fuels raises a to 0.82. Calculate the new surface temperature and the rise. [2] (c) Explain how the increase in a represents the enhanced greenhouse effect. [3] (d) While the surface is warming, a net 0.5 W m⁻² enters a 250 m layer of water (mass 2.5 × 10⁵ kg under each square metre, specific heat capacity 4200 J kg⁻¹ K⁻¹). Estimate its temperature rise in one year of 3.15 × 10⁷ s. [2]

12. The Sun has a surface temperature of 5800 K and the Earth’s surface is at 288 K. (a) Calculate the wavelength of peak emission for each. [2] (b) State the region of the electromagnetic spectrum for each peak. [1] (c) Calculate the energy of an infrared photon of wavelength 15 μm. [2] (d) Use your answers to explain why the atmosphere warms the surface by absorbing outgoing radiation more than incoming radiation. [1]

Answers

1. (a) Albedo is the total scattered power divided by the total incident power. [1] (b) Cloud formations change from day to day. [1] Latitude (surface type such as ice, ocean or forest, and the angle of the rays). [1] Examiner insight: Give two different reasons; “clouds” and “weather” would be the same idea written twice.

2. (a) P = Sπr² = 590 × π × (3.4 × 10⁶)². [1] = 2.14 × 10¹⁶ W. [1] (b) 0.25 × 2.14 × 10¹⁶ = 5.36 × 10¹⁵ W. [1] (c) Mean absorbed = S(1 − α)/4 = 590 × 0.75 / 4. [1] = 111 W m⁻². [1] Examiner insight: Check by dividing the absorbed power 1.61 × 10¹⁶ W by 4πr²: it gives the same 111 W m⁻².

3. (a) σT⁴ = 5.67 × 10⁻⁸ × 400⁴ = 1450 W m⁻². [1] P/A = 620 / 0.50 = 1240 W m⁻². [1] e = 1240 / 1450 = 0.85. [1] (b) T⁴ = 1000 / (0.85 × 5.67 × 10⁻⁸). [1] = 2.07 × 10¹⁰ K⁴. [1] T = 380 K. [1] Examiner insight: Show the substitution into P/A = eσT⁴ before the answer, so a slip in arithmetic does not hide the method.

4. The planet intercepts sunlight over its projected disc of area πR². [1] The energy is shared (and radiated) over the whole surface of area 4πR². [1] Mean intensity = SπR² / 4πR² = S/4. [1] Examiner insight: Name both areas, and say which one receives the sunlight.

5. (a) Absorbed = 1900 × 0.60 / 4 = 285 W m⁻². [1] T⁴ = 285 / 5.67 × 10⁻⁸ = 5.03 × 10⁹ K⁴. [1] T = 266 K. [1] (b) T⁴ = 285 / (0.90 × 5.67 × 10⁻⁸) = 5.58 × 10⁹. [1] T = 273 K. [1] Examiner insight: Round once at the end; a rounded T⁴ can shift the third significant figure after the fourth root.

6. Carbon dioxide: burning fossil fuels (or respiration, volcanoes). [1] Methane: cattle farming or rice paddies (or wetlands). [1] Nitrous oxide: fertilisers (or bacteria in soil). [1] Water vapour: evaporation from oceans (natural) or the warming of the air by other gases. [1] Examiner insight: Match each source to its gas; a vague source such as “pollution” is unlikely to earn the mark.

7. The molecule has discrete vibrational energy levels. [1] An infrared photon is absorbed if its energy hf equals the difference between two levels. [1] Infrared from the cooler surface has photon energies of this size, which sunlight mostly does not. [1] The molecule then falls back to a lower level and emits a photon, or passes energy to other molecules in collisions. [1] The photon goes off in a random direction, so the radiation is re-emitted in all directions and some returns to the surface. [1] Examiner insight: Say “re-emitted in all directions”; “sent back to Earth” misses why only part returns.

