Energy Balance & the Greenhouse Effect

This workbook covers two physics lessons, one independent learning task, and your Collaborative Sciences Project (CSP).

By the end of this workbook you should be able to:

State and apply the conservation of energy to a planet's radiation balance.
Define emissivity and use the Stefan–Boltzmann law, P = eσAT⁴.
Explain the solar constant S, and why the mean incoming intensity on a planet is S/4.
Define albedo, and explain why Earth's albedo varies with cloud cover and latitude.
Explain the greenhouse effect using both the resonance model and molecular energy levels, and name the four main greenhouse gases.
Distinguish the natural greenhouse effect from the human-caused enhanced greenhouse effect.

Name

Part A — Lesson 1: Radiation, Emissivity, the Solar Constant and Albedo

50 minutes pairs with slides 1–19 of “The greenhouse effect and global warming”

All boards · Light

The Inverse Square
Law

Spread the same energy over a sphere and its area grows as r². So whatever you measure at a point — brightness, field strength, force — falls as 1/r². Pick a quantity, then move the detector.

Light intensity I ∝ 1/r²

Where the inverse square law turns up

Any influence that streams outward from a point and isn't absorbed obeys it — across mechanics, fields, waves and nuclear physics, and far beyond the exam spec.

?The law fails when the spreading isn't over a full sphere — a laser beam stays roughly parallel, and a long wire or charged plate spreads over a cylinder or plane, giving 1/r or a constant field instead.

A.1 Starter — conservation of energy 5 min

Energy cannot be created or destroyed; it can only be transferred from one form or place to another. This single idea, the conservation of energy, underpins everything in this unit: if a planet's average temperature is roughly constant from one year to the next, then the rate at which it absorbs energy must equal the rate at which it radiates energy away.

1Explain, in terms of conservation of energy, why a planet with a constant average temperature must be absorbing and emitting energy at the same average rate.

💡 Hint

Think about what it means for a quantity to be “conserved”. If the temperature isn't changing, is energy building up anywhere in the system?

Because the planet's temperature is constant, its internal (thermal) energy is not changing on average. By conservation of energy, energy in (absorbed radiation) must equal energy out (emitted radiation) over any complete cycle (e.g. a year).

1bIf a planet's absorbed power suddenly became greater than its emitted power, what would you expect to happen to its temperature over time? Explain your reasoning.

💡 Hint

Use the Stefan–Boltzmann law: emitted power depends on T⁴. As T rises, what happens to the emitted power, and does that push the system back towards balance?

The planet would warm up: absorbing more energy than it emits means its internal energy — and therefore its temperature — increases, which in turn increases the power it radiates (P ∝ T⁴) until a new, higher-temperature balance is reached.

A.2 Black-body radiators and the Stefan–Boltzmann law 10 min

Every object radiates electromagnetic energy because of its temperature. A black body is an idealised object that absorbs all radiation falling on it, reflecting and transmitting none. Because it doesn't reflect visible light, a cold black body looks black — but hot black bodies glow, and the hotter they are, the more power they radiate.

Stefan–Boltzmann law.

P = e σ A T⁴

P = power radiated (W) A = surface area (m²) T = absolute temperature (K)

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (the Stefan–Boltzmann constant — it's in your data booklet)

e = emissivity of the surface (see A.3)

Worked example

A tungsten filament has a surface area of 5.3 × 10⁻⁵ m², operates at 2500 K, and has emissivity 0.35. Calculate the power it radiates.

Answer:
P = eσAT⁴ = 0.35 × (5.67 × 10⁻⁸) × (5.3 × 10⁻⁵) × (2500)⁴ = 41 W

2[3]A black-body surface (e = 1) of area 0.020 m² is at a temperature of 350 K. Calculate the power it radiates.

💡 Hint

Substitute directly into P = σAT⁴. Work out 350⁴ first, then multiply by σ and A.

A.3 Emissivity 8 min

Definition. Emissivity, e, is the ratio of the power radiated per unit area by a surface to the power radiated per unit area by an ideal black surface at the same temperature.

e = (intensity radiated by the surface) / (intensity radiated by a black body at the same temperature)

e is dimensionless and 0 ≤ e ≤ 1. A perfect black body has e = 1.

