Educerie · IB Diploma · Physics
Theme B The particulate nature of matter · B.2 Greenhouse effect
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Apply conservation of energy to a planet: power absorbed = power emitted in equilibrium | SL, HL | "Explain why the temperature of the Earth's surface is constant" (2 marks) |
| Define emissivity and use emissivity = power per unit area ÷ σT⁴ | SL, HL | "Calculate the emissivity of…" (2 marks) |
| Define albedo, use it as a ratio, and explain why the Earth's albedo varies with cloud and latitude | SL, HL | "Outline two reasons why the albedo of the Earth varies" (2 marks) |
| Use the solar constant S and explain why the mean incoming intensity is S/4 | SL, HL | "Show that the mean intensity is about 340 W m⁻²" (2 marks) |
| Estimate the equilibrium temperature of a planet using albedo, emissivity and S | SL, HL | Paper 2, 3 or 4 marks, working shown |
| Solve energy-balance problems with energy exchanged between the surface and the atmosphere | SL, HL | Paper 2, a labelled flow diagram with one flux missing, 3 to 5 marks |
| Name the four main greenhouse gases and give natural and human sources of each | SL, HL | Paper 1A, or "State one natural and one human source of methane" (2 marks) |
| Explain absorption of infrared using molecular energy levels, and using the resonance model | SL, HL | "Explain, with reference to molecular energy levels, how greenhouse gases warm the surface" (3 or 4 marks) |
| Describe the enhanced greenhouse effect and its main cause | SL, HL | "Describe the enhanced greenhouse effect" (2 marks) |
Before you start
You need three things from B.1. A black body at absolute temperature T radiates a power L = σAT⁴ (the Stefan–Boltzmann law). The peak wavelength of its spectrum obeys Wien's law, λₘₐₓT = 2.9 × 10⁻³ m K, so a hotter body peaks at a shorter wavelength. And intensity is power per unit area, in W m⁻², which is what apparent brightness b = L ÷ 4πd² measures. Temperatures in every formula here are in kelvin.
1The idea in one paragraph
The Earth is heated by sunlight and cools by radiating infrared into space. When the two powers are equal, its temperature stops changing: that is energy balance, which is conservation of energy applied to a planet. Some sunlight never counts, because it is reflected straight back; the fraction reflected is the albedo. Work the balance out for a bare rock at the Earth's distance and you get about 255 K, which is −18 °C. The real average surface is about 288 K. The difference is the greenhouse effect: certain gases absorb the infrared the surface emits and send part of it back down, so the surface must be warmer to get rid of the same energy. Add more of those gases, which is what burning fossil fuels does, and the surface has to warm further. That is the enhanced greenhouse effect.
2A planet in energy balance
Treat the Earth as a system: a region you draw a boundary around so you can count the energy crossing it. Two flows cross the boundary. Sunlight comes in; infrared radiation goes out. Nothing else of any size does.
Conservation of energy says the energy inside changes by whatever comes in minus whatever goes out. So:
- If the power absorbed is greater than the power emitted, the internal energy rises and the planet warms.
- If the power emitted is greater, the planet cools.
- If they are equal, the temperature is steady. The planet is in equilibrium.
The equilibrium is self-correcting, which is why it is stable. Suppose the planet is too cold. It emits less, because emitted power goes as T⁴, so it absorbs more than it loses and warms. Too hot, and it emits more than it absorbs and cools. It settles at the one temperature where the two match, and every calculation in this subtopic is a hunt for that temperature.
In equilibrium: power absorbed = power emitted. Every greenhouse calculation starts by writing this line.
3The solar constant, and why you divide it by four
The solar constant S is the intensity of the Sun's radiation arriving at the top of the Earth's atmosphere, measured on a surface held at right angles to the rays. Its value is about 1.36 × 10³ W m⁻². It is simply the Sun's apparent brightness b = L ÷ 4πd² at the Earth's distance.
Here is the step people find strange. The mean intensity arriving over the whole Earth is not S. It is S ÷ 4. Figure 1 shows why.
