DP Physics SL B.2 Greenhouse Effect

DP Physics SL B.2 Greenhouse Effect - Full notes

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DP Physics SL | B.2 Greenhouse Effect

DP PHYSICS · THEME B: THE PARTICULATE NATURE OF MATTER

B.2

Greenhouse Effect

Full notes for SL · IB-style

Contents

1	Energy balance of the Earth
2	The greenhouse effect
3	The enhanced greenhouse effect and climate change
4	Data-based and graph questions
5	Practical investigations
6	Exam technique and further practice
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
compare solar and terrestrial radiation using Wien's law
use the solar constant, S/4 and albedo to find absorbed intensity
use emissivity and energy balance to calculate equilibrium temperatures
explain the greenhouse effect and the molecular mechanism of infrared absorption
interpret energy-flow diagrams and atmospheric absorption spectra
explain the enhanced greenhouse effect, feedbacks and consequences of climate change

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How does the greenhouse effect explain Earth's surface temperature? How do changes in the atmosphere affect the energy balance of the Earth? What role does physics play in understanding climate change?

How this topic is examined

B.2 appears in Papers 1 and 2: energy balance calculations with albedo and emissivity, explanations of greenhouse gas absorption, energy-flow diagrams and data on climate change.

Show the energy balance equation explicitly and remember the fourth root.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Energy balance of the Earth

Energy in must equal energy out.

Think about it

If the Sun delivers about 1360 W m⁻² to Earth, why isn't Earth's surface as hot as an oven?

Earth's temperature is set by a balance between energy absorbed from the Sun and energy radiated to space. This section uses the laws of black-body radiation from B.1 to model that balance.

1.1 The Sun and the Earth as radiators

The Sun (about 5800 K) emits mostly visible and near-infrared radiation, peaking near 500 nm. Earth (about 288 K) emits long-wavelength infrared, peaking near 10 μm. This difference in wavelength is the key to the greenhouse effect.

Figure 1. Emission spectra of the Sun and the Earth (each normalised to its own peak).

WORKED EXAMPLE 1 Peak wavelengths (standard)

Use Wien's law to find the peak wavelengths for the Sun (5800 K) and Earth (288 K).

Solution

Sun: 2.9 × 10⁻³/5800 = 0.50 μmA1

Earth: 2.9 × 10⁻³/288 = 10 μmA1

1.2 The solar constant

The solar constant S is the intensity of solar radiation at Earth's mean distance from the Sun, measured above the atmosphere: S ≈ 1360 W m⁻². It follows from S = L☉/4πd².

WORKED EXAMPLE 2 From luminosity (standard)

Calculate S from L☉ = 3.83 × 10²⁶ W and d = 1.50 × 10¹¹ m.

Solution

S = 3.83 × 10²⁶/(4π × (1.50 × 10¹¹)²)M1

= 1350 W m⁻²A1

WORKED EXAMPLE 3 Mars (exam level)

Mars is 1.52 times further from the Sun than Earth. Find the solar constant at Mars.

Solution

S ∝ 1/d²: 1360/1.52²M1

= 589 W m⁻²A1

Quick check

1. Why does the intensity at Earth vary slightly through the year?

Answers: 1. Earth's orbit is slightly elliptical, so d varies

1.3 Distance from the Sun

The solar intensity falls as 1/d², so planets further from the Sun receive much less energy per square metre and have lower equilibrium temperatures, T ∝ d−1/2.

Figure 2. Equilibrium temperature of a planet with albedo 0.30 against distance from the Sun.

WORKED EXAMPLE 4 Distance (exam level)

By what factor does the equilibrium temperature change if a planet's distance from its star doubles (same albedo)?

Solution

Intensity falls to 1/4; T ∝ (intensity)1/4M1

T falls by factor 41/4 = 1.41 (to 71%)A1

1.4 Latitude and seasons

Sunlight reaching high latitudes strikes the ground at a shallow angle, so the same beam is spread over a larger area and the intensity on the ground is lower: I = S cos(latitude) when the Sun is overhead at the equator. Earth's 23.4° axial tilt changes this through the year, producing seasons.

Figure 3. Parallel sunlight is spread over a larger area at high latitudes.

Figure 4. Intensity on horizontal ground at noon (no atmosphere, Sun overhead at the equator).

WORKED EXAMPLE 5 Intensity at 60° (standard)

Find the noon intensity on horizontal ground at latitude 60° when the Sun is overhead at the equator.

Solution

I = 1360 cos 60°M1

= 680 W m⁻²A1

Common mistake

Thinking seasons are caused by changes in Earth–Sun distance. They are caused by the tilt of Earth's axis.

