IB DP Physics · Theme B · The particulate nature of matter

Greenhouse effect.

B.2 · The energy balance of a planet, greenhouse gases and the enhanced greenhouse effect
Standard level and higher level

SL + HL · 6 hours B.1 Thermal energy transfers ↗

The journey

Three parts.

Part 01

Energy in, energy out

Emissivity, albedo, the solar constant and a planet with no atmosphere

Part 02

Greenhouse gases

Which gases absorb infrared, and why

Part 03

The balance with an atmosphere

A model of the greenhouse effect, and how human activity enhances it

B.1 gave the tools: \(\sigma T^{4}\), the inverse square law and Wien’s law. Here they are used on a whole planet.

B.2 GREENHOUSE EFFECT · SL + HL

Part 01

Energy in, energy out.

What sets the temperature of a planet

B.2 GREENHOUSE EFFECT · SL + HL

Conservation of energy

The energy balance of a planet.

The only way in or out is radiation
In: sunlight, less what is reflected
Out: infrared from the planet

Steady temperature: power absorbed = power radiated.

A PLANET IN SUNLIGHT

B.2 GREENHOUSE EFFECT · SL + HL

Why radiation only

A planet is surrounded by empty space. Conduction and convection need matter, so they can move energy around inside the planet and its atmosphere but cannot carry any across the gap to or from the Sun.

The balance finds itself

Suppose the planet absorbed more than it radiated. Its temperature would rise, and by the Stefan–Boltzmann law a hotter body radiates more, as \(T^{4}\). The temperature would keep rising until the output had caught up with the input, and then stop. If it radiated more than it absorbed it would cool until the two matched. Whatever the starting point, the planet settles at the one temperature where power out equals power in. Change either side of the balance and the planet moves to a new equilibrium temperature.

Almost every calculation in this topic is one line of physics: write power in = power out, then solve for the unknown.

Real surfaces are not black bodies

Emissivity.

\(\text{emissivity} = \dfrac{\text{power radiated per unit area}}{\sigma T^{4}}\)

Compares a surface with a black body at the same temperature
A number from 0 to 1, with no unit
So the power radiated per unit area is \(\epsilon \sigma T^{4}\)

Surface (approximate values)	Emissivity \(\epsilon\)
Black body	1
Ocean	0.96
Soil, vegetation	0.9 to 0.98
The Earth and its atmosphere, seen from space	0.6
Polished aluminum	0.05

B.2 GREENHOUSE EFFECT · SL + HL

A worked case

A black body at 288 K radiates \(\sigma T^{4} = 5.67 \times 10^{-8} \times 288^{4} = 390 \,\mathrm{W} \,\mathrm{m}^{-2}\). A field at the same temperature that radiates 370 W m⁻² has an emissivity of \(\frac{370}{390} = 0.95\).

Why the Earth as a whole is only 0.6

The ground and sea are nearly black in the infrared. But a satellite looking down measures about 240 W m⁻² leaving the planet, not the 390 W m⁻² that the surface emits. The atmosphere has absorbed part of it on the way up. Treating the whole planet as a single surface at 288 K, its emissivity is \(\frac{240}{390} = 0.6\). That low number is the greenhouse effect expressed as an emissivity.

The data booklet gives emissivity only in words. In symbols, with \(\epsilon\) for emissivity: \(P = \epsilon \sigma A T^{4}\). Albedo has no symbol there either; these slides write \(\alpha\)..

A good absorber is a good emitter: at each wavelength a surface absorbs the same fraction that it emits, compared with a black body.

How much is reflected

Albedo.

\(\text{albedo} = \dfrac{\text{total scattered power}}{\text{total incident power}}\)

From 0 (absorbs everything) to 1 (reflects everything)
The fraction absorbed is \(1 - \alpha\)
The Earth’s albedo varies daily, with cloud cover and with latitude

Surface (approximate values)	Albedo \(\alpha\)
Fresh snow	0.85
Thick cloud	0.7
Desert sand	0.4
Forest	0.15
Ocean	0.06
The Earth, on average	0.30

B.2 GREENHOUSE EFFECT · SL + HL

Why it varies

Clouds are bright from above, so a cloudy day reflects more sunlight than a clear one over the same sea. Near the poles there is ice and snow, and the Sun is low, so light strikes the surface at a glancing angle and more is reflected. The seasons change the snow cover and the leaves on the trees. The figure of 0.30 is an average over the whole globe and the whole year.