Seen from the Sun, the Earth is a flat disc. The rays it intercepts are the ones that would pass through a circle of the Earth's radius R, so the power it catches is the intensity times the area of that disc:
That power is then shared over the whole surface of the sphere, day side and night side, which has area 4πR². Averaged over the surface and over a day:
Two things hide in that factor of four: night, since half the sphere gets nothing at any instant, and slant, since away from the point under the Sun the same beam is spread over tilted ground. The disc is the planet's projected surface along the rays, and it handles both at once.
4Albedo: the energy that never gets in
Not all of the 340 W m⁻² is absorbed. Clouds, ice, snow, deserts and the sea all scatter some of it straight back into space. Albedo measures the fraction scattered:
albedo = total scattered power ÷ total incident power
It is a ratio of two powers, so it has no unit, and it lies between 0 (absorbs everything) and 1 (reflects everything). This page writes it as α. The Earth's average albedo is about 0.30, so roughly 30% of incoming sunlight is reflected and 70% is absorbed. Figure 2 gives rounded typical values for different surfaces.
The guide asks you to explain why the Earth's albedo is not fixed. It changes from day to day and from place to place.
- Cloud. Thick cloud reflects much more than the ground or sea beneath it. Cloud cover changes by the hour, so the Earth's albedo changes by the day.
- Latitude. Near the poles the ground is covered by ice and snow, which have high albedo. The Sun is also low in the sky there, and water reflects far more when light arrives at a glancing angle than when it arrives from overhead. Both push the albedo up at high latitudes.
- Season and surface. Snow cover comes and goes with the seasons; forest, desert and farmland reflect differently; sea ice grows in winter and shrinks in summer.
With albedo α, the mean power absorbed per square metre is:
Keep 238 W m⁻² in your head. It is the number the whole Earth has to get rid of again.
5Emissivity: the energy that gets out
A black body is a perfect emitter, and it radiates σT⁴ from each square metre. Real surfaces emit less than that at the same temperature. Emissivity, e, is the ratio:
emissivity = power radiated per unit area ÷ σT⁴
So a real surface radiates a power
Emissivity has no unit and lies between 0 and 1; a black body has e = 1. Water, soil and vegetation have infrared emissivities close to 1, so a question will often let you treat the surface as a black body. A good emitter at a wavelength is also a good absorber at that wavelength, so the emissivity of a layer of gas also tells you what fraction of incoming infrared it absorbs. Section 8 uses both ideas.
6The temperature of a planet with no atmosphere
Now put sections 3 to 5 together. For a planet with no atmosphere, every square metre on average absorbs (1 − α)S ÷ 4 and emits eσT⁴. In equilibrium the two are equal:
(1 − α) × S ÷ 4 = e σ T⁴
Worked example 1: the Earth without its atmosphere. Take α = 0.30, S = 1.36 × 10³ W m⁻², and treat the surface as a black body.
Figure 3 draws the same balance as arrows.
The measured mean surface temperature of the Earth is about 288 K, or 15 °C. So the bare-planet model is 33 K too cold. Everything that follows in this subtopic is the explanation for those 33 K.
Worked example 2: an invented planet, with emissivity. A rocky planet with no atmosphere receives S = 900 W m⁻². Its albedo is 0.25 and the emissivity of its surface is 0.95. Estimate its equilibrium temperature.
An emissivity below 1 makes the surface warmer than a black body would be, not cooler. It emits less effectively, so it has to be hotter to shed the same power.
Worked example 3: a planet around another star. The guide says questions may use "solar or other constants". A planet orbits a star of luminosity 2.0 × 10²⁶ W at a distance of 1.2 × 10¹¹ m. Its albedo is 0.35. Find S for this planet, then its black-body temperature.
