The greenhouse-effect topic applies energy conservation to a planet. You need emissivity and albedo, the solar constant, and why the mean intensity reaching the surface is S/4. Energy-balance questions ask you to estimate an equilibrium temperature, including energy exchanged between the surface and the atmosphere.
The main greenhouse gases are methane, water vapour, carbon dioxide and nitrous oxide. You should explain their absorption of infrared using both molecular energy levels and the resonance model, and describe the enhanced greenhouse effect caused mainly by burning fossil fuels.
17 questions
103 marks
Paper 1A: 9
Paper 1B: 2
Paper 2: 6
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17 practice questions on B.2 Greenhouse effect
1B-1A-11
Albedo·B.2 Greenhouse effect
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Four regions of the Earth's surface are listed.
Which has the highest albedo?
Show mark scheme
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Mark
Notes
Step 1Albedo is the fraction of incident radiation that is scattered back. Fresh snow is the brightest of these surfaces, so it has the highest albedo.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: fresh snow scatters up to about 0.85 of the radiation falling on it, the highest of any common natural surface.
BWater has one of the lowest albedos, about 0.06, which is why melting sea ice accelerates warming.
CDark vegetation absorbs strongly; its albedo is around 0.15.
DDark soil absorbs most of the radiation, with an albedo of about 0.10.
Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power / total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude Command term: Identify
2B-1A-12
Solar constant & S/4·B.2 Greenhouse effect
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Solar radiation of intensity S is incident on a spherical planet of radius R.
Which row gives the power intercepted by the planet and the mean intensity when this power is averaged over the planet's whole surface?
Power interceptedMean intensity over the whole surface
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Mark
Notes
Step 1The planet intercepts the radiation over its projected area, a disc of area πR², so the power intercepted is πR²S.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Averaged over the whole surface, 4πR², the mean intensity is πR²S / 4πR² = S/4.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe whole surface does not face the Sun. Only the projected disc, πR², intercepts the radiation.
BThe intercepted power is right, but it has been averaged over the sunlit hemisphere, 2πR², instead of the whole surface, 4πR².
CThe sunlit hemisphere, 2πR², has been used as the intercepting area. The power intercepted is set by the projected disc, πR².
DCorrect: the disc πR² intercepts πR²S, and spreading this over 4πR² gives a mean of S/4.
Syllabus understandingB.2 — the solar constant S; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4 Command term: Determine
3B-1A-13
Equilibrium temperature·B.2 Greenhouse effect
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Planets P and Q have no atmosphere and the same albedo, and each radiates as a black body. Q is twice as far from the Sun as P.
S is the solar constant at the position of a planet and T is the planet's equilibrium temperature. Which row gives SQ/SP and TQ/TP?
SQ / SPTQ / TP
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Notes
Step 1The intensity of the Sun's radiation obeys the inverse-square law: SQ/SP = (1/2)² = 0.25.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy balance: (1 − albedo)S/4 = σT⁴, with the same albedo, so T ∝ S1/4.
—
Step 3TQ/TP = 0.251/4 = 1/√2 = 0.71.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe intensity has been taken as inversely proportional to the distance instead of its square; 0.501/4 = 0.84.
BThe fourth root has been left out: T⁴, not T, is proportional to the absorbed intensity.
CA square root has been taken instead of a fourth root: 0.251/2 = 0.50.
DCorrect: the solar constant falls to a quarter, and T ∝ S1/4 gives 0.251/4 = 0.71.
Syllabus understandingB.2 — the solar constant S; problems will include the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants Command term: Determine
4B-1A-14
Emissivity·B.2 Greenhouse effect
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Plates X and Y have equal surface areas. X has an emissivity of 0.40 and is at a temperature of 600 K. Y has an emissivity of 0.80 and is at 300 K.
What is the ratio (power radiated by X) / (power radiated by Y)?
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Notes
Step 1Power radiated = eσAT⁴; the areas are equal, so the ratio = (eX/eY) × (TX/TY)⁴.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = (0.40/0.80) × 2⁴ = 0.5 × 16 = 8.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe temperature ratio has been used to the first power: 0.5 × 2 = 1. Radiated power depends on T⁴.
BThe temperature ratio has been squared instead of raised to the fourth power: 0.5 × 4 = 2.
DThe emissivity ratio has been inverted: 2 × 16 = 32. X is the poorer emitter, so that factor is 0.5.