Dark, dull surfaces tend to have emissivity close to 1 — they are good approximations to a black body. Shiny, silvered surfaces have emissivity close to 0: they reflect rather than absorb (and therefore emit) radiation efficiently.

3[3]Two identical metal spheres are held at the same temperature. One is polished silver (e ≈ 0.05), the other is coated in matt black paint (e ≈ 0.95). Which radiates more power, and by roughly what factor? Explain using the Stefan–Boltzmann law.

💡 Hint

Emissivity is a straight multiplier on P (same A and T for both spheres). Try dividing one emissivity by the other.

The black-painted sphere (e ≈ 0.95) radiates about 19 times more power than the polished sphere (e ≈ 0.05), since P is directly proportional to e for two objects at the same T and A (0.95/0.05 = 19).

A.4 The solar constant, S, and why the mean incoming intensity is S/4 12 min

Stars radiate as black bodies. The Sun radiates a total power of about 3.85 × 10²⁶ W in all directions. By the time this radiation has spread out over a sphere the size of Earth's orbit, its intensity has dropped considerably.

The solar constant, S. The solar constant S is the intensity of solar radiation arriving at the top of a planet's atmosphere, measured on a surface held at right angles to the Sun's rays.

For Earth: S = 1360 W m⁻² (this value is in your data booklet).

At any instant, only the side of the planet facing the Sun receives radiation — and even then, a curved surface is not everywhere perpendicular to the incoming rays. The simplest way to picture the total power intercepted is to imagine the planet's silhouette: a flat disc of radius R, with area πR², blocking the Sun's rays.

Figure A.1 The Earth intercepts sunlight over a projected disc of area πR², but re-radiates that absorbed energy from its entire rotating surface, area 4πR².

Figure A.2 The Sun's radiation spreads outward as an expanding sphere — only the sliver crossing the flat disc directly in front of Earth (red) is absorbed. That sliver is then permanently missing from the wavefront as the rest sweeps past into space.

Figure A.3 Close-up: the full disc of incoming sunlight (the same diameter as Earth) arrives as evenly spaced parallel rays that meet the curved surface. The glowing patch shows that same energy landing on a small, bright area at the equator, then slowly spreading over a much larger, dimmer area nearer the poles — the same energy diluted over more area, so polar regions are colder. (This same effect, combined with Earth's axial tilt, is also what drives the seasons.)

Total power intercepted by the planet: P = S × πR²

Because the planet rotates and its whole surface re-radiates that energy, the incoming power is effectively shared out over the full surface area of the sphere, 4πR², not just the disc that faces the Sun:

Mean incoming intensity = (S × πR²) / (4πR²) = S / 4

4[3]The solar constant depends only on a planet's distance from the Sun. Using P_sun = 3.85 × 10²⁶ W and S = P_sun / (4πd²), calculate the solar constant for Mars, given its average orbital distance is 2.28 × 10¹¹ m.

💡 Hint

Rearrange nothing — just substitute! Work out d² first, then 4πd², then divide P_sun by that.

A.5 Albedo 10 min

Definition. Albedo, α, is a measure of the average energy reflected off a macroscopic system, given by:

α = total scattered power / total incident power

Albedo has no units and ranges from 0 (a perfectly absorbing surface) to 1 (a perfectly reflecting surface). Earth's average albedo is about 0.3.

Earth's albedo is not a fixed number. It varies from day to day and place to place, mainly because of cloud cover and latitude. Fresh cloud tops and polar ice/snow are highly reflective (locally raising the albedo); oceans and dense forest are much darker (locally lowering it). Because global cloud cover is constantly changing, and because the Sun's rays strike different latitudes at very different angles, Earth's instantaneous, whole-planet albedo drifts from day to day even though its long-term average stays close to 0.3.

5[4]Explain why (a) an increase in cloud cover over the oceans, and (b) a decrease in polar sea-ice, would each change Earth's average albedo — and in which direction.