Ice and albedo

Ice reflects about 85% of the sunlight that reaches it; the open sea that replaces it when it melts reflects about 6%. So a warmer planet with less ice absorbs more sunlight, which warms it further. A change that feeds itself in this way is called a positive feedback, and it is one reason the Arctic is warming faster than the rest of the planet.

Multiplying by the albedo to find the power absorbed. The albedo is the part sent back. The part absorbed is \(1 - \alpha\): 70% for the Earth.

What arrives from the Sun

The solar constant.

\(S = 1.36 \times 10^{3} \,\mathrm{W} \,\mathrm{m}^{-2}\)

The intensity of sunlight at the Earth’s distance, above the atmosphere
On a surface at right angles to the rays
It is the Sun’s apparent brightness from B.1

For another planet: the inverse square law \(S = \dfrac{L}{4 \pi d^{2}}\) so \(S \propto \dfrac{1}{d^{2}}\) Venus is at 0.72 of the Earth’s distance \(S_\mathrm{V} = \dfrac{S}{0.72^{2}} = \dfrac{1.36 \times 10^{3}}{0.72^{2}}\) \(S_\mathrm{V} = \class{res}{2.6 \times 10^{3} \,\mathrm{W} \,\mathrm{m}^{-2}}\)

B.2 GREENHOUSE EFFECT · SL + HL

“Constant” is generous

The Earth’s orbit is slightly elliptical, so the intensity is about 7% higher in early January, when the Earth is closest to the Sun, than in early July. The Sun’s own output also rises and falls by about 0.1% over its eleven-year cycle. \(1.36 \times 10^{3} \,\mathrm{W} \,\mathrm{m}^{-2}\) is the average.

Why the ground gets less

About 1000 W m⁻² reaches the ground at noon on a clear day with the Sun overhead, because the atmosphere reflects and absorbs the rest. When the Sun is lower, the same beam is spread over a larger area of ground, so each square meter receives less. That, far more than distance, is why the poles are cold and why winter is cold.

The solar constant is in the data booklet. A question may give a different “constant” for another star or planet: use it in exactly the same way.

Averaging over the whole planet

A quarter of the solar constant.

Power intercepted: a disk of radius \(R\) \(P = S \pi R^{2}\) Shared over the whole sphere \(\text{mean intensity} = \dfrac{S \pi R^{2}}{4 \pi R^{2}} = \dfrac{S}{4}\)

For the Earth: \(\dfrac{1360}{4} = 340 \,\mathrm{W} \,\mathrm{m}^{-2}\).

A DISK AND A SPHERE

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Where the disk comes from

Stand far behind the planet and look towards the Sun. The planet blocks a circular patch of sunlight: its shadow. Every ray that would have passed through that circle has landed on the planet instead, so the power intercepted is the intensity times the area of the circle, \(\pi R^{2}\). The guide calls this the projected surface along the direction of the rays.

Two halvings

You can also see the quarter as two factors of a half. Half the planet is in darkness at any moment. And on the sunlit half most of the surface is tilted to the rays, so the beam is spread out: averaged over a hemisphere, the tilt halves the intensity again.

Why an average is fair

The planet spins, and its winds and oceans move energy around, so the temperature differences between day and night and between the equator and the poles are small compared with 288 K. The model treats the whole surface as being at one temperature.

Using \(S\) itself for the mean intensity, or halving it for “day and night”. Over the whole surface it is \(\frac{S}{4}\).

Worked example

The Earth with no greenhouse effect.