7Why greenhouse gases absorb infrared
The Sun and the Earth radiate in different parts of the spectrum, and the whole greenhouse effect rests on that difference. Wien's law gives the peaks. The Sun's surface is about 5800 K, so λₘₐₓ = 2.9 × 10⁻³ ÷ 5800 = 5.0 × 10⁻⁷ m, which is 0.50 μm, in the visible. The Earth's surface is about 288 K, so λₘₐₓ = 2.9 × 10⁻³ ÷ 288 = 1.0 × 10⁻⁵ m, which is 10 μm, in the infrared. Figure 4 draws both spectra on one wavelength axis.
The atmosphere treats the two very differently. Most of the incoming visible light passes through it to the ground. Much of the outgoing infrared does not get through, because some gases in the air absorb it. Those are the greenhouse gases. The guide names four as the main ones, and each comes from both natural sources and human activity.
| Gas | Natural sources | Human sources |
|---|---|---|
| Carbon dioxide, CO₂ | respiration, decay, volcanoes, wildfires | burning fossil fuels, deforestation, cement making |
| Methane, CH₄ | wetlands, digestion in wild animals, termites | cattle and sheep farming, rice paddies, landfill, leaks from gas and coal extraction |
| Water vapour, H₂O | evaporation from oceans and lakes, transpiration | little directly; it rises because warmer air holds more (section 9) |
| Nitrous oxide, N₂O | bacteria in soil and ocean | nitrogen fertilisers, manure, some industrial processes |
The two gases that make up 99% of dry air, nitrogen and oxygen, are not on the list. Their molecules vibrate in a way that does not shift the charge within them, so infrared radiation has nothing to push against and passes through.
The guide asks you to explain the absorption with two models. You need both. Figure 5 sets them side by side.
The molecular energy level model. A molecule can only hold certain amounts of vibrational energy: its energy levels are discrete. It can absorb a photon only if the photon's energy hf exactly matches the gap between two of its levels. For greenhouse gas molecules, some of those gaps correspond to infrared photons, so the molecule absorbs infrared and moves to an excited state (step ① in Figure 5). Shortly after, it drops back and emits a photon of the same energy, or passes the energy to neighbouring molecules in collisions. The emitted photon goes off in a random direction, not necessarily the direction the original was travelling (step ②). Some of that re-emitted radiation heads back down to the surface.
The resonance model. Picture the atoms in a molecule as masses and the bonds between them as springs. Such a system has natural frequencies at which it vibrates. For CO₂, one of those frequencies is about 2 × 10¹³ Hz, which is infrared radiation of wavelength close to 15 μm. When infrared of that frequency arrives, its oscillating electric field drives the vibration at the natural frequency, the amplitude builds up, and energy is transferred from the radiation to the molecule. That is resonance, the same effect as pushing a swing in time with its own rhythm, and it is why the absorption curve in Figure 5(b) has a sharp peak at f₀. The vibrating molecule then radiates, again in all directions.
Both models say absorption happens only at frequencies the molecule already "has". The resonance model is classical and easy to picture, but a classical spring can hold any energy, so it cannot explain why real absorption comes in discrete lines. The energy level model, a quantum picture, can. That is the limitation the guide's linking question is after.
What does absorption plus re-emission in all directions do to the surface? Radiation that would have gone straight to space is partly sent back down. The surface now receives sunlight and infrared from the atmosphere, so it warms until it emits enough to balance both. The next section puts numbers on that.
8The surface and the atmosphere: a two-part energy balance
The guide says energy-balance problems will include energy exchanged between the surface and the atmosphere. The simplest model that does this treats the atmosphere as a single layer. It is transparent to incoming sunlight, and it absorbs a fraction e of the infrared coming up from the surface, where e is the layer's emissivity. Being at a temperature of its own, it radiates eσT⁴ from both faces: up to space and down to the ground. Figure 6 shows the flows with e = 0.78, a value chosen because it reproduces the measured 288 K.
