B.2 Greenhouse effect

Syllabus
First assessment 2025
Topic
—
Level
HL

Learning objectives

Model Planetary Energy Balance

Treat the planet as a system

Over a long enough time, a planet at steady average temperature receives and emits radiant energy at equal rates:

Pin=PoutP_{\mathrm{in}}=P_{\mathrm{out}}

This is conservation of energy applied to the planet–atmosphere system.

Track every pathway

Incoming solar radiation can be reflected by the planet–atmosphere system, absorbed by the atmosphere or surface, and later emitted as infrared radiation. The energy balance concerns the total absorbed input and total emitted output, not just one arrow in the diagram.

Interpret imbalance

If absorbed power exceeds emitted power, the system’s internal energy and average temperature tend to increase. If emitted power exceeds absorbed power, they tend to decrease. Equal rates mean no net long-term energy accumulation.

Boundary check

A steady temperature does not mean radiation stops. It means the net energy change is zero because input and output balance.

B.2.1 Exam Analysis

1 mark

The energy balance model of a planet's climate is shown. The reflected and radiated intensities are given in terms of the incident incoming intensity I.

What is the radiated intensity from the surface of the planet?

Interpret and Calculate Emissivity

Emissivity

Emissivity, ε\varepsilon, compares the power radiated per unit area by a real surface with that radiated per unit area by an ideal black surface at the same absolute temperature:

ε=P/AσT4\varepsilon=\frac{P/A}{\sigma T^4}

Use the radiation equation

For a surface of area A,

P=εσAT4P=\varepsilon\sigma AT^4

Use T in kelvin. A black body has ε=1\varepsilon=1; real surfaces have emissivity less than or equal to 1 in this model.

What emissivity is not

Emissivity is not the fraction of incoming sunlight reflected; that is albedo. Emissivity concerns emission of thermal radiation by a surface at its temperature.

Worked example from the mapped local textbook

A planet has surface temperature −63 ∘C=210 K-63\,^{\circ}\mathrm{C}=210\,\mathrm{K}, area 1.4×1014 m21.4\times10^{14}\,\mathrm{m^2} and luminosity 1.3×1016 W1.3\times10^{16}\,\mathrm{W}.

ε=P/AσT4=(1.3×1016)/(1.4×1014)(5.67×10−8)(210)4=0.84\varepsilon=\frac{P/A}{\sigma T^4}=\frac{(1.3\times10^{16})/(1.4\times10^{14})}{(5.67\times10^{-8})(210)^4}=0.84

Emissivity is dimensionless, and the kelvin conversion is required by the fourth-power law.

Calculation boundary

Keep surface emission and atmospheric re-radiation as separate energy-flow terms. Do not subtract reflected solar power from the Stefan–Boltzmann emission formula.

B.2.2 Exam Analysis

3 marks

Determine the average intensity re-radiated by the atmosphere towards the surface. Assume that the emissivity of the surface is 0.90 .

Calculate Albedo

Albedo

Albedo is the fraction of incident radiation scattered or reflected by a macroscopic system:

a=PscatteredPincidenta=\frac{P_{\mathrm{scattered}}}{P_{\mathrm{incident}}}

Interpret the value

An albedo near 0 means most incident energy is absorbed; an albedo near 1 means most is reflected. It has no units and is often reported as a decimal or percentage.

Use ratios safely

Calculate each albedo from reflected divided by incident power before comparing surfaces. A snow surface can have a higher albedo than concrete because it reflects a larger fraction of the same incoming intensity.

Worked example from the mapped local textbook

Radiation of intensity 610 W m−2610\,\mathrm{W\,m^{-2}} reaches water with albedo a=0.18a=0.18. The absorbed fraction is 1−a=0.821-a=0.82, so

Iabsorbed=(1−a)Iincident=(0.82)(610)=5.0×102 W m−2I_{\mathrm{absorbed}}=(1-a)I_{\mathrm{incident}}=(0.82)(610)=5.0\times10^2\,\mathrm{W\,m^{-2}}

About 110 W m−2110\,\mathrm{W\,m^{-2}} is reflected; the absorbed and reflected intensities add back to the incident intensity, within rounding.

Common trap

Albedo is a ratio of powers, not the reflected power by itself. Do not compare two reflected intensities unless their incident intensities are the same or you have normalized them.

B.2.3 Exam Analysis

1 mark

Light of intensity 500Wm−2500 \mathrm{Wm}^{-2} is incident on concrete and on snow. 300Wm−2300 \mathrm{Wm}^{-2} is reflected from the concrete and 400Wm−2400 \mathrm{Wm}^{-2} is reflected from the snow.