Albedo 0.30 · the planet radiates as a black body Power absorbed = power radiated, per square meter \((1 - \alpha) \dfrac{S}{4} = \sigma T^{4}\) \(T^{4} = \dfrac{(1 - \alpha) S}{4 \sigma} = \dfrac{0.70 \times 1.36 \times 10^{3}}{4 \times 5.67 \times 10^{-8}}\) \(T^{4} = 4.20 \times 10^{9} \,\mathrm{K}^{4}\) \(T = \class{res}{255 \,\mathrm{K}}\)

That is −18 °C
The measured mean surface temperature is 288 K, +15 °C

The 33 K difference is the greenhouse effect.

B.2 GREENHOUSE EFFECT · SL + HL

The numbers on the way

Of the 340 W m⁻² arriving on average, 30% is reflected and 70% is absorbed: \(0.70 \times 340 = 238 \,\mathrm{W} \,\mathrm{m}^{-2}\). At equilibrium the planet must radiate 238 W m⁻² to space. A black body does that at 255 K.

255 K is a real temperature

Seen from space, the Earth does radiate like a body at 255 K: satellites measure about 240 W m⁻² of infrared leaving it. The balance at the top of the atmosphere is correct. What the calculation gets wrong is where that radiation comes from. Most of it is emitted by the cold upper atmosphere, not by the ground. The ground, underneath the absorbing gases, is 33 K warmer.

A fourth root is forgiving

Because \(T \propto (\text{intensity})^{\frac{1}{4}}\), a 1% change in the absorbed intensity changes \(T\) by only about 0.25%. On 255 K that is 0.6 K, which is still a large change for a climate.

To take a fourth root, press the square-root key twice, or raise to the power 0.25.

Quick check

Energy in, energy out.

01A surface has albedo 0.25. What fraction does it absorb?

0.75

02What is the emissivity of a black body?

1

03Why is the mean intensity \(\frac{S}{4}\), not \(S\)?

A disk \(\pi R^{2}\) intercepts; a sphere \(4 \pi R^{2}\) shares it

04A planet is twice as far from its star as the Earth. Its solar constant?

A quarter of \(S\)

05Ice melts to open sea. What happens to the albedo there?

It falls, so more is absorbed

06A surface at 300 K radiates 413 W m⁻²; a black body would radiate 459. Emissivity?

\(\frac{413}{459} = 0.90\)

PRESS → TO REVEAL EACH ANSWER

Part 02

Greenhouse gases.

Why a few trace gases matter so much more than nitrogen and oxygen

B.2 GREENHOUSE EFFECT · SL + HL

The story · London, 1859

Background story · not assessed

John Tyndall tests the air.

Photograph of John Tyndall, an elderly man with a long white beard, seated in a dark coat — **John Tyndall**, about 1885

Engraving of Tyndall’s apparatus with modern labels: a long tube holding the gas to be tested, a source of heat at each end, a thermopile to compare them, a vacuum pump and a gas supply — His apparatus, drawn in 1861 (labels added later)

Professor at the Royal Institution, he began on 9 May 1859
Radiant heat passed down a tube of gas to a thermopile, which compared it with a second source
He measured nitrogen, oxygen, water vapor, carbon dioxide, methane and more
Water vapor was the strongest absorber in the atmosphere

The gases that make up 99% of the air let the heat through. The trace gases did not.

Portrait: Herbert Rose Barraud, 1880s, public domain. Apparatus: John Tyndall, 1861, annotated, public domain.

B.2 GREENHOUSE EFFECT · SL + HL

Two very different spectra

Sunlight in, infrared out.

From the Sun: peak near 0.5 μm, visible
From the Earth: peak near 10 μm, infrared
The atmosphere is mostly transparent to the first
Greenhouse gases absorb much of the second

TWO BLACK BODIESEACH SCALED TO THE SAME HEIGHT

B.2 GREENHOUSE EFFECT · SL + HL

The two peaks from Wien’s law

For the Sun, \(\lambda_\mathrm{max} = \frac{2.9 \times 10^{-3}}{5800} = 5.0 \times 10^{-7}\) m. For the Earth’s surface, \(\lambda_\mathrm{max} = \frac{2.9 \times 10^{-3}}{288} = 1.0 \times 10^{-5}\) m, twenty times longer. The two curves barely overlap, which is why the incoming and outgoing radiation can be treated so differently.