Syllabus understandingB.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature as given by emissivity = power radiated per unit area / σT⁴ Command term: Determine
5B-1A-15
Greenhouse gases·B.2 Greenhouse effect
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Which of these is not one of the main greenhouse gases?
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Notes
Step 1A greenhouse gas must absorb infrared radiation, which requires a molecule whose vibrations change its dipole moment. Nitrogen, N₂, is symmetrical and does not, so despite its abundance it is not a greenhouse gas.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
ACarbon dioxide is a major greenhouse gas, released naturally by respiration and volcanoes and by human burning of fossil fuels.
BMethane is a powerful greenhouse gas, from wetlands and from agriculture and gas extraction.
CWater vapour is the most abundant greenhouse gas and contributes the largest share of the natural greenhouse effect.
DCorrect: nitrogen is 78 % of the atmosphere, but N₂ is a symmetrical diatomic molecule whose vibration does not change its dipole moment, so it cannot absorb infrared radiation (CO₂ has no permanent dipole either, but some of its vibrations do change it).
Syllabus understandingB.2 — that methane CH₄, water vapour H₂O, carbon dioxide CO₂ and nitrous oxide N₂O are the main greenhouse gases and each of these has origins that are both natural and created by human activity Command term: Identify
6B-1B-04
Equilibrium temperature·B.2 Greenhouse effect
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A student uses a published planetary database to investigate the greenhouse effect. The table gives, for three planets, the mean distance d from the Sun in astronomical units (AU), the albedo α and the measured mean surface temperature Ts. It also gives the equilibrium temperature Teq that each planet would have if its surface radiated as a black body (emissivity 1) and there were no atmosphere.
The solar constant at 1 AU is S = 1361 W m−2. The intensity of solar radiation at a distance d is S/d2, with d in AU.
Planet
d / AU
α
Ts / K
Teq / K
Venus
0.723
0.76
737
229
Earth
1.000
0.30
288
Mars
1.524
0.25
210
210
(a)
Show that the intensity of solar radiation at the distance of Mars is about 590 W m−2.
(1)
(b)
Explain why the mean intensity absorbed by the surface of a planet is (1 − α)S/4, where S is the intensity at the planet.
(2)
(c)
Calculate Teq for Earth.
(2)
(d)
Earth’s albedo varies with cloud cover and is quoted as 0.30 ± 0.02. Determine the absolute uncertainty in your answer to (c).
(2)
(e)
Determine the emissivity that Earth’s surface–atmosphere system would need for an equilibrium temperature of 288 K.
(2)
(f)
Discuss what the data suggest about the greenhouse effect on Mars and on Venus.
(2)
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Notes
Part (a)
1361 / 1.5242 = 586 W m−2
✓ 1
Must see the substitution OR an answer to at least 3 s.f.
Part (b)
a fraction α is reflected/scattered so a fraction (1 − α) is absorbed
✓ 1
the planet intercepts radiation over its cross-section πr2 but the energy is spread over the whole surface 4πr2, a ratio of 1/4
✓ 1
OWTTE.
Part (c)
σT4 = (1 − 0.30) × 1361 / 4
✓ 1
Teq = 255 K
✓ 1
Accept 254–255 K.
Part (d)
fractional uncertainty in (1 − α) = 0.02/0.70 = 2.9 %; T ∝ (1 − α)¼ so 0.71 %
✓ 1
ΔT ≈ ±2 K
✓ 1
Accept ±1.8 K. Allow ECF from (c).
Part (e)
eσ × 2884 = (1 − 0.30) × 1361 / 4
✓ 1
e = 0.61
✓ 1
Accept 0.61–0.62.
Part (f)
Mars: Ts ≈ Teq «210 K», so its thin atmosphere produces very little greenhouse effect
✓ 1
Venus: Ts is far above Teq «737 K ≫ 229 K», so its dense atmosphere «of CO2» absorbs infrared from the surface and re-emits it, a very strong greenhouse effect
✓ 1
OWTTE.
Answers: (a) 586 W m−2 · (c) 255 K · (d) ±2 K · (e) 0.61 (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — problems will include the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants Command term: Determine
7B-2-05
Energy balance of a planet·B.2 Greenhouse effect
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDetermine
An exoplanet P orbits a star. At the distance of P from the star, the intensity of the star's radiation is 1720 W m−2. The albedo of P is 0.35.