1.5 Average incoming intensity: S/4

Earth intercepts sunlight over a disc of area πR² but has surface area 4πR², and rotates so the whole surface receives it on average. So the mean incoming intensity over the surface is S/4 ≈ 340 W m⁻².

Figure 5. Earth intercepts sunlight over a disc of area πR² but radiates from its whole surface.

WORKED EXAMPLE 6 Mean intensity (standard)

Show that the mean intensity is about 340 W m⁻².

Solution

Power intercepted = SπR²; spread over 4πR²M1

1360/4 = 340 W m⁻²A1

Common mistake

Dividing by 2 instead of 4. The factor 4 is the ratio of total surface area to the area of the intercepting disc.

1.6 Albedo

Albedo α is the fraction of incident radiation that is reflected: α = reflected power/incident power. Earth's mean albedo is about 0.30. Fresh snow and ice have high albedo (up to 0.9); oceans and forests have low albedo (about 0.06–0.15). Albedo varies with cloud cover, season and latitude.

Surface	Albedo
fresh snow	0.80–0.90
sea ice	0.50–0.70
clouds (thick)	0.60–0.90
desert sand	0.40
grassland	0.25
forest	0.10–0.15
ocean	0.06

WORKED EXAMPLE 7 Absorbed intensity (standard)

Find the mean intensity absorbed by Earth if its albedo is 0.30.

Solution

Absorbed = (1 − 0.30) × 340M1

= 238 W m⁻²A1

1.7 How albedo affects temperature

Because absorbed power depends on (1 − α), increasing albedo lowers the equilibrium temperature. A completely snow-covered Earth ('snowball Earth') would be far colder than today.

Figure 6. Equilibrium temperature (no greenhouse effect) against planetary albedo.

WORKED EXAMPLE 8 Snowball Earth (exam level)

Estimate the equilibrium temperature of an Earth with albedo 0.65.

Solution

σT⁴ = 0.35 × 340 = 119 W m⁻²M1

T = 214 KA1

1.8 Emissivity

Emissivity e is the ratio of the power radiated by a surface to that radiated by a black body at the same temperature: P = eσAT⁴, with 0 ≤ e ≤ 1. Earth's surface radiates almost as a black body in the infrared, but the planet as seen from space has an effective emissivity of about 0.61 because of the greenhouse effect.

WORKED EXAMPLE 9 Emissivity (standard)

A surface at 300 K radiates 300 W m⁻². Find its emissivity.

Solution

e = 300/(5.67 × 10⁻⁸ × 300⁴)M1

= 0.65A1

1.9 Equilibrium temperature without an atmosphere

In equilibrium, absorbed power equals emitted power: (1 − α)S/4 = σT⁴. With α = 0.30 this gives T ≈ 255 K (−18 °C). The actual mean surface temperature is about 288 K; the 33 K difference is the natural greenhouse effect.

WORKED EXAMPLE 10 Equilibrium temperature (exam level)

Calculate Earth's equilibrium temperature with α = 0.30 and e = 1.

Solution

σT⁴ = 0.70 × 1360/4 = 238 W m⁻²M1

T = (238/5.67 × 10⁻⁸)1/4 = 255 KA1

WORKED EXAMPLE 11 Including emissivity (exam level)

Show that an effective emissivity of 0.61 gives a surface temperature of about 288 K.

Solution

T⁴ = 238/(0.61 × 5.67 × 10⁻⁸)M1

T = 288 KA1

2 The greenhouse effect

How the atmosphere keeps Earth warm.

2.1 The natural greenhouse effect

Short-wavelength solar radiation passes largely through the atmosphere and warms the surface. The surface emits long-wavelength infrared, much of which is absorbed by greenhouse gases. These gases re-emit infrared in all directions, including back towards the surface, raising its temperature above the no-atmosphere value.

Figure 7. Simplified global energy flows in W m⁻² (average values).

Key point

The atmosphere is largely transparent to incoming short-wavelength radiation but absorbs outgoing long-wavelength infrared.

2.2 Greenhouse gases

The main greenhouse gases are water vapour (H₂O), carbon dioxide (CO₂), methane (CH₄) and nitrous oxide (N₂O). Nitrogen and oxygen, which make up 99% of the atmosphere, are not greenhouse gases.

Gas	Main sources	Approx. contribution to natural effect
water vapour	evaporation	about 50%
carbon dioxide	respiration, combustion, volcanoes	about 20%
clouds	condensed water	about 25%
methane	wetlands, livestock, gas leaks	small but growing
nitrous oxide	soils, fertilizers	small

2.3 Molecular mechanism

Greenhouse gas molecules absorb infrared photons whose frequencies match the natural frequencies of their bending and stretching vibrations (a resonance effect). Only vibrations that change the molecule's dipole moment absorb infrared. The excited molecule re-emits infrared in a random direction or transfers energy to other molecules by collisions.