A one-way valve

The whole greenhouse effect rests on this difference. The atmosphere lets most of the short-wavelength sunlight straight through to warm the ground. The ground sends the energy back as long-wavelength infrared, and for that the same atmosphere is partly opaque. Energy gets in more easily than it gets out.

The name is a poor one. A garden greenhouse stays warm mainly because its glass stops warm air from rising away, not because of what the glass does to infrared.

The curves are drawn to the same height so that both can be seen. Sunlight is far more intense at its source.

Four to know

The main greenhouse gases.

Gas	Natural sources	Sources from human activity
Water vapor, H₂O	Evaporation from oceans, lakes and plants	Irrigation; burning fuels; more evaporation as the planet warms
Carbon dioxide, CO₂	Respiration, decay, volcanoes, wildfires	Burning fossil fuels; clearing forests; making cement
Methane, CH₄	Wetlands; termites	Cattle and rice fields; landfill; leaks of natural gas
Nitrous oxide, N₂O	Bacteria in soils and oceans	Nitrogen fertilizers; burning fuels

B.2 GREENHOUSE EFFECT · SL + HL

How little there is

Dry air is 78% nitrogen, 21% oxygen and 0.9% argon. Carbon dioxide is 0.04%, methane 0.0002% and nitrous oxide 0.00003%. Water vapor varies from almost nothing over a cold desert to about 4% in the humid tropics. The gases that control the planet’s loss of infrared add up to well under 1% of the air.

Why water vapor is treated differently

It causes more of the natural greenhouse effect than any other gas. But the amount in the air is set by the temperature: add extra and it rains out within days. Carbon dioxide stays for centuries. So carbon dioxide sets the temperature, and water vapor follows and strengthens the change: warmer air holds more water vapor, which absorbs more infrared.

Molecule for molecule

Methane and nitrous oxide absorb far more strongly than carbon dioxide, but there is much less of them, and methane breaks down in about twelve years. Carbon dioxide dominates the human contribution because so much of it is emitted and it lasts so long.

Be ready to name all four with their formulas, and to give one natural and one human source for each.

First explanation

Molecular energy levels.

A molecule can vibrate only with certain discrete energies
It absorbs a photon whose energy \(hf\) matches a gap
For these gases the gaps match infrared photons
It then emits a photon in a random direction

VIBRATIONAL ENERGY LEVELS

B.2 GREENHOUSE EFFECT · SL + HL

The same idea as atomic spectra

In E.1 an atom absorbs a photon only if its energy equals the difference between two electron energy levels. A molecule has an extra set of levels, for the energy of its vibration. The steps between them are much smaller than the steps between electron levels: about 0.1 eV, against a few eV. A photon of 0.1 eV has a wavelength of about 12 μm, in the middle of the Earth’s infrared emission.

Why sunlight passes

A visible photon carries 2 to 3 eV. That matches none of the vibrational gaps, and it is too little for most of the electron transitions in these molecules. With no gap to fit, it is not absorbed.

What happens to the absorbed energy

The excited molecule may emit a photon, in any direction. More often, in the lower atmosphere, it collides with a nitrogen or oxygen molecule first and passes the energy on as kinetic energy: the air warms. Either way the energy has been stopped on its way to space.

Use \(hc = 1.24 \times 10^{-6}\) eV m from the data booklet: \(E = \frac{hc}{\lambda}\).

Second explanation

Resonance.