In a simple model, the atmosphere of P is a thin layer at a single temperature Ta. The layer is transparent to the radiation from the star. It absorbs a fraction e of the infrared radiation emitted by the surface and transmits the rest. Its emissivity is also e, so it emits an intensity eσTa4 from each of its two faces, upwards and downwards. The surface, at temperature Ts, emits as a black body. The diagram shows the energy flows per unit area of the surface.
Diagram NOT accurately drawn
(a)
State what is meant by the emissivity of a surface.
(1)
(b)
(i)
Show that the mean intensity absorbed by the surface is about 280 W m−2.
(1)
(ii)
Calculate the temperature of the surface if P had no atmosphere.
(1)
(c)
The planet has an atmosphere.
(i)
Annotate the diagram with expressions for the intensities X, Y and Z.
(2)
(ii)
Show that, for the layer to be in equilibrium, σTa4 = ½σTs4.
(2)
(iii)
The mean surface temperature of P is 300 K. Determine e.
(3)
(d)
(i)
Predict and explain the effect on Ts of an increase in the concentration of greenhouse gases in the atmosphere of P.
(3)
(ii)
Outline one limitation of this model.
(1)
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Notes
Part (a)
the power radiated per unit area of the surface divided by the power radiated per unit area by a black body at the same temperature
✓ 1
Accept a ratio of powers for equal areas at the same temperature. Do not accept 'how well it emits'.
Part (b)(i)
(1 − 0.35) × 1720 / 4 = 279.5 W m−2
✓ 1
Must see full substitution OR answer to at least 3 s.f. The factor ¼ «spreading over the whole surface» must be seen.
Part (b)(ii)
σT4 = 279.5 gives T = 265 K
✓ 1
Accept 264–266 K.
Part (c)(i)
X = (1 − e)σTs4
✓ 1
Y = Z = eσTa4
✓ 1
Both Y and Z needed.
Part (c)(ii)
the layer absorbs eσTs4
✓ 1
it emits eσTa4 from each face, 2eσTa4 in total; equating gives σTa4 = ½σTs4
✓ 1
Part (c)(iii)
surface in equilibrium: 279.5 + eσTa4 = σTs4
✓ 1
ALTERNATIVE for MP1: top of the atmosphere, (1 − e)σTs4 + eσTa4 = 279.5.
so 279.5 = σTs4(1 − e/2), with σTs4 = 5.67 × 10−8 × 3004 = 459 W m−2
✓ 1
e = 2(1 − 279.5/459) = 0.78
✓ 1
Allow ECF from (b)(i). Award [3] for CNA.
Part (d)(i)
the layer absorbs a larger fraction of the infrared from the surface, so e increases
because more infrared is re-emitted back to the surface / less escapes directly to space, so the surface must be hotter to emit enough to balance what it absorbs
✓ 1
Accept a statement that Ts cannot exceed 315 K (e = 1) in this model.
Part (d)(ii)
any one: a real atmosphere does not have a single temperature; energy is also carried from the surface by convection and evaporation; the atmosphere absorbs some of the incoming stellar radiation; the albedo and e vary with place and time; the surface is not a perfect black body
✓ 1
Answers: (b)(i) 279.5 W m−2 · (b)(ii) 265 K · (c)(iii) e = 0.78 (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — the conservation of energy; emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4; energy balance problems will include energy exchanged between the surface and the atmosphere of a body Command term: Determine
8B-2-06
Albedo·B.2 Greenhouse effect
Paper 2Easy12 marks
Short answer & extended response12 steps to full marksCalculate
An array of solar panels of total area 12 m2 is mounted on the roof of a house. At noon on a clear day, solar radiation of intensity 900 W m−2 is incident on the array. The array reflects 1.3 kW of the incident power and transfers 20 % of the incident power to electrical energy.
(a)
(i)
State what is meant by the solar constant.
(1)
(ii)
The solar constant is 1360 W m−2. Suggest two reasons why the intensity incident on the array is less than this.
(2)
(b)
(i)
Calculate the albedo of the array.
(2)
(ii)
Calculate the electrical power output of the array.
(1)
(iii)
Determine the rate at which the array absorbs energy that is not transferred to electrical energy, and state what happens to this energy.
(2)
(c)
(i)
The albedo of the Earth varies from day to day. State one factor, other than cloud cover, on which it depends.