Figure 8. Vibration modes of a CO₂ molecule. The bend and the asymmetric stretch absorb infrared.

WORKED EXAMPLE 12 Why not N₂? (exam level)

Explain why nitrogen molecules do not absorb infrared radiation.

Solution

Symmetric diatomic molecule: vibration does not change its dipole momentR1

So it cannot couple to the oscillating electric field of infrared radiationR1

2.4 Absorption spectra of the atmosphere

The atmosphere absorbs strongly at certain infrared wavelengths (water vapour bands, CO₂ at 4.3 and 15 μm). Between 8 and 13 μm there is an 'atmospheric window' where infrared escapes to space, near the peak of Earth's emission.

Figure 9. Simplified transmission of infrared through the atmosphere (low values mean strong absorption).

Quick check

1. Why is the 15 μm absorption band of CO₂ so important?

Answers: 1. it lies near the peak of Earth's thermal emission

2.5 Comparing with a real greenhouse

A glass greenhouse warms mainly because it stops convection: warm air cannot rise and be replaced by cooler air. Glass also absorbs some outgoing infrared. The atmospheric greenhouse effect works by infrared absorption and re-emission, not by blocking convection, so the name is only an analogy.

Quick check

1. Why is the term 'greenhouse effect' slightly misleading?

Answers: 1. real greenhouses warm mainly by preventing convection, not by infrared absorption

2.6 Comparing greenhouse gases

Gases differ in how strongly each molecule absorbs infrared and in how long they stay in the atmosphere. Global warming potential (GWP) compares the warming caused by 1 kg of a gas with 1 kg of CO₂ over 100 years.

Gas	Atmospheric lifetime	GWP (100 years)
CO₂	centuries (part)	1
methane	about 12 years	about 28
nitrous oxide	about 110 years	about 270
CFC/HFC gases	decades to centuries	hundreds to thousands

WORKED EXAMPLE 13 CO₂-equivalent (standard)

A farm emits 2.0 tonnes of methane per year. Find the CO₂-equivalent emission (GWP = 28).

Solution

2.0 × 28M1

= 56 tonnes CO₂eA1

2.7 The atmospheric window

Most of Earth's emitted infrared lies between 5 and 40 μm. The atmosphere absorbs most of it except in the 8–13 μm window, so only part of the surface emission escapes directly to space; the rest is absorbed and re-emitted, much of it back down.

Figure 10. Surface emission and the (simplified) fraction escaping directly to space.

Quick check

1. How would adding a gas that absorbs strongly at 10 μm affect the climate?

Answers: 1. it would close the window, causing strong warming

2.8 A simple one-layer model

Model the atmosphere as a single layer that is transparent to sunlight but absorbs a fraction of outgoing infrared and re-radiates half up and half down. With full absorption, the surface flux is twice the outgoing flux, raising the surface temperature by a factor of 21/4 ≈ 1.19.

WORKED EXAMPLE 14 One-layer atmosphere (exam level)

In the fully absorbing one-layer model, the atmosphere radiates 238 W m⁻² to space. Find the surface temperature.

Solution

Surface emits 2 × 238 = 476 W m⁻²M1

T = (476/5.67 × 10⁻⁸)1/4 = 303 KA1

Quick check

1. Why is the real surface cooler than 303 K?

Answers: 1. the atmosphere does not absorb all infrared (the window), and convection and evaporation also carry energy upward

2.9 A whole-planet energy budget

At the top of the atmosphere, incoming solar (340 W m⁻²) must equal reflected sunlight (100) plus outgoing infrared (240). A Sankey diagram shows this balance clearly.

Figure 11. Sankey-style diagram of Earth's energy budget at the top of the atmosphere (W m⁻²).

WORKED EXAMPLE 15 Imbalance (exam level)

If the outgoing infrared fell to 238 W m⁻² with no other change, state and explain what would happen.

Solution

Net inflow of 2 W m⁻²A1

Earth warms until outgoing infrared rises back to 240 W m⁻²R1

2.10 Energy balance diagrams

Global energy flow diagrams show all the energy transfers in W m⁻²: incoming solar, reflection by clouds and surface, absorption by atmosphere and surface, surface infrared, back radiation, convection and latent heat, and outgoing infrared. Each level must balance.

WORKED EXAMPLE 16 Surface balance (exam level)

The surface absorbs 160 W m⁻² of sunlight and 342 W m⁻² of back radiation. It emits 398 W m⁻² of infrared. Find the energy carried away by convection and evaporation.

Solution

In = 160 + 342 = 502 W m⁻²M1

Other losses = 502 − 398 = 104 W m⁻²A1

3 The enhanced greenhouse effect and climate change

Human influence on Earth's energy balance.