💡 Hint

Think about what colour/brightness does to reflection: a brighter surface reflects a larger fraction of the light hitting it than a darker one.

Figure A.4 What albedo measures: the fraction of incident power that is reflected (scattered) rather than absorbed.

Figure A.5 Typical albedo values for a range of everyday surfaces, compared with Earth's whole-planet average of about 0.3.

Open question. Look at Figure A.5. Towns and cities are dominated by surfaces like dark roads, roofs and car parks. Suggest two ways city planners could change the surfaces in a city to raise its local albedo, and explain the effect this could have on the local climate.

💡 Hint

There's no single correct answer here — discuss it with your group. You could think about materials/colours (e.g. roofing, road surfacing), or about adding features that change what covers the ground (e.g. planting, green roofs, reflective paint).

A.6 Putting it together: the net radiation Earth absorbs 5 min

Not all of the mean incoming intensity S/4 is absorbed — a fraction α is reflected straight back to space. The net intensity absorbed is:

Net absorbed intensity = (S/4) × (1 − α)

For Earth: (1360/4) × (1 − 0.3) = 340 × 0.7 = 238 W m⁻².

6[3]Using your value of S for Mars from Q4, and Mars's average albedo of about 0.25, calculate the net intensity absorbed at Mars's surface.

💡 Hint

You'll need your Q4 answer for S. Then apply Net absorbed intensity = (S/4)(1 − α).

Net absorbed intensity = (S/4)(1−α) = (589/4)(1−0.25) ≈ 147 × 0.75 ≈ 110 W m⁻².

Lesson 1 self-check

Rate yourself honestly before moving on — this is for you, not for marks. Choose a rating from the dropdown.

Rating	I can… statement
	I can state the conservation of energy and apply it to a planet's energy balance.
	I can define emissivity and use the Stefan–Boltzmann law, P = eσAT⁴.
	I can define albedo as total scattered power / total incident power.
	I can explain why Earth's albedo varies daily, with cloud cover and with latitude.
	I can state what the solar constant S represents and calculate it for another planet.
	I can explain why the mean incoming solar intensity on a planet is S/4, using the projected-area argument.

Part B — Lesson 2: The Greenhouse Effect

50 minutes pairs with slides 20–23 of “The greenhouse effect and global warming”

B.1 Starter — recap 5 min

7[3]Without looking back at Part A, write down: (a) the equation for albedo, (b) the equation linking mean incoming intensity to S, and (c) what emissivity measures.

B.2 Earth's energy balance without an atmosphere 10 min

If Earth had no atmosphere, its surface would radiate as an (almost) black body directly to space. At a stable equilibrium temperature, the power absorbed equals the power emitted, so we can set the two Stefan–Boltzmann-based expressions equal to each other:

eσT⁴ = (S/4)(1 − α)
T = ⁴√[ (S/4)(1 − α) / eσ ]

Using S = 1360 W m⁻², α = 0.3 and e ≈ 1: T ≈ 255 K (about −18 °C). Earth's actual average surface temperature is about 288 K (15 °C) — roughly 33 °C warmer. That difference is due to the natural greenhouse effect.

8[4]A hypothetical rocky planet has S = 900 W m⁻², albedo 0.15, and its surface radiates with emissivity 1. Calculate its equilibrium surface temperature assuming no atmosphere.

💡 Hint

Rearrange eσT⁴ = (S/4)(1 − α) to make T the subject: T = ⁴√[(S/4)(1−α)/(eσ)]. Work out the bracket first, then take the fourth root.

B.3 How greenhouse gases actually absorb and re-emit energy 15 min

Greenhouse gas molecules don't just “warm” by magic — IB Physics uses the resonance model to try to explain what is happening. A gas molecule's bonds can be modelled a little like tiny springs: the atoms can vibrate relative to one another, and each vibrational mode has its own natural frequency. When infrared radiation from Earth's surface passes through the atmosphere, it acts like a periodic driving force on these molecular “springs.” When this happens there is an effect known as resonance, which hugely increases the energy of the “springs”: a greenhouse gas molecule absorbs strongly only when the frequency of the incoming IR radiation matches (resonates with) its natural vibration frequency.

