Greenhouse effect

B.2 — Greenhouse Effect

IB Physics HL • Topic B: The particulate nature of matter • SL and HL core content

This lesson develops a quantitative model for how Earth’s temperature is set by an energy balance between incoming solar radiation and outgoing infrared radiation. You will use emissivity, albedo, and the solar constant to explain the natural greenhouse effect and how increasing greenhouse gas concentrations can change the equilibrium temperature.

Big picture

B.2.1 — Energy from the Sun: emissivity, albedo, and the solar constant.
B.2.2 — Energy balance and the greenhouse effect: infrared absorption, resonance, and molecular energy levels.
B.2.3 — Greenhouse gas concentrations: the natural and enhanced greenhouse effects, feedbacks, and equilibrium temperature.

B.2.1 Energy from the Sun

Learning objectives

Define emissivity and apply it in calculations.
Define albedo and explain why Earth’s albedo varies.
Define and calculate the solar constant, $S$ .
Relate emissivity, albedo, and solar radiation to the Earth–atmosphere energy balance.

Energy received from the Sun

The Sun is the primary source of energy for Earth’s climate system. Energy transfers from the Sun to Earth by electromagnetic radiation, mainly in the visible and infrared regions of the spectrum.

To model how much energy reaches and affects Earth, we use three key ideas:

Emissivity — how effectively a body emits thermal radiation.
Albedo — the fraction of incoming radiation that is reflected or scattered.
Solar constant — the solar power per unit area at a specified distance from the Sun, measured perpendicular to the incoming rays.

Emissivity

A real object generally emits less thermal radiation per unit area than a perfect black body at the same temperature.

Definition — emissivity

The emissivity $e$ of an object is the ratio of its emitted power per unit area to the power per unit area emitted by a perfect black body at the same temperature.

$\displaystyle e=\frac{P/A}{\sigma T^4}$

Rearranging gives the Stefan–Boltzmann equation for a real surface:

$\displaystyle \frac{P}{A}=e\sigma T^4$

Here, $T$ is the absolute temperature in kelvin, and $\sigma=5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$ is the Stefan–Boltzmann constant.

Key points:

Emissivity is dimensionless.
A perfect black body has $e=1$ .
Ordinary thermal surfaces have emissivities between zero and one.
At a given wavelength and under thermal-equilibrium conditions, a good emitter is also a good absorber.
Matt black surfaces often have high emissivity for thermal infrared radiation.
Polished metallic surfaces often have low infrared emissivity. Visible colour alone does not determine infrared emissivity.

Worked example 1 — Emissivity

In a simplified model, the surface temperature of Mars is $-63^\circ\text{C}$ , and each square metre of its surface emits $83\,\text{J}$ of energy per second. Calculate the emissivity.

Show answer and full working

Step 1: Convert temperature to kelvin.

$\displaystyle T=-63+273.15=210.15\,\text{K}$

Step 2: Write the Stefan–Boltzmann relation per unit area.

$\displaystyle \frac{P}{A}=e\sigma T^4$

Step 3: Identify the emitted power per unit area.

Emitting $83\,\text{J}$ each second from each square metre means:

$\displaystyle \frac{P}{A}=83\,\text{W m}^{-2}$

Step 4: Calculate emissivity.

$\displaystyle e=\frac{P/A}{\sigma T^4}$

$\displaystyle e=\frac{83}{(5.67\times10^{-8})(210.15)^4}\approx0.75$

Answer: $e\approx0.75$ . Emissivity has no unit.

Albedo

Not all incoming solar radiation is absorbed by Earth. Some is reflected or scattered back into space by the surface, clouds, and atmospheric particles.

Definition — albedo

The albedo $\alpha$ of a macroscopic system is the ratio of total scattered or reflected power to total incident power.

$\displaystyle \alpha=\frac{P_{\mathrm{scattered}}}{P_{\mathrm{incident}}}$

Albedo is dimensionless and lies between zero and one.