Think about it

If CO₂ makes up only about 0.04% of the atmosphere, how can increasing it change the climate?

3.1 The Moon: no atmosphere

The Moon has almost the same intensity of sunlight as Earth but no atmosphere. With no greenhouse effect and slow rotation, its surface swings from about 390 K in daytime to about 100 K at night. Earth's atmosphere and oceans greatly reduce such extremes.

WORKED EXAMPLE 17 Lunar noon (exam level)

Estimate the temperature of the Moon's surface directly below the Sun (albedo 0.12), assuming it radiates only from the sunlit point.

Solution

σT⁴ = 0.88 × 1360 = 1197 W m⁻²M1

T = 381 KA1

3.2 Urban heat islands

Cities are often several kelvin warmer than surrounding countryside. Dark roofs and roads have low albedo, buildings trap infrared, vegetation (which cools by evaporation) is replaced by concrete, and vehicles and buildings release energy. Cool (white) roofs and green spaces reduce the effect.

Quick check

1. Suggest two ways a city could reduce its heat-island effect.

Answers: 1. white or green roofs; more trees and parks; reflective pavements

3.3 Rising greenhouse gas concentrations

Burning fossil fuels, deforestation, agriculture and industry have increased atmospheric CO₂ from about 280 ppm before 1750 to over 420 ppm today. Methane has more than doubled. More infrared is absorbed and re-emitted downward, so the surface warms: the enhanced greenhouse effect.

Figure 12. Atmospheric CO₂ concentration measured at Mauna Loa (rounded values).

WORKED EXAMPLE 18 Rate of increase (standard)

Use the graph to estimate the average rate of increase of CO₂ between 2000 and 2020.

Solution

(414 − 370)/20M1

≈ 2.2 ppm per yearA1

3.4 Evidence for global warming

Global mean surface temperature has risen by about 1.2 K since the late 1800s. Other evidence includes rising sea levels, shrinking glaciers and Arctic sea ice, warming oceans and earlier springs. Ice cores show that CO₂ and temperature have varied together over hundreds of thousands of years.

Figure 13. Global mean temperature anomaly relative to 1951–1980 (decadal averages, rounded).

Quick check

1. Name two lines of evidence for global warming apart from thermometer records.

Answers: 1. sea-level rise; melting glaciers and sea ice; ocean heat content

3.5 Evidence from ice cores

Air bubbles trapped in Antarctic ice preserve samples of past atmospheres going back over 800 000 years. Ratios of oxygen isotopes in the ice reveal past temperatures. CO₂ and temperature rose and fell together through ice-age cycles, but never exceeded about 300 ppm before industrialisation.

Figure 14. Schematic ice-core record: CO₂ and temperature vary together over glacial cycles.

Quick check

1. How do ice cores record past CO₂ levels?

Answers: 1. air bubbles trapped in the ice preserve samples of the atmosphere

3.6 Arctic sea ice

Satellite measurements since 1979 show a decline in Arctic sea ice, especially at the September minimum. Loss of bright ice exposes dark ocean, reinforcing warming.

Figure 15. September Arctic sea-ice extent (rounded values).

WORKED EXAMPLE 19 Trend (standard)

Estimate the average rate of loss of September sea ice between 1980 and 2020.

Solution

(7.8 − 4.6)/40M1

≈ 0.08 million km² per yearA1

3.7 The carbon cycle

Carbon moves between the atmosphere, oceans, land plants, soils and rocks. Burning fossil fuels moves carbon stored for millions of years into the atmosphere much faster than natural processes can remove it. About half of emitted CO₂ remains in the atmosphere; the rest dissolves in oceans (causing acidification) or is taken up by plants.

Figure 16. Simplified carbon cycle showing the human addition from fossil fuels.

3.8 Climate models

Climate models divide the atmosphere and oceans into a 3D grid and apply conservation of energy, momentum and mass to predict future climate. They are tested by checking that they reproduce past climate. Projections depend on future emission scenarios; uncertainties arise mainly from clouds and human behaviour.

Quick check

1. Give one reason for uncertainty in climate projections.

Answers: 1. cloud feedbacks; future emissions; ocean circulation changes

3.9 Feedback mechanisms

Feedbacks amplify (positive) or reduce (negative) an initial change. Positive: ice–albedo feedback; water vapour feedback (warmer air holds more water vapour, a greenhouse gas); methane release from thawing permafrost. Negative: increased infrared emission as T rises (σT⁴); possibly increased low cloud reflecting more sunlight.

Figure 17. The ice–albedo positive feedback loop.

WORKED EXAMPLE 20 Ice–albedo (exam level)

Explain how melting Arctic sea ice amplifies warming.