Figure B.1 Resonance response of a driven molecular oscillator. Absorption is large only near the molecule's natural vibration frequency.

The molecular energy levels model

A more complete quantum picture treats a molecule's vibrational (and rotational) states as a set of discrete, quantised energy levels. A greenhouse gas molecule absorbs an IR photon only if the photon's energy, ΔE = hf, exactly matches the energy gap between two of its allowed vibrational energy levels. Absorbing the photon promotes the molecule to the higher (excited) level.

The molecule doesn't stay excited for long. It relaxes back towards the ground state and, in doing so, re-emits the absorbed energy as infrared radiation. Crucially, this re-emission happens in random directions — up, down, and sideways — not just back the way the original radiation came from. As you might expect, some of the re-emitted radiation heads back down towards the surface, warming it further, rather than continuing out to space.

Figure B.2 Absorption of an IR photon promotes a molecule to a higher vibrational energy level; the energy is later re-emitted in all directions.

Not every gas in the atmosphere behaves this way. Nitrogen (N₂) and oxygen (O₂) make up about 99% of the air, but neither is a greenhouse gas: they are symmetric molecules, so the way they vibrate doesn't change their charge distribution enough to couple to the oscillating electric field of IR radiation and be driven into resonance. Asymmetric molecules like CO₂ and H₂O can, which is why greenhouse gases are able to absorb IR strongly even though they're only present in trace amounts.

9[3]Nitrogen (N₂) and oxygen (O₂) make up about 99% of Earth's atmosphere, yet neither is a greenhouse gas. Using the resonance model, suggest why they don't absorb infrared radiation the way CO₂ and H₂O do.

💡 Hint

A symmetric molecule's vibration doesn't change its charge distribution — so the oscillating electric field of an IR wave has no “handle” to grip and drive the molecule into resonance.

N₂ and O₂ are symmetric diatomic molecules, so when their atoms vibrate the molecule's charge distribution doesn't change. The oscillating electric field of an IR wave has no unequal charge to push and pull on, so it can't drive the molecule's vibration into resonance and no energy is absorbed. CO₂ and H₂O are asymmetric (or have polar bonds), so their vibrations do produce a changing charge distribution that the IR field can couple to and drive resonantly. This is why greenhouse gases can absorb IR strongly even though they make up only a tiny fraction of the atmosphere.

B.4 The four main greenhouse gases 10 min

Water vapour, carbon dioxide, methane and nitrous oxide are the four greenhouse gases you need to know in detail. Every one of them has both natural sources and sources created by human (anthropogenic) activity.

Sorting task. Read the jumbled facts below, then type each one into the correct cell of the blank table. Each fact belongs to exactly one gas and one column (natural or anthropogenic).

• Evaporation from oceans, rivers and lakes • Burning fossil fuels in power stations and cars • Wetlands, oceans, lakes and rivers • Aircraft (flight) emissions • Forest fires and volcanic eruptions • Flooded rice fields, farm animals and termites • Forests, oceans, soils and grasslands • Processing of coal, natural gas and oil, and burning biomass • Manufacture of cement and fertilisers, and deforestation

Greenhouse gas	Formula	Natural sources	Anthropogenic sources
Water vapour	H₂O (g)
Carbon dioxide	CO₂
Methane	CH₄
Nitrous oxide	N₂O

Greenhouse gas	Formula	Natural sources	Anthropogenic sources
Water vapour	H₂O (g)	Evaporation from oceans, rivers and lakes	Flight (aviation emissions)
Carbon dioxide	CO₂	Forest fires, volcanic eruptions, evaporation/outgassing from oceans	Burning fossil fuels in power plants and cars, burning forests
Methane	CH₄	Wetlands, oceans, lakes and rivers	Flooded rice fields, farm animals, termites, processing of coal/natural gas/oil, burning biomass
Nitrous oxide	N₂O	Forests, oceans, soils and grasslands	Burning fossil fuels, manufacture of cement and fertilisers, deforestation

10[2]Which of the four gases is most directly linked to agriculture and land use, through two different sources? Name both sources.