(1)
(ii)
A town plans to cover many pale roofs, of albedo 0.6, with solar arrays like this one. Discuss the effect of the plan on warming, locally and globally.
(3)
Show mark scheme
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Notes
Part (a)(i)
the intensity (power per unit area) of the Sun's radiation at the mean distance of the Earth from the Sun, on a surface perpendicular to the rays, outside the atmosphere
✓ 1
All three ideas needed: intensity/power per unit area; mean Earth–Sun distance; above the atmosphere.
Part (a)(ii)
radiation is absorbed / scattered / reflected (e.g. by clouds, gases, dust) in the atmosphere
✓ 1
the array is not perpendicular to the Sun's rays, so the power is spread over a larger area
✓ 1
Accept 'the Sun is not overhead / the rays pass through a longer path in the atmosphere'.
Part (b)(i)
incident power = 900 × 12 = 1.08 × 104 W
✓ 1
albedo = 1300 / 10800 = 0.12
✓ 1
Award [2] for CNA. No unit.
Part (b)(ii)
0.20 × 10800 = 2.16 × 103 W
✓ 1
Part (b)(iii)
10800 − 1300 − 2160 = 7.34 × 103 W
✓ 1
Allow ECF from (b)(i) and (b)(ii).
it increases the internal energy of the array (warms it) and is transferred to the surroundings by convection, radiation and conduction
✓ 1
Part (c)(i)
latitude / the angle of the Sun's rays; OR the type of surface, e.g. snow and ice cover, ocean, forest, desert
✓ 1
Part (c)(ii)
the arrays have a lower albedo than the pale roofs, so they absorb more solar radiation
✓ 1
this warms the town locally (more energy transferred to the air around the roofs)
✓ 1
but the electricity generated replaces electricity from fossil fuels, reducing CO2 emissions and so the enhanced greenhouse effect: globally the plan reduces warming
✓ 1
Award MP3 only for the global argument linked to reduced greenhouse-gas emissions.
Answers: (b)(i) 0.12 · (b)(ii) 2.16 × 103 W · (b)(iii) 7.34 × 103 W (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power / total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; the solar constant S; that the augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect Command term: Calculate
9B-2-15
Greenhouse gases·B.2 Greenhouse effect
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksExplain
Carbon dioxide molecules absorb infrared radiation strongly in a narrow band of wavelengths centred on 15 μm. Methane molecules absorb strongly near 7.7 μm.
The graph shows the spectrum of the radiation emitted by the Earth's surface, treated as a black body at 288 K.
Graph drawn to scale
(a)
Carbon dioxide and methane are two of the main greenhouse gases. For each gas, state one source that is natural and one source that is created by human activity.
(2)
(b)
Annotate the graph to show the wavelengths at which carbon dioxide and methane absorb strongly. Hence outline why these two gases have a significant effect on the energy radiated by the surface.
(2)
(c)
(i)
Calculate the energy of a photon of wavelength 15 μm.
(1)
(ii)
Carbon dioxide absorbs very little radiation of wavelength 12 μm. Explain, with reference to molecular energy levels, why carbon dioxide absorbs radiation of wavelength 15 μm strongly but not radiation of wavelength 12 μm.
(2)
(d)
Explain how an increase in the concentration of carbon dioxide in the atmosphere leads to an increase in the mean temperature of the Earth's surface.
(3)
(e)
A gas-fired power station has an electrical power output of 450 MW and an overall efficiency of 0.50. The natural gas it burns has an energy density of 55 MJ kg−1, and burning 1.0 kg of the gas releases 2.75 kg of carbon dioxide. Determine the mass of carbon dioxide released by the power station in one day.
methane: natural — wetlands / termites / decay without oxygen; human — livestock / rice paddies / landfill / leaks from gas wells and pipelines
✓ 1
Part (b)
both wavelengths marked on the λ axis at 7.7 μm and 15 μm, one on each side of the peak at about 10 μm
✓ 1
Accept arrows or shaded bands at the correct positions (± 1 μm).
the surface emits strongly at both wavelengths «about 82 % and 72 % of the peak intensity», so the gases can absorb a significant fraction of the surface's infrared radiation
a photon is absorbed only if its energy is equal to the difference between two «vibrational» energy levels of the molecule
✓ 1
the photon energy at 15 μm matches such a difference; at 12 μm the photon energy «1.7 × 10−20 J» does not match any energy difference, so it is not absorbed
✓ 1
OWTTE. Do not accept 'the molecule resonates' without reference to energy levels.