Solution

Ice (high albedo) replaced by ocean (low albedo)R1

More solar energy absorbed, causing further warming and more meltingR1

3.10 Effect of changing albedo

A small change in albedo has a large effect on the equilibrium temperature because the absorbed power depends on (1 − α).

WORKED EXAMPLE 21 Albedo change (exam level)

Earth's albedo falls from 0.30 to 0.28. Calculate the change in equilibrium temperature (no atmosphere model).

Solution

T ∝ (1 − α)1/4: T₂ = 255 × (0.72/0.70)1/4M1

= 256.8 K: increase of about 1.8 KA1

3.11 Radiative forcing

Radiative forcing is the change in net incoming energy (W m⁻²) caused by a change such as increased CO₂. Doubling CO₂ gives a forcing of about 3.7 W m⁻². The surface warms until outgoing radiation rebalances the energy budget.

WORKED EXAMPLE 22 Forcing estimate (exam level)

Without feedbacks, a forcing of 3.7 W m⁻² at an effective emitting temperature of 255 K requires a temperature rise ΔT such that 4σT³ΔT ≈ 3.7. Estimate ΔT.

Solution

4 × 5.67 × 10⁻⁸ × 255³ = 3.76 W m⁻² K⁻¹M1

ΔT ≈ 1.0 K (feedbacks roughly triple this)A1

3.12 Sea-level rise

Sea level rises because seawater expands as it warms and because land ice (glaciers and ice sheets) melts. Melting floating sea ice does not raise sea level directly, because floating ice already displaces its own weight of water.

Figure 18. Global mean sea-level rise since 1900 (rounded values).

WORKED EXAMPLE 23 Rate (standard)

Compare the average rate of rise from 1900 to 1980 with that from 1990 to 2020.

Solution

1900–1980: 13 cm/80 yr = 0.16 cm per yearA1

1990–2020: 9 cm/30 yr = 0.30 cm per year: almost doubleA1

Common mistake

Saying melting Arctic sea ice raises sea level. It is the melting of land ice (Greenland, Antarctica, glaciers) and thermal expansion that do.

3.13 Consequences of climate change

Consequences include sea-level rise (thermal expansion of seawater and melting land ice), more frequent heatwaves and extreme rainfall, droughts, ocean acidification, coral bleaching and shifts in ecosystems and agriculture.

WORKED EXAMPLE 24 Thermal expansion (exam level)

The top 700 m of ocean warms by 0.50 K. The volume expansivity of seawater is 2.0 × 10⁻⁴ K⁻¹. Estimate the sea-level rise.

Solution

Δh = hγΔT = 700 × 2.0 × 10⁻⁴ × 0.50M1

= 0.07 m (7 cm)A1

In the real world

Global mean sea level has risen by about 20 cm since 1900, and the rate has more than doubled since the 1990s.

3.14 Future scenarios

Climate projections depend on future emissions. Under a low-emission scenario, warming could be limited to about 1.5–2 K above pre-industrial levels by 2100; under high emissions it could reach 4 K or more.

Figure 19. Illustrative warming projections under different emission scenarios (approximate central values).

Quick check

1. Why do projections differ so much by 2100?

Answers: 1. they depend on the amount of greenhouse gas emitted in future

3.15 Reducing emissions

Mitigation strategies include switching to low-carbon energy (renewables, nuclear), improving efficiency, carbon capture and storage, reforestation and reducing methane leaks. Adaptation includes flood defences and drought-resistant crops.

Strategy	Physics link
solar and wind power	photovoltaics; ½ρAv³
nuclear power	fission energy, E = Δmc²
better insulation	thermal conductivity, U-values
electric vehicles	motor efficiency, batteries
carbon capture	energy cost of separating and storing CO₂

4 Data-based and graph questions

Interpreting climate data.

4.1 Reading energy-flow diagrams

Questions give a diagram of energy flows and ask you to apply conservation of energy at the top of the atmosphere, in the atmosphere and at the surface.

WORKED EXAMPLE 25 Top of atmosphere (standard)

Incoming solar = 340 W m⁻²; reflected = 100 W m⁻². Find the outgoing infrared needed for balance and the planetary albedo.

Solution

Outgoing IR = 340 − 100 = 240 W m⁻²A1

Albedo = 100/340 = 0.29A1

4.2 Spectra and wavelengths

Use Wien's law to explain which radiation the atmosphere absorbs. Compare the wavelength ranges of incoming solar and outgoing terrestrial radiation with the absorption bands of greenhouse gases.

WORKED EXAMPLE 26 Matching bands (exam level)

CO₂ absorbs strongly at 15 μm. Calculate the temperature of a black body that peaks at this wavelength and comment.