Check that each part balances, because that is exactly what an exam question asks you to do with one number missing. All values are in W m⁻².
| Part | In | Out |
|---|---|---|
| Surface | 238 from the Sun + 152 from the atmosphere = 390 | σT⁴ = 390 |
| Atmosphere | 0.78 × 390 = 304 from the surface | 152 up + 152 down = 304 |
| Whole planet, seen from space | 238 from the Sun | 86 passing through + 152 from the atmosphere = 238 |
The last row is worth seeing. From space, the planet still emits 238 W m⁻², exactly what the bare planet did. The atmosphere has not changed how much energy leaves; it has changed the surface temperature needed for that much to leave.
Worked example 4: find the surface temperature. In the model of Figure 6, the surface absorbs 238 W m⁻² of sunlight and 152 W m⁻² of radiation from the atmosphere. Treating the surface as a black body, determine its temperature.
Worked example 5: the emissivity of the Earth seen from space. A classic short question: the Earth's surface is at 288 K and the planet emits 238 W m⁻² to space. Treating the Earth as a single emitter at the surface temperature, determine its effective emissivity.
This effective 0.61 is not a property of any one surface. It is a way of saying that only 61% of what a black body at 288 K would radiate actually escapes.
How the model gives the surface temperature for any e. You will not be asked to derive this, but it shows why more absorption means a warmer surface. The atmosphere balances when eσTₛ⁴ = 2eσTₐ⁴, so it radiates half of what it absorbs in each direction. Putting that into the balance at the top of the atmosphere gives:
With e = 0 this is the bare planet, 255 K. With e = 0.78 it is 288 K. With e = 1 the surface reaches 303 K. Figure 7 plots the whole curve.
9The enhanced greenhouse effect
The natural greenhouse effect is good for life. Without it the mean surface temperature would be around −18 °C and most of the ocean would be frozen. The guide's term for the part added by people is the enhanced greenhouse effect: the augmentation of the greenhouse effect due to human activities.
The mechanism follows from Figure 7. Human activity adds greenhouse gases. The burning of fossil fuels is the primary cause, because it releases carbon that had been locked underground for millions of years as CO₂. Farming and landfill add methane; fertilisers add nitrous oxide. The concentration of CO₂ in the air has risen from about 280 parts per million before industrialisation to above 420 parts per million in the 2020s. More greenhouse gas means the atmosphere absorbs a larger fraction of the surface's infrared, so its effective emissivity e rises. More radiation is sent back down, and the surface warms until it again emits enough for the planet to balance. Figure 8 lays out the chain.
Two feedbacks make the warming larger than the gas alone would.
- Ice–albedo feedback. Warming melts ice and snow. The darker ground or sea underneath has a lower albedo, so more sunlight is absorbed and the surface warms further.
- Water vapour feedback. Warmer air can hold more water vapour, and water vapour is itself a greenhouse gas, so e rises again.
The way electricity is generated matters here: a coal or gas power station adds CO₂ with every joule it delivers, while wind, solar, hydroelectric and nuclear generation add almost none while running.
The one-layer model is only a model. The real atmosphere has many layers at different temperatures, clouds that reflect sunlight and absorb infrared, and heat carried by convection and evaporation. It is used because it captures the one idea that matters with arithmetic you can do in an exam.
10Where marks are lost
- Using S instead of S/4. S is measured facing the Sun. The average over a rotating sphere is S ÷ 4, because the planet intercepts a disc (πR²) and spreads it over a sphere (4πR²).
- Applying albedo to the emitted radiation. Albedo reduces the sunlight absorbed: absorbed = (1 − α)S ÷ 4. It does not appear on the emission side, which is eσT⁴.
- Using degrees Celsius in σT⁴. Stefan–Boltzmann needs kelvin. 15⁴ and 288⁴ differ by a factor of about 135,000.
- Forgetting the fourth root. After T⁴ = …, take the fourth root. An answer of 4 × 10⁹ K is not a temperature anyone should hand in.
- "Greenhouse gases trap heat like the glass of a greenhouse." Not accepted. They absorb infrared and re-emit it in all directions, so some returns to the surface. (A real greenhouse stays warm mainly by stopping warm air escaping, which is a different mechanism.)
- Saying greenhouse gases absorb the incoming sunlight. Mostly they do not. The incoming radiation is mainly visible and passes through; the absorbed radiation is the outgoing infrared.