Part (d)
more carbon dioxide molecules absorb more of the infrared radiation emitted by the surface «exciting them to higher energy levels»
✓ 1
the excited molecules re-emit infrared radiation in all/random directions, so a larger intensity of infrared radiation is directed back down to the surface
✓ 1
Do not accept 'reflected'.
the surface now absorbs more power than it emits, so its temperature rises until it again radiates as much as it absorbs «enhanced greenhouse effect»
✓ 1
Do not accept 'traps heat' alone.
Part (e)
power input = 450/0.50 = 900 MW
✓ 1
mass of gas burned per second = 9.0 × 108/5.5 × 107 = 16.4 kg s−1
✓ 1
Allow ECF.
mass of CO2 per day = 16.36 × 2.75 × 86 400 = 3.9 × 106 kg
✓ 1
Accept 3.8 × 106–3.9 × 106 kg. Award [2 max] for 1.9 × 106 kg «electrical output used as the input». Award [3] for CNA.
Answers: (c)(i) 1.3 × 10−20 J · (e) 3.9 × 106 kg (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — that methane CH4, water vapour H2O, carbon dioxide CO2 and nitrous oxide N2O are the main greenhouse gases and each of these has origins that are both natural and created by human activity; the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; that the augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect (the burning of fossil fuels is a primary cause) (linked: E.1 — photons of energy E = hf; A.3 — energy density of fuels and efficiency) Command term: Explain
10B-1A-32
Greenhouse gases·B.2 Greenhouse effect
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Which statement about the enhanced greenhouse effect is correct?
Show mark scheme
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Notes
Step 1The natural greenhouse effect keeps the Earth habitable. The enhanced greenhouse effect is the extra warming that follows from the rise in greenhouse gas concentrations caused by human activity, chiefly the burning of fossil fuels.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: burning fossil fuels and other human activities have raised greenhouse gas concentrations, adding to the natural effect.
BThat describes the ordinary greenhouse effect. The enhanced effect is the additional warming caused by human activity.
COzone depletion is a separate problem, concerning ultraviolet radiation reaching the surface, not infrared radiation leaving it.
DThe measured change in solar output is far too small to account for the observed warming, and the effect concerns outgoing radiation rather than incoming.
Syllabus understandingB.2 — that the augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect Command term: Identify
11B-1A-33
Greenhouse gases·B.2 Greenhouse effect
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksIdentify
Molecules of carbon dioxide in the atmosphere absorb some of the infrared radiation emitted by the Earth's surface.
Which statement about this absorption is correct?
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Notes
Step 1An infrared photon is absorbed when its energy equals the spacing of the molecule's vibrational energy levels (in the resonance model, when its frequency matches a natural frequency of the molecule). The excited molecule later emits infrared radiation in a random direction, so only part of it returns to the surface.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: absorption needs a photon energy matching a gap between molecular energy levels, and the re-emission is in all directions — part of it back towards the surface.
BInfrared photons have far too little energy to ionise a molecule; the absorbed energy excites the molecule's vibrations.
CAbsorption is selective. Only photons whose energy matches a gap between discrete energy levels (a natural frequency of vibration) are strongly absorbed, so each gas absorbs in particular bands.
DRe-emission is in all directions. Some of the energy returns to the surface, but the rest goes upwards or sideways.
Syllabus understandingB.2 — the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; that the greenhouse effect can be explained in terms of both a resonance model and molecular energy levels Command term: Identify
12B-1A-34
Equilibrium temperature·B.2 Greenhouse effect
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
In a simple model of a planet, the atmosphere is a single layer that is transparent to solar radiation but absorbs all of the infrared radiation emitted by the surface. The atmosphere emits infrared radiation of equal intensity upwards and downwards.
The surface absorbs a mean solar intensity I and radiates as a black body at temperature Ts. The surface and the atmosphere are in equilibrium.
What is σTs⁴?
Show mark scheme
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Mark
Notes
Step 1Atmosphere: it absorbs σTs⁴ and emits an intensity D upwards and D downwards, so σTs⁴ = 2D.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Surface: it absorbs I from the Sun and D from the atmosphere, so σTs⁴ = I + D.
—
Step 3Eliminate D: σTs⁴ = I + σTs⁴/2, so σTs⁴ = 2I.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe radiation from the atmosphere has been treated as a loss from the surface (σTs⁴ = I − D). The surface absorbs it, so it adds to I.