Solution

T = 2.9 × 10⁻³/15 × 10⁻⁶ = 193 KA1

Earth's emission (peak 10 μm) extends strongly to 15 μm, so CO₂ absorbs much of itR1

4.3 Equilibrium temperatures of other planets

The same balance (1 − α)S/4 = σT⁴ gives equilibrium temperatures for other planets, which can be compared with measured surface temperatures to reveal greenhouse effects.

Planet	S / W m⁻²	Albedo	T (no greenhouse) / K	Actual T / K
Venus	2600	0.77	227	737
Earth	1360	0.30	255	288
Mars	590	0.25	210	210

WORKED EXAMPLE 27 Venus (exam level)

Show that Venus's equilibrium temperature with no greenhouse effect is about 227 K, and comment on the actual 737 K.

Solution

σT⁴ = 0.23 × 2600/4 = 150 W m⁻²: T = 227 KM1 A1

Thick CO₂ atmosphere produces an extreme greenhouse effectR1

5 Practical investigations

Modelling the greenhouse effect in the lab.

5.1 Bottle model of the greenhouse effect

Two identical sealed bottles, one containing air and one enriched with CO₂, are placed under a lamp with thermometers inside. The temperature of each is recorded every minute. The CO₂ bottle usually warms slightly more because CO₂ absorbs infrared emitted by the inner surfaces.

Figure 20. Comparing the warming of air and CO₂-enriched air under a lamp.

WORKED EXAMPLE 28 Evaluating (exam level)

Suggest two limitations of the bottle model.

Solution

Bottles warm partly because they stop convection (like a real greenhouse)R1

Small path length and high CO₂ concentration are unlike the real atmosphereR1

5.2 Measuring albedo

A light meter pointed at the sky measures incoming light; pointed at the ground (at the same height) it measures reflected light. The ratio gives the albedo of the surface. Compare grass, concrete, sand and snow.

WORKED EXAMPLE 29 Albedo reading (standard)

A meter reads 52 000 lux facing the sky and 13 000 lux facing a concrete path. Find the albedo of the concrete.

Solution

13 000/52 000M1

= 0.25A1

Try this

Use a phone light-meter app to estimate the albedo of different surfaces around your school. Rank them and relate the results to surface temperatures on a sunny day.

5.3 Infrared absorption demonstration

An infrared camera viewing a candle through a tube shows the flame clearly with air in the tube but fainter when the tube is filled with CO₂, which absorbs the infrared.

Quick check

1. What does the demonstration show about CO₂?

Answers: 1. it absorbs infrared radiation

6 Exam technique and further practice

Putting it together.

6.1 Method checklist

•Use S/4 for the mean intensity over Earth's surface.
•Multiply by (1 − α) for the absorbed intensity.
•Use T⁴ and kelvin in all radiation laws.
•State the wavelength regions: incoming short-wavelength (visible), outgoing long-wavelength (infrared).
•Explain absorption using molecular vibrations at resonant frequencies.

6.2 Multi-step problem: planetary temperature

WORKED EXAMPLE 30 Exoplanet (exam level)

A planet orbits a star of luminosity 1.5 × 10²⁶ W at 1.2 × 10¹¹ m, with albedo 0.40. Find (a) the intensity at the planet; (b) its equilibrium temperature.

Solution

(a) S = 1.5 × 10²⁶/(4π × (1.2 × 10¹¹)²) = 829 W m⁻²M1 A1

(b) σT⁴ = 0.60 × 829/4 = 124 W m⁻²: T = 216 KM1 A1

6.3 Graded practice

WORKED EXAMPLE 31 Snow cover (standard)

A field of area 1.0 × 10⁴ m² receives 500 W m⁻². Compare the power absorbed when covered with grass (α = 0.25) and snow (α = 0.85).

Solution

Grass: 0.75 × 500 × 10⁴ = 3.75 MWA1

Snow: 0.15 × 500 × 10⁴ = 0.75 MWA1

WORKED EXAMPLE 32 Emitting temperature (exam level)

Earth emits 240 W m⁻² to space. Find the effective emitting temperature and explain why it is lower than the surface temperature.

Solution

T = (240/5.67 × 10⁻⁸)1/4 = 255 KA1

Most emission to space comes from cold upper layers of the atmosphereR1

6.4 Further practice: energy balance

WORKED EXAMPLE 33 Planet X (exam level)

A planet receives 900 W m⁻² at its orbit, has albedo 0.35 and effective emissivity 0.80. Find its surface temperature.