- Explaining with energy levels but leaving out the re-emission direction. Absorption alone is only half the mark. The effect needs the re-emission in all directions, part of it back towards the surface.
11Draw it right
- An energy-balance diagram has three parts drawn distinctly: surface, atmosphere, space. Each arrow carries a label and a value in W m⁻².
- Sunlight arrows are drawn straight; infrared arrows are drawn differently (wavy, or in another colour), and the difference is stated in a key.
- The atmosphere emits from both faces. An atmosphere with only an upward arrow, or only a downward one, loses the mark.
- Check every box balances before you finish: in = out for the surface, for the atmosphere, and for the planet as a whole.
- When you sketch a spectrum, put the Sun's peak at about 0.5 μm and the Earth's at about 10 μm, and label the axes with quantity and unit.
- An energy level diagram shows horizontal levels, an upward arrow for absorption labelled with hf = ΔE, and re-emission in more than one direction.
12Try it
Marks in brackets. Take σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Answers and marker's notes are at the end.
Q1. (Paper 1A style) The solar constant at a planet is S and the planet's albedo is α. What is the mean intensity absorbed by the planet's surface, averaged over the whole planet? 1 mark
A. (1 − α)S B. (1 − α)S ÷ 2 C. (1 − α)S ÷ 4 D. αS ÷ 4
Q2. An invented planet has no atmosphere. The intensity of its star's radiation at the planet is 2.0 × 10³ W m⁻² and the planet's albedo is 0.40.
(a) Show that the mean intensity absorbed is 300 W m⁻². 2 marks
(b) Determine the equilibrium temperature of the surface, treating it as a black body. 2 marks
(c) The planet later gains an atmosphere that returns 180 W m⁻² of infrared to the surface. Calculate the new surface temperature. 2 marks
Q3. Explain, with reference to molecular energy levels, how the presence of carbon dioxide in the atmosphere raises the temperature of the Earth's surface. 4 marks
Q4. In a simple model of a planet, the surface absorbs 245 W m⁻² from the star and 165 W m⁻² from the atmosphere. The surface behaves as a black body.
(a) Determine the temperature of the surface. 2 marks
(b) The atmosphere absorbs 80% of the radiation emitted by the surface. Calculate the power per unit area emitted by the surface that reaches space directly. 2 marks
Q5. (Paper 1B style) A student places two identical sealed glass flasks under the same lamp. Flask X contains air; flask Y contains air enriched with carbon dioxide. The thermometers have an uncertainty of ±0.5 °C. The data below are invented for this question.
| Time / min | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| Flask X (air) / °C | 20.0 | 21.4 | 22.6 | 23.6 | 24.4 | 25.0 |
| Flask Y (CO₂-enriched) / °C | 20.0 | 21.9 | 23.5 | 24.8 | 25.8 | 26.5 |
(a) Calculate the mean rate of temperature rise of each flask over the first 6 minutes. 2 marks
(b) Discuss whether the data support the claim that flask Y warms faster than flask X. 3 marks
(c) Suggest one reason why this experiment is a poor model of the Earth's atmosphere. 1 mark
Q6. Outline two reasons why the albedo of the Earth is not constant. 2 marks
13In one breath
In equilibrium a planet absorbs as much power as it emits. It catches sunlight on a disc, πR², and spreads it over a sphere, 4πR², so the mean incoming intensity is S ÷ 4, about 340 W m⁻². Albedo, about 0.30, is the fraction scattered straight back; it is higher over cloud, ice and snow and at high latitudes, and it changes daily. The rest, (1 − α)S ÷ 4 ≈ 238 W m⁻², leaves as eσT⁴, which gives 255 K for a bare Earth against a real 288 K. The gap is the greenhouse effect: CO₂, CH₄, H₂O and N₂O absorb the surface's infrared because the photon energy matches a gap between molecular energy levels (or, in the resonance picture, a natural frequency of the bonds), then re-emit it in all directions, some back down, so the surface must be warmer to balance. Burning fossil fuels adds these gases and warms the surface further: the enhanced greenhouse effect, amplified by ice–albedo and water vapour feedback.
