IB DP Physics: The greenhouse effect and planetary energy balance (DP Physics)

This study guide follows the IB Physics guide, first assessment 2025, and covers all of topic B.2 (Greenhouse effect) in the theme on the particulate nature of matter. It applies to Standard Level and Higher Level alike: the guide sets no extra content for Higher Level alone in B.2. B.2 builds on the radiation ideas in B.1, so read the B.1 study guide on the molecular model and heating first if the Stefan-Boltzmann law is new to you; the B.1 study guide on conduction, convection and thermal radiation covers it.

Pair this page with the revision notes and the practice set. The course page is at IB DP Physics, the printable list is the IB DP Physics checklist, and the free 10-minute diagnostics help you find weak spots. For the overall course layout see the syllabus guide.

What this unit covers

B.2 understanding	What you must be able to do	Level
Conservation of energy	Use it to balance power in and power out	SL and HL
Emissivity	Compare a surface with an ideal black surface at the same temperature	SL and HL
Albedo	Calculate it from scattered and incident power; explain how it varies	SL and HL
Solar constant S	Use it with the planet’s projected area	SL and HL
Mean incoming intensity S/4	Explain where the factor of 4 comes from	SL and HL
Main greenhouse gases	Name CH4, H2O, CO2 and N2O with natural and human origins	SL and HL
Absorption and re-emission	Explain with molecular energy levels, emission in all directions	SL and HL
Resonance model	Explain the greenhouse effect as a resonance	SL and HL
Enhanced greenhouse effect	Link extra human-made warming to fossil fuel burning	SL and HL
Equilibrium temperature	Solve energy balance problems, including surface-atmosphere exchange	SL and HL

Constants such as σ (the Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴) should be supplied or recalled; the worked examples below state every value they use.

Conservation of energy and the idea of balance

Energy cannot be created or destroyed, only transferred. For a planet this gives a simple rule: if the power absorbed is larger than the power radiated away, the planet’s thermal energy rises, and so does its temperature. If the two are equal, the temperature stays constant. That steady state is the equilibrium temperature.

A small imbalance matters over long times. Suppose a 100 m deep layer of ocean (1.0 × 10⁵ kg under each square metre, specific heat capacity 4200 J kg⁻¹ K⁻¹) absorbs a net 0.8 W m⁻² more than it loses.

Energy needed per kelvin under 1 m²: 1.0 × 10⁵ × 4200 = 4.2 × 10⁸ J K⁻¹. Rate of warming: 0.8 / 4.2 × 10⁸ = 1.9 × 10⁻⁹ K s⁻¹. In one year (3.15 × 10⁷ s): 0.060 K.

A tiny imbalance therefore adds up to a measurable warming over decades.

Emissivity

A black surface is an ideal emitter: at temperature T it radiates σT⁴ for each square metre. Real surfaces radiate less. Emissivity is the ratio of the power per unit area a surface radiates to the power per unit area an ideal black surface radiates at the same temperature:

emissivity e = (P/A) / σT⁴

It is a pure ratio, so it has no unit, and its value cannot exceed 1. Rearranged, a real surface radiates P = eσAT⁴.

Worked example 1. A rock face of area 2.0 m² is at 330 K and radiates 1.24 kW. Find its emissivity.

P/A = 1240 / 2.0 = 620 W m⁻². σT⁴ = 5.67 × 10⁻⁸ × 330⁴ = 672 W m⁻². e = 620 / 672 = 0.92.

Albedo

Albedo measures how much of the incident radiation a body scatters (reflects) back to space:

albedo α = total scattered power / total incident power

It is a ratio, so it has no unit. A fresh snowfield has a high albedo; open ocean and dark forest have low values. The fraction absorbed is 1 − α.

Earth’s albedo is not fixed. It changes through the day and with the season as cloud cover moves. It depends strongly on cloud formations, because thick cloud scatters a large fraction of sunlight. It also depends on latitude, because ice and snow cover polar regions while ocean and vegetation dominate elsewhere, and because the angle at which sunlight meets the surface changes. A convenient average for Earth is about 0.30.

Worked example 2. A planet receives 1.7 × 10¹⁷ W of sunlight and scatters 5.1 × 10¹⁶ W. Find its albedo and the power absorbed.

α = 5.1 × 10¹⁶ / 1.7 × 10¹⁷ = 0.30. Absorbed = 0.70 × 1.7 × 10¹⁷ = 1.2 × 10¹⁷ W.

The solar constant and the factor of one quarter

The solar constant S is the intensity of solar radiation at the top of a planet’s atmosphere, measured on a surface facing the Sun, at the planet’s distance from the star. For Earth S is about 1360 W m⁻².

Sunlight arrives as parallel rays. The planet intercepts only those rays that hit its projected disc, of area πR². But the planet radiates from its whole surface, of area 4πR². Sharing the intercepted power over the whole surface gives the mean incoming intensity:

mean intensity = S πR² / 4πR² = S/4

So for Earth the mean is 1361/4 = 340 W m⁻² (using S = 1361 W m⁻²). Including albedo, the mean absorbed intensity is S(1 − α)/4.

Worked example 3. Find the power Earth intercepts, using S = 1361 W m⁻² and R = 6.37 × 10⁶ m.

P = S × πR² = 1361 × π × (6.37 × 10⁶)² = 1.73 × 10¹⁷ W. Check: dividing by 4πR² = 5.10 × 10¹⁴ m² gives 340 W m⁻².