Solution

Absorbed = 0.65 × 900/4 = 146 W m⁻² = 0.80σT⁴M1

T = 238 KA1

WORKED EXAMPLE 34 Emissivity from data (exam level)

Earth's surface emits 398 W m⁻² and 240 W m⁻² leaves the top of the atmosphere. Find the effective emissivity of the Earth–atmosphere system.

Solution

e = 240/398M1

= 0.60A1

6.5 Further practice: radiation and gases

WORKED EXAMPLE 35 Photon energy (standard)

Find the energy of a 15 μm infrared photon absorbed by CO₂.

Solution

E = hc/λ = 6.63 × 10⁻³⁴ × 3.0 × 10⁸/15 × 10⁻⁶M1

= 1.3 × 10⁻²⁰ J (0.083 eV)A1

WORKED EXAMPLE 36 Frequency (standard)

Find the frequency of 15 μm radiation and state what this matches in the CO₂ molecule.

Solution

f = 3.0 × 10⁸/15 × 10⁻⁶ = 2.0 × 10¹³ HzA1

The natural frequency of the bending vibration (resonance)R1

6.6 Further practice: climate data

WORKED EXAMPLE 37 Energy imbalance (exam level)

Earth currently absorbs about 0.8 W m⁻² more than it emits. Find the extra energy absorbed per year by the whole planet (R = 6.37 × 10⁶ m).

Solution

Area = 4πR² = 5.1 × 10¹⁴ m²; P = 4.1 × 10¹⁴ WM1

E = 4.1 × 10¹⁴ × 3.16 × 10⁷ = 1.3 × 10²² JA1

WORKED EXAMPLE 38 Ocean warming (exam level)

If this energy all warmed the top 700 m of ocean (mass 2.5 × 10²⁰ kg, c = 4000 J kg⁻¹ K⁻¹), find the temperature rise per year.

Solution

ΔT = 1.3 × 10²²/(2.5 × 10²⁰ × 4000)M1

= 0.013 K per yearA1

6.7 Further practice: albedo and surfaces

WORKED EXAMPLE 39 Solar panel field (exam level)

A 2.0 km² desert (albedo 0.40) is covered by solar panels with albedo 0.10, which convert 20% of incident energy to electricity. For a mean intensity of 250 W m⁻², find the change in power absorbed as heat at the surface.

Solution

Before: 0.60 × 250 × 2.0 × 10⁶ = 300 MWM1

After: (0.90 − 0.20) × 250 × 2.0 × 10⁶ = 350 MW: increase of 50 MWA1

WORKED EXAMPLE 40 Polar vs equator (exam level)

At noon, sunlight of 1000 W m⁻² strikes ground at the equator (vertical) and at 70° latitude (Sun 20° above horizon). The polar ground is snow (α = 0.85); the equatorial ground is forest (α = 0.12). Compare the absorbed intensities.

Solution

Equator: 0.88 × 1000 = 880 W m⁻²A1

Pole: 0.15 × 1000 × sin 20° = 51 W m⁻²M1 A1

6.8 Further practice: wavelengths and spectra

WORKED EXAMPLE 41 Solar photon (standard)

Calculate the energy of a 500 nm photon from the Sun and compare it with a 10 μm photon from Earth.

Solution

500 nm: 4.0 × 10⁻¹⁹ J; 10 μm: 2.0 × 10⁻²⁰ JM1 A1

Solar photon has 20 times the energyA1

WORKED EXAMPLE 42 Star temperature (standard)

A star's spectrum peaks at 350 nm. Find its surface temperature.

Solution

T = 2.9 × 10⁻³/3.5 × 10⁻⁷M1

= 8300 KA1

WORKED EXAMPLE 43 Cold planet (exam level)

A planet has equilibrium temperature 120 K. Find its peak emission wavelength and the region of the spectrum.

Solution

λ = 2.9 × 10⁻³/120 = 24 μmA1

Far infraredA1

6.9 Key terms

Term	Meaning
solar constant S	intensity of sunlight at Earth's distance, above the atmosphere (about 1360 W m⁻²)
albedo α	fraction of incident radiation reflected
emissivity e	power emitted compared with a black body at the same temperature
equilibrium temperature	temperature at which absorbed and emitted power are equal
greenhouse gas	gas whose molecules absorb and re-emit infrared
enhanced greenhouse effect	extra warming from human-produced greenhouse gases
radiative forcing	change in net energy flow (W m⁻²) caused by a change in the climate system
positive feedback	a response that amplifies the original change
negative feedback	a response that reduces the original change

6.10 Self-test

Try these without notes, then check the answers.