The atoms of a molecule vibrate at a natural frequency
Infrared is an oscillating electric field: it drives the molecule
When the two frequencies match, the amplitude grows: resonance
Energy passes from the radiation to the molecule

A MOLECULE DRIVEN BY INFRARED

B.2 GREENHOUSE EFFECT · SL + HL

A mass on a spring

The bonds in a molecule behave like springs joining the atoms. From C.1, a mass on a spring has one natural frequency, set by the mass and the stiffness. Atoms are very light and bonds are very stiff, so the natural frequencies are enormous: about \(2 \times 10^{13}\) Hz for the bending of carbon dioxide. That is the frequency of infrared of wavelength 15 μm. C.4 treats resonance fully: a system driven at its natural frequency takes in energy most readily.

Why nitrogen and oxygen do not absorb

The field of the wave can only push on charge. In carbon dioxide the oxygen atoms are slightly negative and the carbon slightly positive, so when the molecule bends, its charge moves and the field can drive it. N₂ and O₂ are two identical atoms. However they vibrate, there is no separation of charge for the field to act on, so infrared passes straight through them.

Where the model falls short

It does not explain why absorption happens only in definite amounts of energy. For that you need the energy levels of the last slide. The two explanations describe the same thing: the level spacing is \(hf\), where \(f\) is the natural frequency.

The guide expects both models. If asked to “explain how a greenhouse gas absorbs infrared”, either is accepted; name the one you are using.

The result

Absorbed, then emitted in all directions.

The surface radiates infrared upwards
Greenhouse gases absorb it and emit in all directions
Part comes back down and warms the surface

The surface receives sunlight and infrared from the atmosphere.

INFRARED IN THE ATMOSPHEREDOTS: GREENHOUSE GAS MOLECULES

B.2 GREENHOUSE EFFECT · SL + HL

Why the surface must be warmer

The surface now has two sources of energy: the sunlight it absorbs, and the infrared sent down by the atmosphere. To be in balance it must radiate away the sum of the two, which is more than it would radiate with no atmosphere. By \(\sigma T^{4}\), radiating more means being hotter.

Nothing is trapped for ever

Words such as “trapped” suggest that energy cannot get out. It always does, and at equilibrium exactly as much leaves the planet as arrives. The greenhouse gases delay it: the energy is absorbed and emitted many times on the way up, and the surface below has to be warmer to push the same power through.

Not reflection

The gases do not reflect the infrared. Each molecule absorbs a photon and later emits a new one in a direction that has nothing to do with where the first came from.

Saying that greenhouse gases “reflect heat back” or “trap sunlight”. They absorb infrared from the surface and re-radiate it in all directions.

Quick check

Greenhouse gases.

01Name the four main greenhouse gases.

H₂O, CO₂, CH₄, N₂O

02Which part of the spectrum do they absorb?

Infrared

03Why does sunlight pass through them?

Its photon energies match no gap

04In the resonance model, what must match?

The infrared and natural frequencies

05In which direction is the absorbed energy re-emitted?

All directions

06Give a human source of methane.

Cattle, rice fields, landfill or gas leaks

PRESS → TO REVEAL EACH ANSWER

Part 03

The balance with an atmosphere.

A model of the greenhouse effect, and what happens when the atmosphere changes

B.2 GREENHOUSE EFFECT · SL + HL

Surface and atmosphere

A one-layer model of the atmosphere.

The layer lets sunlight through
It absorbs all the infrared from the surface
It radiates equally up and down

Each part is in balance: what it absorbs, it radiates.

THREE BALANCESINTENSITIES IN W m⁻²

B.2 GREENHOUSE EFFECT · SL + HL

Start at the top

The planet as a whole must still send 238 W m⁻² to space, to match the sunlight it absorbs. In this model the only radiation that reaches space comes from the upper face of the layer, so the layer radiates 238 W m⁻² upwards.

Then the layer

A layer has two faces and radiates the same from both: 238 W m⁻² down as well as up, 476 W m⁻² in all. Its only income is the infrared from the surface. For the layer to be in balance, the surface must be radiating 476 W m⁻².

Check the surface

It absorbs 238 W m⁻² of sunlight and 238 W m⁻² of infrared from the layer: 476 W m⁻² in, 476 W m⁻² out. All three balances hold together. The surface emits twice what it would with no atmosphere.