BThis ignores the atmosphere: it is the energy balance of a surface with no atmosphere above it.
CThe atmosphere has been taken to send down half of the solar intensity. It absorbs the surface's infrared, not sunlight, so it sends down σTs⁴/2.
DCorrect: σTs⁴ = I + D with D = σTs⁴/2 gives 2I (and the atmosphere's upward emission, D = I, balances the absorbed sunlight).
Syllabus understandingB.2 — the conservation of energy; energy balance problems will include energy exchanged between the surface and the atmosphere of a body Command term: Deduce
13B-1A-35
Emissivity·B.2 Greenhouse effect
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A flat plate in space faces the Sun. Its front face absorbs solar radiation at a rate I per unit area, and the plate radiates only from this face. The emissivity of the front face is e.
What is the equilibrium temperature of the plate?
Show mark scheme
Marking point
Mark
Notes
Step 1At equilibrium the power emitted per unit area equals the power absorbed per unit area: eσT⁴ = I.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2T = (I/eσ)1/4.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe emissivity has been multiplied into the absorbed intensity. A poorer emitter (e < 1) must be hotter to shed the same power, so e belongs in the denominator.
BThis treats the plate as a black body; the emissivity e has been left out.
CCorrect: eσT⁴ = I gives T = (I/eσ)1/4.
DThe emissivity has been applied to T instead of to T⁴.
Syllabus understandingB.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature; problems will include the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants Command term: Determine
14B-1B-14
Albedo·B.2 Greenhouse effect
Paper 1BEasy9 marks
Data-based question9 steps to full marksDetermine
An albedometer consists of two radiation sensors, one facing up and one facing down, mounted 1.5 m above a flat area of dry sand. The upward sensor measures the incident solar intensity Iin and the downward sensor measures the intensity Ir scattered from the sand. Readings were taken at intervals on a day with passing clouds. Each intensity has an uncertainty of ±10 W m−2.
Iin / W m−2
205
310
395
480
560
655
740
830
Ir / W m−2
82
113
153
183
209
255
278
317
Graph drawn to scale
(a)
Calculate the albedo of the sand using the reading with Iin = 480 W m−2.
(1)
(b)
Determine the absolute uncertainty in your answer to (a).
(2)
(c)
Draw the line of best fit for the data.
(1)
(d)
Using the graph, determine the albedo of the sand.
(2)
(e)
Outline one advantage of using the graph rather than a single reading to find the albedo.
(1)
(f)
Published albedos of dry sand lie between 0.35 and 0.45. Comment on the student’s result.
(1)
(g)
The albedo of the whole Earth is not constant. State one reason why it varies from day to day.
single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (d)
albedo = gradient «= Ir/Iin» from a large triangle
✓ 1
0.38
✓ 1
Accept 0.36–0.40. No unit.
Part (e)
uses all the readings so the effect of random errors is reduced / shows up a systematic error «e.g. a sensor offset» as a non-zero intercept
✓ 1
OWTTE.
Part (f)
the value «0.38» lies inside the published range, so it is consistent / typical for dry sand
✓ 1
Allow ECF from (d).
Part (g)
cloud cover changes «clouds have a high albedo» / snow and ice cover changes
✓ 1
Accept changes in the part of the Earth that is sunlit, since albedo depends on latitude.
Answers: (a) 0.381 · (b) ±0.03 · (d) 0.38 (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power / total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude Command term: Determine
15B-2-24
Equilibrium temperature·B.2 Greenhouse effect
Paper 2Hard15 marks
Short answer & extended response15 steps to full marksDeduce
The Moon has no atmosphere and is, on average, at the same distance from the Sun as the Earth. The solar constant is S = 1360 W m−2. The albedo of the lunar surface is 0.12 and its emissivity is 0.95.
(a)
Outline what is meant by an emissivity of 0.95 for the lunar surface, and why a surface with an emissivity less than 1 reaches a higher equilibrium temperature than a black body that absorbs the same intensity.
(2)
(b)
(i)
Show that the mean intensity absorbed by the lunar surface is about 300 W m−2.
(1)
(ii)
Determine the mean equilibrium temperature of the lunar surface predicted by this model.
(2)
(c)
At the subsolar point the Sun is directly overhead, so the radiation arrives perpendicular to the surface. Assume that each square metre of the surface there emits all the energy it absorbs.