💡 Hint

Look back at the sorting-task facts — which gas's anthropogenic sources are farming-related in two separate ways?

B.5 The enhanced greenhouse effect 5 min

The natural greenhouse effect keeps Earth roughly 33 °C warmer than it would otherwise be — without it, Earth would likely be too cold to support life as we know it. The enhanced greenhouse effect refers specifically to the additional warming caused by human activity increasing the atmospheric concentration of greenhouse gases beyond their natural levels.

11[4]Classify each statement as describing the natural greenhouse effect (N) or the enhanced greenhouse effect (E): (a) Water vapour evaporating from the ocean absorbs outgoing IR radiation. (b) Rising CO₂ concentrations from fossil fuel combustion since 1850 have increased radiative forcing. (c) Termites release methane as part of the carbon cycle. (d) Deforestation reduces a carbon sink while also releasing stored carbon.

💡 Hint

Ask yourself: would this still be happening if humans had never industrialised? If yes, it's natural (N). If it's an increase caused by human activity, it's enhanced (E).

Lesson 2 self-check

Rate yourself honestly before moving on — this is for you, not for marks. Choose a rating from the dropdown.

Rating	I can… statement
	I can calculate a planet's no-atmosphere equilibrium temperature from S, albedo and emissivity.
	I can explain the greenhouse effect using the resonance model, including re-emission in all directions.
	I can explain why N₂ and O₂ are not greenhouse gases, but CO₂ and H₂O are.
	I can name the four main greenhouse gases and at least one natural and one anthropogenic source for each.
	I can distinguish the natural greenhouse effect from the enhanced (human-caused) greenhouse effect.

Part C — Synthesis Task: Reading Real Planetary Energy Budgets

Sylvia Knight's “Planetary energy budgets” infographic (Physics Review, February 2016) shows measured energy-flow data for Mars, Venus, Titan and Jupiter, each drawn the same way as the Earth energy-balance diagram on slide 20. Ask your teacher to display it so you can work through the questions below — the key numbers for Venus are reproduced in the table so you can calculate with them directly.

Venus energy budget (selected values)	Intensity / W m⁻²
Incoming solar radiation (short-wave)	656
Total short-wave reflected to space	496
Net energy absorbed by surface	22
Long-wave infrared radiation emitted to space	161
Long-wave infrared radiation emitted by surface, absorbed by atmosphere	17 154
Long-wave infrared radiation emitted by atmosphere back down to surface	17 132
Net energy emitted by surface	22

(Data adapted from Knight, S., “Planetary energy budgets,” Physics Review, February 2016, Hodder Education.)

12[3]Calculate Venus's albedo from the flux values in the table. Compare it to Earth's average albedo of about 0.3, and suggest why Venus's is so different.

💡 Hint

Albedo = total scattered power ÷ total incident power. Use the two intensity values in the top two rows of the table.

α = 496/656 ≈ 0.76. This is roughly two and a half times Earth's average albedo of 0.3, because Venus is almost completely covered by thick, highly reflective sulfuric-acid clouds, which scatter most incoming sunlight before it ever reaches the surface.

13[3]The “net energy absorbed by surface” and “net energy emitted by surface” values are both 22 W m⁻². Explain what this tells you, using the conservation of energy.

💡 Hint

If energy in and energy out are equal, what does that tell you about whether the surface's temperature is changing?

14[4]Notice that the long-wave flux emitted by Venus's surface (17 154 W m⁻²) is over 25 times larger than the incoming solar radiation (656 W m⁻²). Using ideas from B.3, explain qualitatively why Venus's thick CO₂ atmosphere allows such an enormous amount of infrared energy to circulate between the surface and the atmosphere before a comparatively tiny net amount finally escapes to space.

💡 Hint

Think about the re-emission idea from B.3: after a molecule re-emits a photon in a random direction, what happens if the atmosphere above and below it is also full of other absorbing molecules?