Quick check

1. State the mean intensity of sunlight over Earth's surface.

2. Calculate the absorbed intensity for albedo 0.40.

3. Why does N₂ not absorb infrared?

4. Name one positive and one negative climate feedback.

5. Find the equilibrium temperature for an absorbed intensity of 200 W m⁻² (e = 1).

6. Why does melting land ice raise sea level but melting sea ice does not?

Answers: 1. about 340 W m⁻² 2. 204 W m⁻² 3. its vibration does not change its dipole moment 4. ice–albedo (positive); increased σT⁴ emission (negative) 5. 243 K 6. sea ice already displaces its own weight of water

6.11 Common exam errors

•Dividing S by 2 instead of 4.
•Using α instead of (1 − α) for absorbed power.
•Saying greenhouse gases 'trap heat' without explaining absorption and re-emission of infrared.
•Confusing the greenhouse effect with ozone depletion.
•Forgetting to take the fourth root when finding T.

7 Common misconceptions and links

Students often think…	Actually…
greenhouse gases reflect heat back	they absorb and re-emit infrared in all directions
the greenhouse effect is entirely harmful	the natural effect keeps Earth habitable
ozone depletion causes global warming	they are largely separate problems
albedo is the fraction absorbed	albedo is the fraction reflected
all atmospheric gases absorb infrared	N₂ and O₂ do not

How B.2 connects to the rest of the course

Topic	Connection
B.1 Thermal energy transfers	black-body radiation, Wien and Stefan–Boltzmann
A.3 Work, energy and power	energy resources and efficiency
C.1 Simple harmonic motion	molecular vibration and resonance
E.4 Fission	low-carbon power generation

8 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

The solar constant is 1360 W m⁻² and Earth's mean albedo is 0.30.

(a) Explain why the mean intensity received over Earth's surface is S/4. [2]

(b) Calculate the mean intensity absorbed by Earth. [2]

(d) The actual mean surface temperature is 288 K. Explain the difference. [3]

(e) Explain how melting polar ice could lead to further warming. [2]

Markscheme

Part	Answer	Marks
(a)	Earth intercepts a disc of area πR²	R1
	but radiates/receives over total surface 4πR²	R1
(b)	0.70 × 340	M1
	238 W m⁻²	A1
(c)	T = (238/σ)1/4	M1
	255 K	A1
(d)	greenhouse gases absorb outgoing infrared	R1
	and re-emit some back towards the surface	R1
	surface must be warmer to restore balance	R1
(e)	ice replaced by darker land/ocean; albedo falls	R1
	more energy absorbed: positive feedback	R1

Examiner's comments. In (d), answers that said gases 'trap heat' without explaining absorption and re-emission of infrared did not gain full marks.

9 Check your understanding

Section A: multiple choice

1. Earth's albedo is approximately

A. 0.30 B. 0.70 C. 0.03 D. 1.0

2. Greenhouse gases absorb mainly

A. visible light B. ultraviolet C. infrared radiation D. radio waves

3. The mean solar intensity over Earth's surface is

A. S B. S/4 C. S/2 D. 4S

4. Which is not a greenhouse gas?

A. carbon dioxide B. methane C. water vapour D. nitrogen

5. Melting ice reduces albedo, causing

A. negative feedback B. positive feedback C. no change D. cooling

6. Earth's emission peaks at about

A. 10 μm B. 500 nm C. 1 mm D. 100 nm

7. The equilibrium temperature of Earth without greenhouse gases is about

A. 288 K B. 300 K C. 200 K D. 255 K

8. Emissivity of a perfect black body is

A. 0 B. 0.5 C. 1 D. 0.3

Section B: structured

9. Explain, in terms of molecular behaviour, how CO₂ absorbs infrared. [3]

10. Calculate the change in equilibrium temperature if Earth's albedo fell to 0.25. [3]

11. Describe two positive feedbacks in the climate system. [4]

12. Explain why the solar spectrum and Earth's emission spectrum peak at different wavelengths. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. IR frequency matches natural frequency of bending/asymmetric stretching (resonance); dipole changes; energy absorbed and re-emitted.

10. T = (0.75 × 340/σ)1/4 = 259 K: about 4 K higher.

11. Ice–albedo; water vapour; permafrost methane release.

12. Wien's law: Sun about 5800 K peaks in visible; Earth about 288 K peaks in infrared.

10 Summary and topic map

Idea	Remember
Solar constant	S ≈ 1360 W m⁻²; mean S/4
Albedo	fraction reflected; Earth about 0.30
Emissivity	P = eσAT⁴
Energy balance	(1 − α)S/4 = eσT⁴
Greenhouse effect	IR absorbed and re-emitted by gases
Climate change	enhanced effect, feedbacks, consequences

Before the exam, make sure you…

•use S/4 and (1 − α)
•take the fourth root
•explain absorption by molecular resonance
•distinguish natural and enhanced greenhouse effects

Greenhouse Effect on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 21. Topic map for Greenhouse Effect.

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