(i)
Determine the equilibrium temperature at the subsolar point.
(2)
(ii)
Deduce that the ratio of the subsolar temperature to the mean temperature does not depend on the albedo or on the emissivity.
(2)
(d)
Measured temperatures on the Moon range from about 390 K at the subsolar point to about 100 K during the lunar night.
(i)
Explain why the night side of the Moon becomes far colder than the night side of the Earth.
(3)
(ii)
Suggest why the mean equilibrium temperature found in (b)(ii) is a poor guide to the temperature at any particular place on the Moon.
(1)
(e)
The Earth receives the same solar constant but has a mean albedo of 0.30. Ignoring the effect of its atmosphere on outgoing infrared radiation and taking its emissivity as 1, state and explain whether the mean equilibrium temperature of the Earth is higher or lower than that of the Moon.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the surface radiates 95 % of the power per unit area that a black body at the same temperature would radiate
✓ 1
OWTTE.
so it must be at a higher temperature before the power it radiates equals the power it absorbs
✓ 1
Part (b)(i)
(1360/4) × (1 − 0.12) = 299 W m−2
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
0.95 × 5.67 × 10−8 × T4 = 299
✓ 1
Allow ECF from (b)(i).
T = 273 K
✓ 1
Award [1 max] for 270 K «emissivity omitted».
Part (c)(i)
0.95 × 5.67 × 10−8 × T4 = 1360 × (1 − 0.12)
✓ 1
T = 386 K
✓ 1
Accept 385–387 K. Award [1 max] for 381 K «emissivity omitted».
Part (c)(ii)
in both cases εσT4 = (1 − α) × incident intensity, so T ∝ (incident intensity)1/4 with the same factor (1 − α)/εσ
✓ 1
ratio = (S/(S/4))1/4 = 41/4 = 1.41 «= 386/273», in which α and ε cancel
✓ 1
Accept a numerical demonstration that 386/273 = √2 together with the cancelling argument.
Part (d)(i)
the Moon has no atmosphere to absorb infrared radiation from the surface and re-emit some of it back «no greenhouse effect»
✓ 1
there are no winds or ocean currents «convection» to carry energy from the day side to the night side
✓ 1
a lunar night lasts about two weeks, so the surface radiates energy away for a much longer time
✓ 1
Accept: the lunar dust is a poor conductor, so little energy is conducted up from below. Any three distinct points.
Part (d)(ii)
the model assumes that absorbed energy is shared evenly over the whole surface, but the Moon has no way of redistributing energy, so the local temperature depends on the local absorbed intensity
✓ 1
OWTTE.
Part (e)
lower
✓ 1
No mark for a bare statement without reasoning.
a larger fraction of the radiation is reflected, so less is absorbed «238 W m−2 rather than 299 W m−2» AND with an emissivity of 1 the absorbed power is radiated at a lower temperature
✓ 1
Accept a calculation giving about 255 K.
Answers: (b)(i) 299 W m−2 · (b)(ii) 273 K · (c)(i) 386 K · (c)(ii) 41/4 = 1.41 (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4; emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature; albedo as a measure of the average energy reflected off a macroscopic system; problems will include the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants Command term: Deduce
16B-2-25
Albedo·B.2 Greenhouse effect
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksEstimate
Satellite measurements show that the area of Arctic sea ice remaining at the end of summer has decreased over recent decades. Sea ice has an albedo of 0.60 and open ocean has an albedo of 0.06. During the three summer months (92 days) the mean intensity of the solar radiation reaching the surface of the Arctic Ocean is 220 W m−2.
Specific latent heat of fusion of ice = 3.34 × 105 J kg−1. Density of ice = 920 kg m−3.
(a)
(i)
State what is meant by albedo.
(1)
(ii)
Outline why the albedo of the same region of the Arctic, measured by a satellite, can change from one day to the next.
(1)
(b)
(i)
Calculate the increase in the intensity absorbed by the surface when sea ice is replaced by open ocean.
(2)
(ii)
In one year 1.5 × 1012 m² of sea ice is replaced by open ocean. Estimate the thickness of ice at 0 °C, over this area, that could be melted by the extra energy absorbed during the three summer months.
(3)
(c)
Explain why the loss of sea ice is described as a positive feedback in climate change.