Draw the diagram and write a balance for each box: the surface, the atmosphere, and the planet as a whole. Any two of the three are enough to solve the problem.

Worked example

The surface under one layer.

Sunlight absorbed: 238 W m⁻² Layer at \(T_\mathrm{a}\) · surface at \(T_\mathrm{s}\) Whole planet \(\sigma T_\mathrm{a}^{4} = 238 \,\mathrm{W} \,\mathrm{m}^{-2}\) so \(T_\mathrm{a} = 255 \,\mathrm{K}\) Layer: two radiating faces \(\sigma T_\mathrm{s}^{4} = 2 \sigma T_\mathrm{a}^{4} = 476 \,\mathrm{W} \,\mathrm{m}^{-2}\)

\(T_\mathrm{s}^{4} = \dfrac{476}{\sigma} = \dfrac{476}{5.67 \times 10^{-8}}\) \(T_\mathrm{s}^{4} = 8.40 \times 10^{9} \,\mathrm{K}^{4}\) \(T_\mathrm{s} = \class{res}{303 \,\mathrm{K}}\)

Too warm: the real atmosphere does not absorb all the infrared.

B.2 GREENHOUSE EFFECT · SL + HL

A neat result

\(T_\mathrm{s}^{4} = 2 T_\mathrm{a}^{4}\), so \(T_\mathrm{s} = 2^{\frac{1}{4}} T_\mathrm{a} = 1.19 \times 255 = 303\) K. A perfectly absorbing layer raises the surface temperature by 19%, whatever the planet.

What the model gets right

No atmosphere gives 255 K. A layer that absorbs everything gives 303 K. The real Earth, at 288 K, lies between the two, as it should for an atmosphere that absorbs most but not all of the infrared. The model also shows the mechanism clearly: the upper atmosphere, at 255 K, is what space “sees”, while the surface underneath is warmer.

What it leaves out

Convection and evaporation carry a large amount of energy up from the surface without radiation. The real atmosphere also absorbs some sunlight directly, has many layers at different temperatures, and has clouds.

Giving the layer only one radiating face. It emits upwards and downwards, so the factor of 2 is essential.

Worked example

An atmosphere that absorbs 78%.

The layer absorbs a fraction \(\epsilon = 0.78\) of the infrared so its emissivity is also \(\epsilon\) Layer \(\epsilon \sigma T_\mathrm{s}^{4} = 2 \epsilon \sigma T_\mathrm{a}^{4}\) so \(\sigma T_\mathrm{a}^{4} = \tfrac{1}{2} \sigma T_\mathrm{s}^{4}\) Surface: sunlight + radiation from the layer \(238 + \epsilon \sigma T_\mathrm{a}^{4} = \sigma T_\mathrm{s}^{4}\)

\(238 = \sigma T_\mathrm{s}^{4} \left(1 - \dfrac{\epsilon}{2}\right)\) \(\sigma T_\mathrm{s}^{4} = \dfrac{238}{1 - 0.39} = 390 \,\mathrm{W} \,\mathrm{m}^{-2}\) \(T_\mathrm{s}^{4} = \dfrac{390}{5.67 \times 10^{-8}} = 6.88 \times 10^{9} \,\mathrm{K}^{4}\) \(T_\mathrm{s} = \class{res}{288 \,\mathrm{K}}\)

B.2 GREENHOUSE EFFECT · SL + HL

Why the same \(\epsilon\) twice

A layer that absorbs only 78% of the infrared passing through it is not a black body, and a poor absorber is an equally poor emitter. So it radiates \(\epsilon \sigma T_\mathrm{a}^{4}\) from each face, not \(\sigma T_\mathrm{a}^{4}\). The \(\epsilon\) cancels in the layer’s own balance but not in the surface’s.

Checking the top

Space receives the 22% of the surface radiation that gets through, \(0.22 \times 390 = 86\), plus the layer’s upward emission, \(0.78 \times 195 = 152\). Together that is 238 W m⁻², equal to the sunlight absorbed, as it must be.