For introductory calculations, Earth’s average albedo is commonly approximated as $\alpha=0.30$ . This means:

About 30% of incoming solar radiation is reflected or scattered back to space.
About 70% is absorbed by the Earth–atmosphere system.

Why Earth’s albedo varies

Earth’s albedo varies with location, time, and several physical factors:

Cloud thickness: thick clouds generally reflect more sunlight than thin clouds.
Cloud properties: cloud type, droplet size, and ice content influence reflection. Clouds also absorb and emit infrared radiation.
Latitude: higher latitudes often have more reflective snow and ice.
Surface type: snow and ice generally have high albedo, deserts have moderate albedo, and forests and oceans often have lower albedo.
Angle of illumination: the fraction reflected can depend on the angle at which sunlight strikes the surface.

Worked example 2 — Albedo

The global average incoming solar flux at the top of Earth’s atmosphere is $340\,\text{W m}^{-2}$ , and the reflected flux is $102\,\text{W m}^{-2}$ . Calculate Earth’s albedo.

Show answer and full working

The two fluxes are averaged over the same area, so their ratio is the same as the ratio of the corresponding powers.

$\displaystyle \alpha=\frac{P_{\mathrm{reflected}}}{P_{\mathrm{incident}}}=\frac{102}{340}=0.30$

Answer: $\alpha=0.30$ , or 30%.

The solar constant

Definition — solar constant

The solar constant $S$ is the solar power per unit area at the top of a planet’s atmosphere, measured on a surface perpendicular to the incoming rays at the specified distance from the Sun.

It is equal to the apparent brightness of the Sun at that distance.

A representative value at Earth is $S\approx1.36\times10^3\,\text{W m}^{-2}$ . It varies with orbital distance and slightly with solar activity.

Important: this is not the solar power per unit area averaged over Earth’s entire surface. The global average incoming flux is $S/4$ .

Calculating the solar constant

If the Sun has luminosity $L$ , its radiation is spread over a sphere of area $4\pi r^2$ at distance $r$ from its centre.

$\displaystyle S=\frac{L}{4\pi r^2}$

Worked example 3 — Solar irradiance at Mercury

For this calculation, the distance from the centre of the Sun to Mercury is $4.6\times10^{10}\,\text{m}$ . The Sun emits $3.9\times10^{26}\,\text{J}$ of energy every second. Calculate the solar irradiance at Mercury at this distance.

Show answer and full working

Step 1: Use the inverse-square relation.

$\displaystyle S=\frac{L}{4\pi r^2}$

Step 2: Express the luminosity in watts.

$\displaystyle L=3.9\times10^{26}\,\text{W}$

Step 3: Substitute.

$\displaystyle S=\frac{3.9\times10^{26}}{4\pi(4.6\times10^{10})^2}$

$\displaystyle S\approx1.47\times10^4\,\text{W m}^{-2}$

Answer: approximately $1.5\times10^4\,\text{W m}^{-2}$ to two significant figures.

The calculation applies at the stated orbital distance; Mercury’s solar irradiance varies as its distance from the Sun changes.

Key ideas — B.2.1

Emissivity measures how effectively a body emits thermal radiation compared with a black body at the same temperature.
Albedo measures the fraction of incoming solar radiation that is reflected or scattered.
Earth’s albedo varies with clouds, surface type, and illumination.
Solar irradiance follows the inverse-square relation: $S\propto1/r^2$ .

B.2.2 Energy balance in the Earth surface–atmosphere system

Learning objectives

Explain why some atmospheric gases absorb infrared radiation strongly while others do not.
Use resonance and molecular energy-level models to explain selective absorption.
Apply conservation of energy to exchanges between Earth’s surface, atmosphere, and space.
Solve energy-balance problems involving incoming and outgoing radiation.
Explain the role of the atmosphere and compare the thermal conditions on Earth and the Moon.

The greenhouse effect

The greenhouse effect is the warming of a planet’s surface and lower atmosphere caused by atmospheric absorption and emission of outgoing thermal infrared radiation.