8. The bonds in the molecule behave as oscillators with a natural frequency in the infrared. [1] Radiation at that frequency drives the oscillation to a large amplitude, so energy is absorbed strongly. [1] The oscillating charges then radiate, in all directions. [1] Examiner insight: Link resonance to the frequency match; “the molecule vibrates” alone is too vague.

9. (a) α = 95 / 340 = 0.28. [1] (b) In: 170 + 330 = 500 W m⁻². [1] Out: 380 + 120 = 500 W m⁻², so the surface is in balance. [1] (c) Atmosphere in: 75 + 350 + 120 = 545 W m⁻². [1] Out: 330 + X, so X = 215 W m⁻². [1] (d) To space: 30 + 215 = 245 W m⁻², equal to the absorbed solar 340 − 95 = 245 W m⁻². [1] (e) T⁴ = 380 / 5.67 × 10⁻⁸, so T = 286 K. [1] Examiner insight: For a “show that” balance, write both totals and the comparison; a bare statement that they are equal does not demonstrate it.

10. (a) Absorbed = 3000 × 0.80 / 4 = 600 W m⁻². [1] T⁴ = 600 / 5.67 × 10⁻⁸ = 1.06 × 10¹⁰ K⁴. [1] T = 321 K. [1] (b) Top of atmosphere: σTA⁴ = 600. Surface: 600 + σTA⁴ = σTs⁴, so σTs⁴ = 1200 W m⁻². [1] Ts = (1200 / 5.67 × 10⁻⁸)^(1/4) = 381 K. [1] (c) Atmosphere: 0.60σTs⁴ = 2 × 0.60σTA⁴, so TA⁴ = Ts⁴/2. [1] Top of atmosphere: 0.40σTs⁴ + 0.60σTA⁴ = 600, so 0.70σTs⁴ = 600. [1] Ts⁴ = 1.51 × 10¹⁰, so Ts = 351 K. [1] (d) Less of the surface infrared is absorbed, so less is returned to the surface and the surface does not need to be as hot. [1] Examiner insight: Write one balance equation per layer before substituting.

11. (a) Ts⁴ = 238 / [5.67 × 10⁻⁸ × (1 − 0.39)] = 6.88 × 10⁹. [1] Ts = 288 K. [1] (b) Ts⁴ = 238 / [5.67 × 10⁻⁸ × 0.59] = 7.11 × 10⁹, so Ts = 290 K. [1] Rise = 2.4 K. [1] (c) Burning fossil fuels raises the concentration of greenhouse gases such as carbon dioxide. [1] The atmosphere takes in a larger share of the surface infrared and sends more of it down again, so a is larger. [1] The surface warms until the power escaping to space again equals the absorbed solar power. [1] (d) Rate = 0.5 / (2.5 × 10⁵ × 4200) = 4.8 × 10⁻¹⁰ K s⁻¹. [1] Over one year: 0.015 K. [1] Examiner insight: Take the difference from unrounded temperatures, not from rounded ones.

12. (a) Sun: 2.9 × 10⁻³ / 5800 = 5.00 × 10⁻⁷ m. [1] Earth: 2.9 × 10⁻³ / 288 = 1.01 × 10⁻⁵ m. [1] (b) Sun: visible. Earth: infrared. [1] (c) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 15 × 10⁻⁶. [1] = 1.33 × 10⁻²⁰ J. [1] (d) Sunlight is mostly visible and passes through; surface infrared matches molecular energy level gaps, so it is absorbed and some is returned. [1] Examiner insight: Convert 15 μm to metres before using E = hc/λ.

Where marks are usually lost

• Using S where S/4 is needed, or leaving out the factor 1 − α.
• Using °C in σT⁴, or forgetting the fourth root.
• Describing greenhouse gases as reflecting or trapping radiation instead of absorbing and re-emitting it.
• Letting the atmosphere radiate only upward.

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.2, Greenhouse effect.