Venus's atmosphere is extremely opaque to infrared radiation (very high CO₂ concentration and pressure). Once the surface emits IR radiation, it is absorbed and re-emitted by atmosphere molecules many times over — in all directions, including back down to the surface — before a net amount can finally escape to space. This repeated absorption/re-emission cycle allows the internal long-wave flux to build up to a value many times larger than the original incoming solar flux, which is why Venus's actual surface temperature (about 737 K) is vastly higher than the simple no-atmosphere blackbody estimate.

Part D — Your Collaborative Sciences Project

The Collaborative Sciences Project (CSP) gives you the chance to work in an interdisciplinary team on a real scientific question, using the physics of energy balance and the greenhouse effect you've just met — but seen through the lens of a genuine investigation rather than a textbook problem. The purpose of the CSP is for students to collaborate to identify a problem using science-based reasoning, leading to action. The focus is on the process of collaborative problem-solving, not on producing a single “correct” product, and the CSP itself is not formally assessed.

Sample outcomes for your project

Your outcome shouldn't be a physics experiment alone — data is used to inform actions. What action would you plan or suggest? Almost any local action can work, provided it's grounded in the physics of energy balance and clearly linked to either changing the albedo of the area, reducing carbon dioxide emissions, or the tree/shrub coverage of the area.

A local rubbish-picking or clean-up drive that also promotes walking dogs locally instead of driving, cutting local vehicle emissions.
A fashion show promoting second-hand clothing, to reduce the energy cost of producing new clothes.
A publicity campaign encouraging people to delete unused files from cloud storage, to reduce the energy used by data centres.
Designing simple sensors that help reduce a building's energy use.
Modelling and proposing the most effective recycling system for your school.
A campaign for reflective, brightly-coloured roofs and “white roads” (light-coloured road surfacing), to raise the local albedo.
A campaign promoting more solar panels on roofs.

The product could be a scientific poster, a planning document, or something to inspire action such as visual art (e.g. painting, sculpture, installation), performing art (e.g. theatre, film, music, dance), or literature (e.g. poetry, short story). Whatever the medium, it must connect to both a local context and a global issue, be documented in a process journal (in a medium of your choice — video, narrated presentation, booklet, website), and be shared with the school community.

D.1 Project proposal (complete in Session 1, then get mentor sign-off)

Team members & roles

Investigation question

Link to a global issue (e.g. climate change, urban heat islands, land-use change)

Independent variable

Dependent variable

Controlled variables

Outline of method (bullet points)

Equipment list

Risk assessment

Hazard	Who might be harmed?	Control measures

Mentor approval — digital signature & date

D.2 Sample experiments

Sample experiments

These two options are illustrative starting points for your investigation — your own project doesn't have to be exactly either one, provided it's grounded in the same physics.

Option A — Measuring the albedo of different surfaces
Links directly to Understandings: albedo formula; Earth's albedo varies with cloud/latitude; conservation of energy

Aim. To measure and compare the albedo of a range of real surfaces (e.g. grass, tarmac/concrete, bare soil, white card, aluminium foil, water, gravel), and relate your results to why Earth's own albedo varies with cloud cover, ice cover and latitude.

Background theory. Albedo α = total scattered power ÷ total incident power. A real satellite-based albedometer uses a matched pair of sensors — one facing the sky to record incoming (incident) intensity, one facing the ground to record reflected intensity — and takes the ratio. You will build a simplified version of the same method.

Analysis & evaluation prompts:

Rank your surfaces from lowest to highest albedo. Does the order match what you'd predict from colour/texture?
How would your results change on a cloudy day compared with a clear day? Link your answer to why Earth's whole-planet albedo varies daily.
What is the main source of uncertainty in your method? How could a real satellite-based measurement avoid it?
If your school is at a high or low latitude, discuss how the angle of the Sun's rays might affect your measured values at different times of day.

Option B — Energy transfer under different foliage/canopy coverings
Links directly to Understandings: solar constant S and S/4; conservation of energy; absorption/emission of radiation

Aim. To investigate how different types or densities of foliage/canopy covering affect the amount of solar energy reaching the ground beneath them, as a simple physical model for how vegetation cover and land use affect a local (and, in aggregate, global) energy budget.