A substantial fraction of incoming solar radiation passes through the atmosphere and is absorbed by the surface. Some is reflected to space or absorbed in the atmosphere. The warmed surface emits mainly infrared radiation, which greenhouse gases absorb at selected wavelengths.

The atmosphere emits infrared radiation both upward and downward. Downward atmospheric radiation reduces the surface’s net radiative energy loss and allows a higher equilibrium surface temperature.

Energy is not permanently trapped. The Earth–atmosphere system continues to radiate energy to space, and energy must still be conserved.

Earth–Moon comparison

The Moon has no significant atmosphere and experiences large surface-temperature differences between day and night.
Earth’s atmosphere exchanges energy with the surface and helps redistribute energy.
Earth’s natural greenhouse effect raises its surface temperature above the value predicted by a simple atmosphere-free radiation model with the same assumed albedo.

Earth and the Moon receive similar solar irradiance, but their albedos, rotation rates, heat storage, and atmospheres differ. These differences must be considered when comparing their temperatures.

Incoming and outgoing radiation

Incoming: solar radiation includes visible light, near-infrared, and ultraviolet. It may be transmitted, reflected, or absorbed by the surface and atmosphere.
Outgoing: thermal infrared radiation. Radiation escaping to space comes from both the atmosphere and the surface.

Worked example 4 — Peak wavelength emitted by Earth

Using $T=254\,\text{K}$ as an illustrative effective radiating temperature for Earth, calculate the wavelength at which the maximum spectral intensity is emitted. State the corresponding region of the electromagnetic spectrum.

Show answer and full working

Step 1: Use Wien’s displacement law.

$\displaystyle \lambda_{\mathrm{max}}T=b_W$

Here, $b_W=2.90\times10^{-3}\,\text{m K}$ .

Step 2: Substitute.

$\displaystyle \lambda_{\mathrm{max}}=\frac{b_W}{T}=\frac{2.90\times10^{-3}}{254}$

$\displaystyle \lambda_{\mathrm{max}}\approx1.14\times10^{-5}\,\text{m}$

$\displaystyle \lambda_{\mathrm{max}}\approx11.4\,\mu\text{m}$

Answer: $11.4\,\mu\text{m}$ , which lies in the infrared region.

The effective radiating temperature describes emission to space in a simplified model. It is not the same as the average temperature of Earth’s surface.

Greenhouse gases

The main components of Earth’s atmosphere are nitrogen, $\text{N}_2$ , and oxygen, $\text{O}_2$ . These absorb thermal infrared radiation only weakly under ordinary atmospheric conditions.

Important greenhouse gases include:

Carbon dioxide, $\text{CO}_2$ .
Methane, $\text{CH}_4$ .
Water vapour, $\text{H}_2\text{O}$ .
Nitrous oxide, $\text{N}_2\text{O}$ .

Why do some gases absorb infrared strongly?

Two complementary models explain selective absorption.

Model 1 — Resonance model

Molecules have natural frequencies of vibration.
An electromagnetic wave can transfer energy efficiently when its frequency matches an infrared-active vibrational mode.
In the classical description, this is associated with resonance and an increased vibration amplitude.
For strong electric-dipole infrared absorption, the vibration must change the molecule’s electric dipole moment, meaning its separation of positive and negative charge.
Some vibrational modes of greenhouse gases meet this condition and occur in the infrared range.
The ordinary vibrations of $\text{N}_2$ and $\text{O}_2$ do not change their electric dipole moment, so they do not absorb strongly through these transitions.
Infrared-active vibrations occur at much lower frequencies than visible light. Many greenhouse gases therefore transmit much of the visible sunlight while absorbing selected infrared wavelengths.

Model 2 — Molecular energy-level model

Light can be treated as photons with energy:

$\displaystyle E=hf=\frac{hc}{\lambda}$

Molecules have discrete rotational and vibrational energy levels.
An infrared photon can be absorbed when its energy matches an allowed molecular transition, within the absorption line’s width.
Infrared photon energies match allowed transitions in greenhouse gases.
Absorbed energy can be shared with neighbouring molecules through collisions.
The atmosphere also emits infrared radiation in all directions. An absorbed photon need not be immediately re-emitted by the same molecule.
$\text{N}_2$ and $\text{O}_2$ have vibrational energy levels, but their ordinary vibrational transitions are electric-dipole forbidden.