Answers
Q1. C. The planet intercepts power on a disc and spreads it over a sphere four times the area, and a fraction α is reflected. C only. A forgets the factor of 4; B divides by 2 for night alone; D is the reflected intensity, not the absorbed one.
Q2. (a) Mean incident intensity = 2.0 × 10³ ÷ 4 = 500 W m⁻². Absorbed = (1 − 0.40) × 500 = 300 W m⁻². (b) σT⁴ = 300, so T⁴ = 300 ÷ (5.67 × 10⁻⁸) = 5.29 × 10⁹ and T = 270 K. (c) Total absorbed = 300 + 180 = 480 W m⁻², so T⁴ = 480 ÷ (5.67 × 10⁻⁸) = 8.47 × 10⁹ and T = 303 K. (a) M1 for dividing by 4, A1 for applying (1 − α) to reach 300 (a "show that" needs both steps visible). (b) M1 for σT⁴ = 300, A1 for 270 K. (c) M1 for adding the two inputs, A1 for 303 K. ECF from (a). An answer in °C with no kelvin value loses the A1.
Q3. The Earth's surface emits infrared radiation. CO₂ molecules have discrete energy levels, and the energy of some infrared photons matches the difference between two of them, so those photons are absorbed and the molecule is excited. The molecule then de-excites and re-emits infrared photons in random directions, so part of the radiation that would have escaped to space is sent back towards the surface. The surface therefore receives extra energy and must reach a higher temperature before it emits enough to balance what it absorbs. 1 for the surface emitting infrared, 1 for photon energy matching the gap between energy levels, 1 for re-emission in all/random directions with some returning to the surface, 1 for the surface warming until a new balance. "CO₂ traps heat" with no mechanism scores 0.
Q4. (a) σT⁴ = 245 + 165 = 410 W m⁻², so T⁴ = 410 ÷ (5.67 × 10⁻⁸) = 7.23 × 10⁹ and T = 292 K. (b) The surface emits 410 W m⁻²; 20% passes through, so 0.20 × 410 = 82 W m⁻² reaches space directly. (a) M1 for summing the inputs and equating to σT⁴, A1 for 292 K. (b) M1 for the surface emission taken as 410 and multiplied by 0.20, A1 for 82 W m⁻². Using 0.80 in (b) gives 328, the absorbed power, and scores M0 A0.
Q5. (a) Flask X: (23.6 − 20.0) ÷ 6 = 0.60 °C min⁻¹. Flask Y: (24.8 − 20.0) ÷ 6 = 0.80 °C min⁻¹. (b) Each temperature rise is the difference of two readings, so its uncertainty is ±1.0 °C. At 6 minutes the rises are 3.6 ± 1.0 °C and 4.8 ± 1.0 °C; the ranges (2.6 to 4.6 and 3.8 to 5.8) overlap, so a single reading cannot separate the flasks. But Y is higher at every later reading and the gap widens, which suggests a real difference: supported, not proven; repeats with finer thermometers would settle it. (c) Any one: the glass itself absorbs infrared; the lamp's spectrum is not the Sun's; a sealed flask stops convection; the CO₂ concentration is far above the atmosphere's. (a) 1 for each rate with unit. (b) 1 for combining the uncertainties to ±1.0 °C or showing error ranges, 1 for noting the overlap, 1 for a conclusion that weighs the consistent trend against the overlap. (c) 1 for any sensible difference. "It supports the claim because Y is hotter" with no reference to uncertainty scores 0 in (b).
Q6. Any two: cloud cover changes from day to day, and cloud reflects more than the surface beneath; snow and ice cover varies with season and has a high albedo; at high latitudes the Sun is low and water reflects more at glancing angles; the type of surface (ocean, forest, desert) changes from place to place. 1 for each distinct reason with its effect on reflection. "Weather" alone scores 0.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section B.2 Greenhouse effect. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.