(2)
(d)
State one assumption made in your estimate in (b)(ii).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the ratio of the total scattered «reflected» power to the total incident power
✓ 1
Do not accept “the amount of energy reflected” without a ratio.
Part (a)(ii)
the amount of cloud cover changes from day to day, and clouds have a high albedo
extra energy = 119 × 1.5 × 1012 × 92 × 86 400 = 1.4 × 1021 J
✓ 1
Allow ECF from (b)(i).
mass melted = 1.4 × 1021/3.34 × 105 = 4.2 × 1015 kg
✓ 1
thickness = 4.2 × 1015/(920 × 1.5 × 1012) ≈ 3 m
✓ 1
Accept 3.0–3.1 m. ALTERNATIVE: energy per square metre = 119 × 7.95 × 106 = 9.4 × 108 J ✓, mass per square metre = 2.8 × 103 kg ✓, thickness = 2.8 × 103/920 ≈ 3 m ✓.
Part (c)
less ice means a lower albedo, so more solar radiation is absorbed
✓ 1
this causes further warming, which melts more ice, so the change reinforces its own cause
✓ 1
Part (d)
all of the extra energy absorbed is used to melt ice «none is radiated, conducted away or used to warm water» OR the extra intensity is the same over the whole area for all 92 days OR all the ice is at 0 °C
✓ 1
Accept any one valid assumption.
Answers: (b)(i) 119 W m−2 · (b)(ii) 1.4 × 1021 J; ≈ 3 m (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; the conservation of energy (linked: B.1 — Q = mL) Command term: Estimate
17B-2-26
Greenhouse gases·B.2 Greenhouse effect
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksExplain
Nitrous oxide (N₂O) is one of the main greenhouse gases. In a simple resonance model of one vibration of the N₂O molecule, the oxygen atom, of mass 16 u, oscillates on a spring of spring constant 330 N m−1 attached to the rest of the molecule, which is treated as fixed.
The mean temperature of the Earth's surface is 288 K. The temperature of the Sun's surface is 5800 K.
Axes for (c)
(a)
Show that the natural frequency of this vibration is about 1.8 × 1013 Hz.
(1)
(b)
Calculate the wavelength of electromagnetic radiation of this frequency and state the region of the electromagnetic spectrum to which it belongs.
(2)
(c)
Electromagnetic radiation acts as a periodic driving force on the oxygen atom. On the axes, sketch a graph to show how the amplitude of the vibration varies with the frequency of the radiation.
(2)
(d)
Explain, using the resonance model, why N₂O absorbs some of the radiation emitted by the Earth's surface but very little of the radiation from the Sun.
(3)
(e)
In some N₂O molecules the oxygen atom is the isotope oxygen-18 (mass 18 u). The spring constant is unchanged. Deduce the wavelength of the radiation that these molecules absorb most strongly.
amplitude small at frequencies well above and well below the peak, approaching zero at high frequency
✓ 1
Do not accept a curve with more than one peak.
Part (d)
the Earth's surface emits infrared with a peak near 10 μm «2.9 × 10−3/288», and its spectrum includes much radiation at about 17 μm, i.e. near the natural frequency
✓ 1
at resonance the amplitude of vibration is large, so energy is absorbed strongly from radiation of this frequency
✓ 1
solar radiation peaks at about 0.5 μm «2.9 × 10−3/5800», at frequencies far above the natural frequency, where the amplitude «and absorption» is very small
✓ 1
OWTTE.
Part (e)
f ∝ 1/√m, so λ ∝ √m: λ increases by a factor √(18/16) = 1.06
✓ 1
λ = 1.8 × 10−5 m «18 μm»
✓ 1
Allow ECF from (b). Award [1 max] for 1.9 × 10−5 m «λ ∝ m used».
Answers: (a) 1.77 × 1013 Hz · (b) 1.7 × 10−5 m, infrared · (e) 1.8 × 10−5 m (the remaining parts are explanations — see the table above)
Syllabus understandingB.2 — that the greenhouse effect can be explained in terms of both a resonance model and molecular energy levels; that methane CH₄, water vapour H₂O, carbon dioxide CO₂ and nitrous oxide N₂O are the main greenhouse gases (linked: C.1 — T = 2π√(m/k); C.4 — the nature of resonance including natural frequency and amplitude of oscillation based on driving frequency; C.2 — v = fλ; B.1 — Wien's displacement law) Command term: Explain
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