Important: matching the photon energy to an energy gap is necessary but is not, by itself, sufficient for strong absorption. The transition must also be allowed.

Atmospheric emission and reduced net energy loss

Infrared radiation emitted by the atmosphere travels in different directions:

Some travels upward and can eventually escape to space.
Some travels downward and is absorbed by Earth’s surface.

Downward atmospheric radiation reduces the surface’s net radiative loss. The surface consequently reaches a higher equilibrium temperature than it would under solar heating alone.

Energy balance and conservation of energy

At radiative equilibrium, the total solar power absorbed by the Earth–atmosphere system equals its total thermal power emitted to space.

A change in absorption or emission produces an energy imbalance. The system’s temperature changes as energy is stored or lost, and a new equilibrium may eventually be reached.

Simplified energy-balance geometry

Earth intercepts sunlight over its projected disk area, $\pi R^2$ , while the global average is taken over its full spherical surface area, $4\pi R^2$ .

The total incident solar power is:

$\displaystyle P_{\mathrm{incident}}=S\pi R^2$

Dividing by the spherical surface area gives:

$\displaystyle F_{\mathrm{in}}=\frac{S\pi R^2}{4\pi R^2}=\frac{S}{4}$

Here, $F$ denotes power per unit area, also called energy flux, measured in $\text{W m}^{-2}$ .

Accounting for albedo, the average absorbed solar flux is:

$\displaystyle F_{\mathrm{abs}}=(1-\alpha)\frac{S}{4}$

Worked example 5 — Absorbed solar power

Calculate the average solar power absorbed per square metre by the Earth–atmosphere system if its albedo is $\alpha=0.30$ . The average incident solar flux is $340\,\text{W m}^{-2}$ .

Show answer and full working

$\displaystyle F_{\mathrm{abs}}=(1-\alpha)(340)$

$\displaystyle F_{\mathrm{abs}}=(1-0.30)(340)=238\,\text{W m}^{-2}$

Answer: $238\,\text{W m}^{-2}$ .

This is the absorption by the combined surface–atmosphere system, not solely by the ground.

Key ideas — B.2.2

Greenhouse gases absorb and emit infrared radiation at selected wavelengths.
Resonance and allowed molecular transitions explain selective absorption.
Downward atmospheric radiation reduces net radiative energy loss from the surface.
The global average incoming solar flux is $S/4$ .
At equilibrium, absorbed solar power equals thermal power emitted to space.

B.2.3 The effect of greenhouse gas concentrations

Learning objectives

Identify important greenhouse gases and their natural and human-related sources.
Explain how increasing greenhouse gas concentrations affect Earth’s energy balance.
Distinguish between the natural and enhanced greenhouse effects.
Analyse how changes in model parameters alter equilibrium temperature.
Explain positive feedbacks and the stabilising response of increased thermal emission.
Recognise the assumptions and limitations of simplified climate models.

Sources and sinks of greenhouse gases

Carbon dioxide — $\text{CO}_2$

Natural sources: respiration, decomposition, volcanic activity, and release from oceans.
Natural sinks: photosynthesis and ocean uptake.
Human sources: fossil fuel combustion, cement production, and deforestation.
Ocean uptake also changes with temperature and chemistry.

Methane — $\text{CH}_4$

Natural sources: wetlands and anaerobic decomposition.
Sinks: chemical reactions in the atmosphere and uptake by some soils.
Human sources: livestock farming, fossil fuel extraction and processing, rice cultivation, and landfill decomposition.

Water vapour — $\text{H}_2\text{O}$

Water enters and leaves the atmosphere through the water cycle.
Warmer air can sustain more water vapour before saturation, and evaporation generally increases as the climate warms.
Combustion also produces water, but ordinary surface emissions are not the main control on global atmospheric water vapour.
Water vapour mainly acts as a feedback that amplifies warming initiated by other changes.