The enhanced greenhouse effect in one line

Add greenhouse gases and \(\epsilon\) rises. Put \(\epsilon = 0.80\) into the same working and the surface comes out at 289 K: about one kelvin warmer, for a change of two parts in a hundred in the absorption.

Exam questions give the fraction absorbed or the emissivity of the atmosphere. The method never changes: one balance per box.

Worked example

The same physics on Venus.

Venus: solar constant \(2.6 \times 10^{3} \,\mathrm{W} \,\mathrm{m}^{-2}\) · albedo 0.76 With no greenhouse effect \((1 - \alpha) \dfrac{S_\mathrm{V}}{4} = \sigma T^{4}\) \(T^{4} = \dfrac{(1 - \alpha) S_\mathrm{V}}{4 \sigma} = \dfrac{0.24 \times 2.6 \times 10^{3}}{4 \times 5.67 \times 10^{-8}}\) \(T^{4} = 2.75 \times 10^{9} \,\mathrm{K}^{4}\) \(T = \class{res}{229 \,\mathrm{K}}\)

Colder than the Earth’s 255 K, although Venus is closer to the Sun: its clouds reflect three quarters of the sunlight
The measured surface temperature is about 735 K

An atmosphere of 96% carbon dioxide, at ninety times the pressure of ours.

B.2 GREENHOUSE EFFECT · SL + HL

Two effects pulling opposite ways

Venus receives almost twice the intensity that the Earth does, but absorbs less of it: \(0.24 \times \frac{2600}{4} = 156\) W m⁻², against the Earth’s 238 W m⁻². On the balance at the top of the atmosphere alone, Venus should be the colder planet. Its surface is instead hot enough to melt lead, about 500 K above that estimate. The whole difference is its greenhouse effect.

Why one layer is not enough

A single perfectly absorbing layer would give \(1.19 \times 229 = 272\) K. Venus’s atmosphere is so thick that infrared is absorbed and re-emitted again and again on the way up, as if there were many layers stacked on each other, each warming the one below.

Mars has an atmosphere of carbon dioxide too, but less than a hundredth as dense as the Earth’s. Its greenhouse effect is only a few kelvin.

Human activity

The enhanced greenhouse effect.

The natural effect keeps the Earth at 288 K, not 255 K
Human activity adds greenhouse gases, mainly by burning fossil fuels
More infrared is absorbed, so the surface warms to restore the balance

The extra warming is the enhanced greenhouse effect.

CARBON DIOXIDE AT MAUNA LOAA FIT TO THE RECORD

B.2 GREENHOUSE EFFECT · SL + HL

Keep the two effects apart

The greenhouse effect itself is natural and necessary: without it the mean surface temperature would be below freezing. The enhanced greenhouse effect is the additional warming caused by the gases people have added. Exam answers lose marks by describing the natural effect as a problem.

The chain of reasoning

More greenhouse gas means that more of the surface’s infrared is absorbed and more is radiated back down. For a time the planet radiates less to space than it absorbs from the Sun. The surface warms, and goes on warming until the planet is once more sending out as much as it takes in. The new equilibrium is at a higher surface temperature.

How we know the extra carbon dioxide is ours

Carbon from coal, oil and gas contains no carbon-14, which has long since decayed, and is poor in carbon-13. The carbon in the air has been shifting towards exactly that signature, and the oxygen in the air has fallen by the amount that burning would use.

“The burning of fossil fuels is a primary cause of the enhanced greenhouse effect” is the sentence in the guide. Be able to say it and to explain the mechanism.

The story · Hawaii, 1958

Background story · not assessed

A measurement that never stopped.

Aerial photograph of a small group of white buildings and a tall mast on a bare slope of dark lava, under a clear blue sky — The **Mauna Loa Observatory**, high on a Hawaiian volcano, far from cities and forests

Charles David Keeling began measuring the carbon dioxide in the air here in March 1958: about 316 parts per million
The record has run ever since. It passed 400 on 9 May 2013. Its yearly ripple is the northern forests growing each summer
Air trapped in ice shows 275 to 285 for thousands of years before the nineteenth century

Photograph: Brian Vasel, NOAA, public domain.