Background theory. A canopy of leaves — or an artificial substitute such as shade netting of different grades — intercepts and scatters incoming short-wave solar radiation before it reaches the ground, in much the same way that clouds intercept sunlight over the oceans (A.5). Comparing the energy reaching the ground with and without different coverings lets you quantify a “local albedo/shielding effect.” Note: this experiment models shielding of incoming shortwave radiation, which is not the same mechanism as the atmospheric greenhouse effect (which traps outgoing longwave radiation) — a good discussion point for your evaluation.

Analysis & evaluation prompts:

Which covering blocked the most energy? Does this match the density/thickness you observed?
Estimate, using your control reading and S = 1360 W m⁻², what fraction of the actual solar constant your open-sky reading represents (accounting for cloud cover, angle of the Sun, and atmospheric losses).
Explain, in your own words, why this experiment models shielding of incoming shortwave radiation rather than the trapping of outgoing longwave radiation that defines the atmospheric greenhouse effect.
What real-world land-use decision could your data inform (e.g. urban tree planting, greenhouse/polytunnel design, solar panel siting)?

D.3 Sharing the outcome of your project

Your data informed something — but what? What will you now do? That outcome or action needs to be shared alongside your scientific data, not treated as an afterthought. Choose whichever medium suits your team, your data and your action best: a scientific poster, a short video, a podcast, a slide deck, a web page, or another format agreed with your mentor. Whatever the format, make sure it includes:

The investigation question and its link to a local context and a global issue.
Your method, in enough detail that someone else could repeat it.
Your data and at least one clear graph or table.
Your conclusion and evaluation, and how they informed the action you took (or plan to take).
A description of the action itself — what you did, planned or executed, and why.

D.4 Individual 100-word reflection

Write this individually, honestly, and specifically — it should be a genuine commentary on your experience, not a list of what you did. Consider:

A challenge and a success in collaborating with your team.
A challenge and a success in communicating your ideas or findings.
One thing you now understand about energy balance or the greenhouse effect that you didn't before.
One or two salient successes, and one or two challenges and how your team overcame them.
A brief reflection on how your approach to learning skills — particularly collaboration and communication — developed over the project.

Part E — Understanding Coverage Checklist

For teacher (and student) reference: exactly where each required Understanding is introduced, practised and evidenced.

#	Understanding	Introduced in	Workbook Qs	Further evidence
1	The conservation of energy.	A.1 (starter); B.2 (energy-balance derivation)	Q1, Q1b, Q13	CSP data analysis — checking absorbed ≈ emitted
2	Emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature.	A.2–A.3	Q2, Q3	Lesson 1 self-check
3	Albedo as a measure of the average energy reflected off a macroscopic system: albedo = total scattered power / total incident power.	A.5, Figures A.4–A.5	Q5, Q12	CSP Option A directly measures albedo
4	Earth's albedo varies daily and is dependent on cloud formations and latitude.	A.5	Q5, Q5b (open, city albedo)	CSP Option A (weather-condition notes); Lesson 1 self-check
5	The solar constant, S.	A.4	Q4	CSP Option B (comparison to S)
6	Incoming radiative power depends on the planet's projected surface along the direction of the rays, giving a mean incoming intensity of S/4.	A.4 (with Figure A.1 diagram)	Q4, Q6	Synthesis Part C
7	Methane, water vapour, carbon dioxide and nitrous oxide are the main greenhouse gases, each with both natural and human (anthropogenic) origins.	B.4	Q10 & sorting table	Lesson 2 self-check
8	Absorption of infrared radiation by the main greenhouse gases in terms of molecular energy levels, and subsequent emission of radiation in all directions.	B.3 (The molecular energy levels model), Figure B.2	Q9	Part C, Q14 (Venus)
9	The greenhouse effect can be explained in terms of both a resonance model and molecular energy levels.	B.3, Figures B.1–B.2	Q9	—
10	The augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect.	B.5	Q11	CSP reflection (D.4)