Nitrous oxide — $\text{N}_2\text{O}$

Natural sources: microbial activity in soils and oceans.
Human sources: agricultural fertilisers, manure management, industrial processes, and combustion.

Natural and enhanced greenhouse effects

Natural greenhouse effect: atmospheric absorption and emission of infrared radiation help maintain a surface temperature suitable for life.
Enhanced greenhouse effect: human activities increase greenhouse gas concentrations, changing outgoing infrared radiation and producing additional warming.

A simplified uniform-temperature model with Earth’s present albedo and no greenhouse warming gives approximately $255\,\text{K}$ , or $-18^\circ\text{C}$ . This is an effective radiating temperature rather than a detailed prediction of the average surface temperature of an atmosphere-free Earth.

Energy balance and greenhouse gas concentration

In a simplified explanation, incoming solar irradiance and albedo are held fixed when greenhouse gas concentration increases. At the initial temperature, the increase reduces the outgoing infrared radiation to space.

Absorbed power temporarily exceeds outgoing power.
The Earth–atmosphere system gains energy.
The surface and lower atmosphere warm.
Thermal emission increases as temperature rises.
A new equilibrium can be reached at a higher temperature if the external conditions and greenhouse gas concentrations stabilise.

Worked example 6 — Conceptual energy balance

Explain qualitatively how an increase in atmospheric carbon dioxide concentration leads to global warming.

Show answer and explanation

Increasing $\text{CO}_2$ changes the atmosphere’s absorption and emission of thermal infrared radiation. At the initial temperature, less energy escapes to space, while absorbed solar power is approximately unchanged in this explanation.

The resulting positive energy imbalance causes the Earth–atmosphere system to gain energy and warm. As temperatures rise, thermal emission increases. A higher equilibrium temperature is needed to restore the balance between absorbed solar power and outgoing thermal power.

Feedback mechanisms

A feedback occurs when an initial change causes another change that influences the original effect.

Positive feedback — water vapour

An initial warming allows more water vapour to remain in the atmosphere.
Additional water vapour increases infrared absorption.
This amplifies the initial warming.

Positive feedback — ice and albedo

Warming melts snow and ice.
Darker surfaces are exposed, reducing albedo.
More solar radiation is absorbed.
This produces further warming.

Stabilising radiative response

As temperature rises, thermal emission increases. This tends to oppose the original warming and helps the system approach a new equilibrium. Positive feedbacks amplify a change; they do not automatically imply unlimited warming.

Quantitative equilibrium model

A simple model treats Earth as a uniform-temperature sphere. Let $e_{\mathrm{eff}}$ describe the outgoing radiation to space relative to a black body at the model’s surface temperature.

At equilibrium:

$\displaystyle (1-\alpha)\frac{S}{4}=e_{\mathrm{eff}}\sigma T^4$

Rearranging:

$\displaystyle T=\left(\frac{(1-\alpha)S}{4e_{\mathrm{eff}}\sigma}\right)^{1/4}$

Increasing greenhouse gases can be represented in this model by a decrease in $e_{\mathrm{eff}}$ : less infrared energy escapes to space at the same model surface temperature. To restore balance, $T$ must increase.

Surface emissivity and effective emissivity

Surface emissivity, $e$ , describes the emission of a material surface.

Effective emissivity, $e_{\mathrm{eff}}$ , is a simplified parameter describing the outgoing radiation of the whole surface–atmosphere system relative to the model surface temperature.

A reduction in effective emissivity does not mean that greenhouse gases have changed the ground’s material emissivity.

Worked example 7 — Effect of effective emissivity

Assume the Earth–atmosphere system absorbs $238\,\text{W m}^{-2}$ . Calculate the model equilibrium temperatures when the effective emissivity changes from $e_{\mathrm{eff}}=1.0$ to $e_{\mathrm{eff}}=0.95$ .

Show answer and full working

Step 1: Write the equilibrium relation.

$\displaystyle F_{\mathrm{abs}}=e_{\mathrm{eff}}\sigma T^4$

Step 2: Rearrange.