What is  albedo of concrete  albedo of snow \frac{\text { albedo of concrete }}{\text { albedo of snow }} ?

Explain Why Earth’s Albedo Varies

Earth’s albedo is variable

Earth’s average albedo is not a universal fixed property. It changes with the surfaces and clouds that are illuminated and with the angle at which radiation arrives.

Daily variation

Cloud cover changes with weather and the position of the Sun changes during the day. Both alter the fraction of incident radiation reflected by a region.

Latitude and incidence angle

At different latitudes, sunlight arrives at different angles to the surface normal. The effective surface and reflected fraction therefore vary; snow, ice, ocean, land and clouds also contribute different albedos.

Climate link

If snow or ice melts, a darker surface may reflect less and absorb more incoming energy. This is a consequence of changing albedo, not a change in the definition of albedo.

B.2.4 Exam Analysis

1 mark

A student makes three statements about Earth's albedo.

I. It varies daily.
II. It depends on latitude.
III. It depends on cloud formation.

Which of the statements are correct?

Define the Solar Constant

Solar constant

The solar constant, SS, is the solar-radiation intensity received per unit area at the Earth’s orbital distance, with the surface perpendicular to the incoming rays. Its units are W m⁻².

It is not the global mean

The solar constant describes the incident intensity on a surface facing the Sun. A planet’s spherical geometry spreads the intercepted power over a larger total surface, so the planet-wide mean incoming intensity is smaller.

Use it as an input

In an energy-balance problem, S is the incoming solar intensity before accounting for the planet’s projected area, albedo, atmospheric absorption or averaging over the whole sphere.

Common trap

Do not call the solar constant the total solar power intercepted by Earth. It is an intensity: power per unit area.

B.2.5 Exam Analysis

1 mark

State what is meant by the solar constant.

Derive the Mean Solar Intensity S/4

Why the factor is 1/4

A planet intercepts incoming sunlight over its projected disk, area πr2\pi r^2. The intercepted power is then averaged over the planet’s whole spherical surface, area 4πr24\pi r^2.

Mean incoming intensity

If the solar constant is S,

I‾=Sπr24πr2=S4\overline I=\frac{S\pi r^2}{4\pi r^2}=\frac S4

This is the mean intensity before accounting for reflection or atmospheric absorption.

Add albedo when required

If the planetary albedo is a, the globally averaged absorbed intensity is

Iabs=(1−a)S4I_{\mathrm{abs}}=(1-a)\frac S4

provided the problem’s model treats the planet as a uniform system.

Worked example from local Question Bank row 30218

For S=1400 W m−2S=1400\,\mathrm{W\,m^{-2}} and atmospheric albedo a=0.30a=0.30, the transmitted incident intensity is

I=(1−a)S=(0.70)(1400)=980 W m−2I=(1-a)S=(0.70)(1400)=980\,\mathrm{W\,m^{-2}}

Averaging the intercepted power over the full sphere gives

I‾=9804=245 W m−2\overline I=\frac{980}{4}=245\,\mathrm{W\,m^{-2}}

This result combines reflection with geometry: (1−a)S/4(1-a)S/4.

Common trap

Do not divide S by 4 because sunlight is four times weaker at every point. The factor comes from intercepted disk area divided by total spherical area.

B.2.6 Exam Analysis

2 marks

Show that the average global intensity of radiation absorbed by the surface is about 240Wm−2240 \mathrm{Wm}^{-2}.

Identify the Main Greenhouse Gases

Main gases

The syllabus identifies water vapour (H₂O), carbon dioxide (CO₂), methane (CH₄) and nitrous oxide (N₂O) as the main greenhouse gases.

Natural and human origins

Each of these gases has natural sources and sources affected by human activity. For example, water vapour participates in the natural water cycle, while combustion, agriculture and land-use changes can alter atmospheric concentrations of several greenhouse gases.

Abundance is not the only factor

A gas’s contribution depends on both its atmospheric abundance and how strongly it absorbs infrared radiation. Do not rank gases using concentration alone.

Common trap

O₂, N₂ and Ar are not treated as the main greenhouse gases in this syllabus objective. The question may test recognition of the listed gases, not whether a molecule is simply present in the atmosphere.

B.2.7 Exam Analysis

1 mark

Which of the following is not considered to be a greenhouse gas?

Explain Infrared Absorption and Re-emission

Absorb at molecular frequencies

Greenhouse-gas molecules can absorb infrared radiation when its frequency matches an allowed molecular vibration or transition. The molecule moves to a higher energy state.

Re-emit in all directions

The excited molecule soon returns to a lower energy state and emits infrared radiation. The emission is in random directions, so some radiation continues upward and some is directed back toward the surface.