B.2 GREENHOUSE EFFECT · SL + HL

Worked example

What a small change in albedo does.

The Earth’s albedo falls from 0.30 to 0.29 as ice melts \(\sigma T^{4} = (1 - \alpha) \dfrac{S}{4}\) so \(T \propto (1 - \alpha)^{\frac{1}{4}}\) \(\dfrac{T_{2}}{T_{1}} = \left(\dfrac{1 - \alpha_{2}}{1 - \alpha_{1}}\right)^{\frac{1}{4}} = \left(\dfrac{0.71}{0.70}\right)^{\frac{1}{4}}\) \(\dfrac{T_{2}}{T_{1}} = 1.0036\) \(\Delta T = 0.0036 \times 255 = \class{res}{0.9 \,\mathrm{K}}\)

An extra \(0.01 \times 340 = 3.4 \,\mathrm{W} \,\mathrm{m}^{-2}\) absorbed
Warmer, so more ice melts and the albedo falls again

A positive feedback: the change strengthens itself.

B.2 GREENHOUSE EFFECT · SL + HL

Using a ratio

When one quantity in a balance changes slightly, there is no need to work out both temperatures in full. Write how \(T\) depends on the quantity, here \(T \propto (1 - \alpha)^{\frac{1}{4}}\), and take the ratio. Everything that has not changed cancels.

Feedback both ways

The same loop runs in reverse. A cooler planet grows more ice, reflects more sunlight and cools further, which is part of how ice ages deepen. Not all feedbacks are positive: a warmer body radiates more, as \(T^{4}\), and that is the negative feedback that brings any planet to a steady temperature in the end.

0.9 K sounds small. The whole of the warming since the nineteenth century is a little over 1 K.

Quick check

The balance with an atmosphere.

01The Earth absorbs 238 W m⁻² of sunlight. How much must it radiate to space?

238 W m⁻²

02Why does the layer in the model radiate twice \(\sigma T_\mathrm{a}^{4}\)?

It has two faces: up and down

03Which two inputs warm the surface?

Sunlight, and infrared from the atmosphere

04What is the enhanced greenhouse effect?

Extra warming from gases added by human activity

05Its primary cause?

Burning fossil fuels

06More greenhouse gas is added. What happens to the emissivity of the atmosphere?

It rises

PRESS → TO REVEAL EACH ANSWER

Summary

The greenhouse effect on one slide.

Emissivity

\(\frac{\text{power per unit area}}{\sigma T^{4}}\)

1 for a black body
Power per unit area \(= \epsilon \sigma T^{4}\)

Albedo

\(\frac{\text{scattered power}}{\text{incident power}}\)

0.30 for the Earth
Varies with cloud, ice and latitude

Incoming

Mean intensity \(\frac{S}{4}\)

A disk intercepts, a sphere shares
Absorbed: \((1 - \alpha) \frac{S}{4}\)

Balance

Power in = power out

For the surface, the atmosphere and the whole planet

Greenhouse gases

H₂O, CO₂, CH₄, N₂O

Absorb infrared: energy levels, or resonance
Re-emit in all directions

Enhanced effect

More gas, warmer surface

Mainly from burning fossil fuels

B.2 GREENHOUSE EFFECT · SL + HL

Summary

Key words.

EmissivityPower radiated compared with a black body at the same temperature

AlbedoFraction of incident power scattered back

Solar constantIntensity of sunlight above the atmosphere: \(1.36 \times 10^{3} \,\mathrm{W} \,\mathrm{m}^{-2}\)

Equilibrium temperatureWhere power absorbed = power radiated

Greenhouse gasAbsorbs infrared and re-emits it in all directions

ResonanceLarge response when the driving and natural frequencies match

Greenhouse effectThe natural warming of the surface by the atmosphere

Enhanced greenhouse effectThe extra warming from human activity

Positive feedbackA change that strengthens itself

B.2 GREENHOUSE EFFECT · SL + HL