$\displaystyle T=\left(\frac{F_{\mathrm{abs}}}{e_{\mathrm{eff}}\sigma}\right)^{1/4}$

Step 3: Calculate the initial temperature.

$\displaystyle T_1=\left(\frac{238}{(1.0)(5.67\times10^{-8})}\right)^{1/4}$

$\displaystyle T_1\approx254.54\,\text{K}$

Step 4: Calculate the new temperature.

$\displaystyle T_2=\left(\frac{238}{(0.95)(5.67\times10^{-8})}\right)^{1/4}$

$\displaystyle T_2\approx257.82\,\text{K}$

Step 5: Find the temperature increase.

$\displaystyle \Delta T=257.82-254.54\approx3.3\,\text{K}$

Answer: the equilibrium temperature increases from approximately $255\,\text{K}$ to $258\,\text{K}$ , with an increase of about $3.3\,\text{K}$ calculated before rounding the temperatures.

This illustrates the model’s response to an assumed parameter change. It is not a forecast of Earth’s future temperature.

Worked example 8 — Comparative reasoning

Two planets receive the same solar irradiance and have the same albedo. Planet A has a higher concentration of greenhouse gases than Planet B. In the simplified model, assuming other relevant atmospheric properties are comparable, which planet has the higher equilibrium surface temperature?

Show answer and explanation

Planet A has the higher equilibrium surface temperature in this model.

Its stronger greenhouse effect reduces outgoing infrared radiation at a given surface temperature, corresponding to a smaller effective emissivity. The surface must reach a higher temperature for the outgoing thermal power to balance the absorbed solar power.

Evidence for the enhanced greenhouse effect

Measurements show that atmospheric carbon dioxide has increased substantially since the Industrial Revolution.
Laboratory and atmospheric spectra show absorption at characteristic greenhouse-gas wavelengths.
Temperature records show long-term warming.
Climate attribution studies find that natural influences alone do not explain the observed long-term warming. Human influences, especially greenhouse-gas emissions, are necessary to explain it.

Key terms

Greenhouse effect, greenhouse gases, infrared radiation, resonance, molecular energy levels, electric dipole moment, conservation of energy, natural greenhouse effect, enhanced greenhouse effect, emissivity, effective emissivity, albedo, solar constant, equilibrium temperature, and feedback mechanism.

IB exam tips

Define symbols clearly when introducing an equation.
Remember the geometry factor: the global average incoming solar flux is $S/4$ .
Use kelvin for the Stefan–Boltzmann and Wien equations.
Distinguish total power, measured in watts, from power per unit area, measured in $\text{W m}^{-2}$ .
For greenhouse explanations, mention infrared absorption, atmospheric emission, reduced net surface energy loss, and a higher equilibrium temperature.
Explain that molecular transitions must be allowed, as well as having suitable energy differences.
Distinguish material emissivity from the effective emissivity used in a simplified climate model.
Keep extra digits in intermediate calculations and round the final answer appropriately.

Essential equations — B.2

Emissivity and albedo are dimensionless. Use kelvin for all radiation calculations.

$\displaystyle e=\frac{P/A}{\sigma T^4}$

$\displaystyle \alpha=\frac{P_{\mathrm{scattered}}}{P_{\mathrm{incident}}}$

$\displaystyle S=\frac{L_{\odot}}{4\pi r^2}$

$\displaystyle F_{\mathrm{abs}}=(1-\alpha)\frac{S}{4}$

$\displaystyle T=\left(\frac{(1-\alpha)S}{4e_{\mathrm{eff}}\sigma}\right)^{1/4}$

Two distinct quantities: $e$ measures emission at thermal wavelengths. $\alpha$ measures reflection of incoming solar radiation. They generally refer to different wavelength ranges and must not be treated as interchangeable.