Why Earth’s radiation matters

The cooler Earth emits mainly longer-wavelength infrared radiation. Greenhouse gases absorb part of this outgoing radiation; they are less likely to absorb most of the Sun’s shorter-wavelength incoming radiation.

Model boundary from local Question Bank row 30018

Mars's atmosphere is mainly carbon dioxide, but its pressure is less than 1%1\% of Earth's. The low pressure means far fewer absorbing molecules per unit volume, so much less outgoing infrared radiation is absorbed and re-emitted toward the surface.

The presence of a greenhouse-gas species alone does not determine the size of the effect; the amount of gas also matters.

Common trap

The greenhouse effect is not mainly reflection of incoming sunlight. The key process is absorption and re-emission of outgoing infrared radiation.

B.2.8 Exam Analysis

2 marks

Explain the effect of an increase in the concentration of greenhouse gases in the atmosphere on I2I_{2}.

The following data are given.

I0=240Wm−2I2=150Wm−2\begin{aligned} I_{0} & =240 \mathrm{Wm}^{-2} \\ I_{2} & =150 \mathrm{Wm}^{-2} \end{aligned}

Explain the Greenhouse Effect with Two Models

Resonance model

A greenhouse-gas molecule can absorb infrared radiation when the radiation frequency matches one of the molecule’s natural vibrational frequencies. The molecule is driven into a larger-amplitude vibration: this is the resonance description.

Molecular energy-level model

The same absorption can be described as a photon whose energy matches the gap between molecular energy levels. The molecule is excited, then returns to a lower level and emits infrared radiation.

Complete mechanism

Earth’s surface emits long-wave infrared radiation. Greenhouse gases absorb part of it and re-emit radiation in random directions; some is directed back toward the surface, raising the surface temperature compared with an atmosphere-free model.

Frequency test from local Question Bank row 31337

If a carbon dioxide vibration has period 5×10−14 s5\times10^{-14}\,\mathrm{s}, its natural frequency is

f0=1T=15×10−14=2×1013 Hz=20 THzf_0=\frac1T=\frac1{5\times10^{-14}}=2\times10^{13}\,\mathrm{Hz}=20\,\mathrm{THz}

Earth's outgoing infrared is around 30 THz30\,\mathrm{THz}, much closer to this molecular frequency than the cited solar infrared near 300 THz300\,\mathrm{THz}. The resonance model therefore predicts stronger absorption of part of Earth's outgoing infrared; the energy-level model describes the same selectivity as matched photon-energy gaps.

Exam boundary

Do not write only “greenhouse gases trap heat”. Name infrared absorption, molecular resonance or energy-level matching, and re-emission in all directions.

B.2.9 Exam Analysis

2 marks

Outline the physical mechanism by which some of the radiation emitted by the surface is absorbed by greenhouse gases in the atmosphere and re-radiated towards the surface.

Distinguish the Enhanced Greenhouse Effect

Natural greenhouse effect

The natural greenhouse effect is the normal warming produced when atmospheric gases absorb and re-emit some of Earth’s outgoing infrared radiation. It helps keep Earth’s surface suitable for life.

Enhanced greenhouse effect

The enhanced greenhouse effect is the augmentation of that effect due to human activity. Increased concentrations of greenhouse gases can increase absorption and downward re-radiation of infrared energy.

Primary cause in the syllabus

Burning fossil fuels is identified as a primary cause because it increases atmospheric carbon dioxide and contributes to changes in the Earth–atmosphere energy balance. Other human activities can affect greenhouse-gas concentrations too.

Common trap

Do not say that the greenhouse effect itself is caused only by humans. Human activity enhances a naturally occurring effect.

B.2.10 Exam Analysis

1 mark

What is a primary cause of the enhanced greenhouse effect?

Synthesize B.2 Greenhouse Effect

Start with the energy balance

For a planet at steady average temperature, absorbed incoming radiant power equals emitted outgoing radiant power. Albedo controls the reflected fraction; emissivity controls thermal emission relative to a black body.

Average incoming solar energy

The solar constant S is an intensity on a surface perpendicular to the rays. A spherical planet averages the intercepted power over four times the projected area, giving S/4S/4; with albedo a, the simple globally averaged absorbed intensity is (1−a)S/4(1-a)S/4.

Atmospheric mechanism

Earth emits infrared radiation. Greenhouse molecules absorb selected wavelengths through molecular resonance or energy-level transitions, then re-emit in all directions, including back toward the surface.

Human enhancement

The natural greenhouse effect supports a habitable surface temperature. Human-driven increases in greenhouse-gas concentration augment the effect; fossil-fuel burning is a primary cause of this enhanced greenhouse effect.