Equilibrium temperature with no atmosphere

For a planet with no greenhouse effect, balance the absorbed and emitted intensities (both per square metre of the whole surface):

S(1 − α)/4 = eσT⁴

T = [ S(1 − α) / (4eσ) ]^(1/4)

Worked example 4. A planet has S = 2150 W m⁻², α = 0.28 and e = 1. Find its equilibrium temperature.

Absorbed: 2150 × 0.72 / 4 = 387 W m⁻². T⁴ = 387 / 5.67 × 10⁻⁸ = 6.83 × 10⁹ K⁴. T = 287 K (about 14 °C).

Worked example 5. Earth, S = 1361 W m⁻², α = 0.30. With e = 1 the absorbed intensity is 238 W m⁻² and T = 255 K, about −18 °C. With e = 0.95 the same absorbed intensity gives T = 258 K, because a less efficient emitter must be hotter to lose the same power.

Earth’s actual mean surface temperature is nearer 288 K. The 33 K gap is the greenhouse effect at work: the greenhouse effect helps maintain life by keeping the surface warm enough for liquid water.

The main greenhouse gases

Gas	Natural origins	Human-created origins
Carbon dioxide CO2	Respiration, decay, volcanoes, release from oceans	Burning fossil fuels, cement making, clearing forests
Methane CH4	Wetlands, termites, ocean sediments	Cattle and sheep farming, rice paddies, landfill, gas and coal extraction
Nitrous oxide N2O	Bacteria in soils and oceans	Nitrogen fertilisers, some industrial processes, burning fuels
Water vapour H2O	Evaporation from oceans, lakes and plants	Mainly an indirect effect: a warmer atmosphere holds more water vapour

The guide expects you to know that each gas has both natural and human-created sources. The guide singles out the burning of fossil fuels as the main human cause of the enhanced greenhouse effect.

How a greenhouse gas absorbs infrared

The Sun is hot (about 5800 K) so its radiation peaks in the visible region. By Wien’s displacement law, λmax = 2.9 × 10⁻³ / 5800 = 5.0 × 10⁻⁷ m. The Earth’s surface is much cooler: at 288 K, λmax = 2.9 × 10⁻³ / 288 = 1.0 × 10⁻⁵ m, which is infrared. The atmosphere lets most sunlight through but absorbs much of the infrared the surface sends back.

Molecular energy levels. A molecule can only hold certain vibrational energies. A photon is absorbed when its energy hf equals the gap between two levels. For the main greenhouse gases the gaps match infrared photons. Take a photon of wavelength 15 μm (an infrared band absorbed by CO2):

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 15 × 10⁻⁶ = 1.33 × 10⁻²⁰ J.

The molecule is left in a higher level. It later returns to a lower level and emits a photon, in a random direction. So the radiation is re-emitted in all directions: some goes out to space, and about half goes back down towards the surface. The molecule can also pass its energy to other molecules in collisions, warming the air.

Resonance model. Think of the bonds in a molecule as tiny oscillators with natural frequencies. When the electromagnetic wave has the same frequency as a natural frequency, the molecule is driven at resonance. It absorbs energy efficiently, oscillates with a large amplitude, and radiates in all directions. Both models give the same result: infrared at particular frequencies is absorbed strongly, other radiation is not.

The enhanced greenhouse effect

The natural greenhouse effect exists without human action. The enhanced greenhouse effect is the extra warming caused by human activities that raise greenhouse gas concentrations. More molecules absorb more of the outgoing infrared, more is re-emitted downward, and the surface warms until the power escaping to space again matches the power absorbed.

Energy balance with the atmosphere

When the problem includes energy exchanged between surface and atmosphere, balance each layer in turn. The question gives you the model. Here are two standard ones, per square metre, with I(in) = S(1 − α)/4 the absorbed intensity.

Model A: one layer that absorbs all the surface infrared. Sunlight passes through. The atmosphere radiates σTA⁴ upward and the same downward.

Top of atmosphere: σTA⁴ = I(in). Surface: I(in) + σTA⁴ = σTs⁴, so σTs⁴ = 2I(in). Ts = 2^(1/4) × Te, where Te is the no-atmosphere temperature.

For Earth: Ts = 1.189 × 254.6 = 303 K. This is higher than the real 288 K, because the model absorbs all the infrared.

Model B: the layer absorbs a fraction a of the surface infrared. The layer radiates with emissivity a.

Atmosphere: aσTs⁴ = 2aσTA⁴, so TA⁴ = Ts⁴/2. Top of atmosphere: (1 − a)σTs⁴ + aσTA⁴ = I(in), so σTs⁴(1 − a/2) = I(in).

Worked example 6. Earth with I(in) = 238 W m⁻² and a = 0.78.

Ts⁴ = 238.2 / [5.67 × 10⁻⁸ × (1 − 0.39)] = 6.89 × 10⁹ K⁴. Ts = 288 K. TA = Ts / 2^(1/4) = 242 K.

Now raise a to 0.82 as the concentration of greenhouse gases grows: Ts = 290 K, a rise of 2.4 K. That is the enhanced greenhouse effect in a model.

Common errors

• Using S instead of S/4 for the mean intensity of the whole planet.
• Forgetting the (1 − α) factor, or using α where 1 − α belongs.
• Using °C in σT⁴. Convert to kelvin.
• Taking the fourth root of the wrong quantity, or forgetting to take it at all.
• Saying greenhouse gases “trap” heat or “reflect” infrared, rather than absorb and re-emit in all directions.
• Treating albedo as a percentage of what is emitted, when it is a fraction of what is incident.
• Treating emissivity above 1 as possible.

Official syllabus

Source: the Physics guide for the IB Diploma Programme (first assessment 2025), International Baccalaureate Organization. Topic B.2, Greenhouse effect.