Comparing equilibrium temperatures

For two states of the same simplified planet model:

$\displaystyle \frac{T_2}{T_1}=\left[\frac{(1-\alpha_2)S_2e_{\mathrm{eff},1}}{(1-\alpha_1)S_1e_{\mathrm{eff},2}}\right]^{1/4}$

At fixed albedo and effective emissivity, $T\propto S^{1/4}$ .
At fixed irradiance and albedo, $T\propto e_{\mathrm{eff}}^{-1/4}$ .
A lower albedo means more absorbed sunlight and therefore a higher model equilibrium temperature, with other parameters fixed.

Model assumptions and limitations

The planet is spherical and represented by one uniform temperature.
Albedo and effective emissivity are treated as prescribed average values.
The model describes radiative equilibrium, not how quickly the planet warms.
It does not explicitly resolve clouds, atmospheric layers, convection, evaporation, or geographical temperature differences.
Real climate models represent wavelength-dependent absorption, atmospheric structure, feedbacks, and ocean heat storage in much more detail.

Practice — check your understanding

1. Why divide the solar constant by four?

Explain why the global average incoming solar power per unit area is $S/4$ , rather than $S$ or $S/2$ .

Show answer

The planet intercepts sunlight over its projected disk area, $\pi R^2$ , while the global average is taken over its spherical surface area, $4\pi R^2$ .

$\displaystyle F_{\mathrm{in}}=\frac{S\pi R^2}{4\pi R^2}=\frac{S}{4}$

The geometry includes both the dark hemisphere and the varying angles of illumination on the sunlit hemisphere.

2. Reflected and absorbed solar radiation

A planet has solar irradiance $S=1200\,\text{W m}^{-2}$ and albedo $\alpha=0.25$ . Calculate the global average incident, reflected, and absorbed solar fluxes.

Show answer and working

Incident flux:

$\displaystyle F_{\mathrm{in}}=\frac{1200}{4}=300\,\text{W m}^{-2}$

Reflected flux:

$\displaystyle F_{\mathrm{reflected}}=(0.25)(300)=75\,\text{W m}^{-2}$

Absorbed flux:

$\displaystyle F_{\mathrm{abs}}=(0.75)(300)=225\,\text{W m}^{-2}$

The reflected and absorbed fluxes add to the incident flux.

3. Thermal emission

A surface at $300\,\text{K}$ has emissivity $e=0.80$ . Calculate its emitted power per unit area. Use $\sigma=5.67\times10^{-8}\,\text{W m}^{-2}\text{K}^{-4}$ .

Show answer and working

$\displaystyle \frac{P}{A}=e\sigma T^4$

$\displaystyle \frac{P}{A}=(0.80)(5.67\times10^{-8})(300)^4$

$\displaystyle \frac{P}{A}\approx367\,\text{W m}^{-2}$

This is the emitted flux. Net radiative loss also depends on radiation received from the surroundings.

4. Ice–albedo feedback

In a simplified equilibrium model, albedo decreases from $0.30$ to $0.20$ . Solar irradiance and effective emissivity remain constant. Calculate the ratio of the new temperature to the original temperature and explain the direction of the change.

Show answer and reasoning

$\displaystyle \frac{T_2}{T_1}=\left(\frac{1-0.20}{1-0.30}\right)^{1/4}$

$\displaystyle \frac{T_2}{T_1}=\left(\frac{0.80}{0.70}\right)^{1/4}\approx1.034$

The absolute temperature increases by approximately 3.4% in this model. Less reflection means more absorbed solar energy.

This is an illustrative change in a simplified model, not a prediction for Earth.

5. Infrared absorption and energy conservation

A student says: “Nitrogen cannot absorb infrared because it has no vibrational energy levels. Greenhouse gases create extra energy by sending radiation back to the ground.” Explain the errors.

Show answer

Nitrogen molecules do have vibrational energy levels. Their ordinary vibrational transitions do not change the electric dipole moment, so strong electric-dipole infrared absorption is forbidden. Both the energy match and the transition’s selection rules matter.

Greenhouse gases do not create energy. Absorption, collisions, and emission redistribute energy within the surface–atmosphere system. Downward atmospheric radiation reduces the surface’s net radiative loss. At equilibrium, absorbed solar power still equals outgoing thermal power at the top of the atmosphere.

Faith connection — caring for the world entrusted to us

“The earth is the LORD’S, and the fulness thereof; the world, and they that dwell therein.”
Psalm 24:1 — KJV

Creation — a world that belongs to God

The Bible presents the Earth as God’s creation. Its atmosphere, water, and sunlight support life through processes that we can study with care and curiosity. The natural greenhouse effect is part of the physical system that makes Earth habitable. As Christians, we respond with gratitude and recognise our responsibility to care for what God has entrusted to us, as expressed in Genesis 2:15.

The Fall — broken relationships and careless choices

Sin has damaged our relationship with God, one another, and the world around us. Selfishness and waste can influence how we use resources and how we treat people affected by environmental change. Learning the science carefully can help us make informed decisions and serve our neighbours with compassion.

Redemption — new life through Jesus Christ

Jesus Christ came to bring forgiveness and reconciliation through His death and resurrection. Salvation is God’s gift of grace, received through faith, as taught in Ephesians 2:8–10. Turn to Jesus in repentance and faith, and receive Him as your personal God, Saviour, and Lord. His love can shape both our hearts and our daily actions.

Restoration — hope expressed through faithful care

God promises a renewed creation in which suffering and death will end, as described in Revelation 21:4. That hope encourages us to live responsibly today: use energy thoughtfully, avoid needless waste, seek truth, and care for vulnerable people. Connect with other believers, read and study the Bible, and spend time in prayer so that your faith and knowledge of God continue to grow.

Reflect and pray

Reflect: What practical step could you take this week to use resources more carefully or help someone in need? How can you discuss scientific evidence with honesty, humility, and respect?

Prayer: Lord Jesus, thank You for the world You have made and for Your saving grace. Forgive my sins and help me follow You as my God, Saviour, and Lord. Give me wisdom to understand Your creation, use resources responsibly, and love my neighbours. Help me grow through Your Word, prayer, and fellowship with other believers. Amen.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Greenhouse effect

B.2 — Greenhouse Effect

Big picture

B.2.1 Energy from the Sun

Learning objectives

Energy received from the Sun

Emissivity

Definition — emissivity

Worked example 1 — Emissivity

Albedo

Definition — albedo

Why Earth’s albedo varies

Worked example 2 — Albedo

The solar constant

Definition — solar constant

Calculating the solar constant

Worked example 3 — Solar irradiance at Mercury

Key ideas — B.2.1

B.2.2 Energy balance in the Earth surface–atmosphere system

Learning objectives

The greenhouse effect

Earth–Moon comparison

Incoming and outgoing radiation

Worked example 4 — Peak wavelength emitted by Earth

Greenhouse gases

Why do some gases absorb infrared strongly?

Model 1 — Resonance model

Model 2 — Molecular energy-level model

Atmospheric emission and reduced net energy loss

Energy balance and conservation of energy

Simplified energy-balance geometry

Worked example 5 — Absorbed solar power

Key ideas — B.2.2

B.2.3 The effect of greenhouse gas concentrations

Learning objectives

Sources and sinks of greenhouse gases

Natural and enhanced greenhouse effects

Energy balance and greenhouse gas concentration

Worked example 6 — Conceptual energy balance

Feedback mechanisms

Stabilising radiative response

Quantitative equilibrium model

Surface emissivity and effective emissivity

Worked example 7 — Effect of effective emissivity

Worked example 8 — Comparative reasoning

Evidence for the enhanced greenhouse effect

Key terms

IB exam tips

Essential equations — B.2

Comparing equilibrium temperatures

Model assumptions and limitations

Practice — check your understanding

1. Why divide the solar constant by four?

2. Reflected and absorbed solar radiation

3. Thermal emission

4. Ice–albedo feedback

5. Infrared absorption and energy conservation

Faith connection — caring for the world entrusted to us

Creation — a world that belongs to God

The Fall — broken relationships and careless choices

Redemption — new life through Jesus Christ

Restoration — hope expressed through faithful care

Further reading

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