116 free, high-quality past-paper-style questions on Greenhouse effect for IB Physics SL, each with a full mark scheme. Covers: Albedo; Greenhouse gases; Emissivity; and more. For when you have finished the past papers, or want to practise one topic at a time.B.2 at Higher Level →Read the B.2 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.546 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1B-1A-11AlbedoPaper 1AEasy1 mark
Four regions of the Earth's surface are listed.
Which has the highest albedo?
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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.
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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.
Which of these is not one of the main greenhouse gases?
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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.
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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).
Which statement about the enhanced greenhouse effect is correct?
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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.
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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.
Which row correctly pairs one of the main greenhouse gases with a source of that gas that is due to human activity?
Greenhouse gasSource due to human activity
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Step 1Each of CH₄, H₂O, CO₂ and N₂O has natural and human sources. Nitrogen fertilisers are broken down by soil bacteria, releasing N₂O: a source created by human activity (agriculture).
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
APhotosynthesis takes in CO₂ and releases oxygen; it does not produce methane, and it is a natural process. Human sources of methane include livestock farming, rice paddies, landfill and leaks from gas extraction.
BCorrect: N₂O is a main greenhouse gas, and fertiliser use on farmland is its largest human source.
CCarbon dioxide is a greenhouse gas, but volcanoes are a natural source. The main human source is the burning of fossil fuels.
DWater vapour is a greenhouse gas, but evaporation from the oceans is a natural source, not one due to human activity.
Solar radiation of intensity 900 W m−2 is incident on the top of a forest canopy. Of this, 110 W m−2 is scattered back upwards, 700 W m−2 is absorbed by the leaves and the rest is transmitted to the forest floor.
What is the albedo of the canopy?
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Step 1Albedo = total scattered power / total incident power = 110/900 = 0.12. The absorbed and transmitted radiation are not scattered back, so neither belongs in the numerator.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis is the transmitted fraction, 90/900: radiation that passes down to the floor is not scattered back.
A satellite measures the albedo of the same region of open ocean at local noon on two days. The albedo on the second day is much greater than on the first day.
What is the most likely explanation?
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Step 1Albedo is the ratio scattered power / incident power, so it depends on what the radiation meets, not on how much arrives. Clouds scatter a far larger fraction of sunlight than open water, so more cloud means a larger albedo.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: cloud tops scatter a large fraction of the incident sunlight back to space, whereas open ocean scatters only a few per cent, so extra cloud raises the albedo of the region.
BAlbedo is a ratio. A greater incident intensity increases the scattered power in proportion, so the albedo is unchanged.
CThis confuses emission with scattering. Albedo concerns the fraction of incident solar radiation scattered back, not the infrared radiation the surface emits.
DCarbon dioxide absorbs and re-emits infrared radiation from the surface; it does not scatter a significant fraction of incoming sunlight, so it has almost no effect on albedo.
Step 1Emissivity = power radiated per unit area / σT⁴, i.e. the power per unit area compared with that of a black body at the same temperature: here 40 %.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis describes an albedo of 0.40, the fraction of incident radiation scattered. Emissivity concerns the radiation the surface emits.
BAt equilibrium a surface radiates all of the energy it absorbs, whatever its emissivity; the emissivity compares its emission with that of a black body.
CCorrect: emissivity is the ratio of the power radiated per unit area to σT⁴ for a black surface at the same temperature.
DThe factor has been applied to T instead of T⁴, and in the wrong direction: to radiate the same power per unit area the panel must be hotter, by a factor 0.40−1/4 ≈ 1.26.
The absorption of infrared radiation by greenhouse-gas molecules can be explained by a resonance model and by a model based on molecular energy levels.
Which row gives the condition for strong absorption in each model?
Resonance modelEnergy-level model
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Step 1Resonance: the radiation drives the molecule's vibration, and the amplitude (and energy absorbed) is large only when the driving frequency is close to a natural frequency. Energy levels: a photon is absorbed only if its energy hf equals the difference between two energy levels. Both conditions pick out the same frequencies.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: matching a natural frequency (resonance) and matching an energy difference (hf = ΔE) are the two descriptions of the same selective absorption.
BBoth conditions are wrong: driving far above the natural frequency gives a small amplitude, and a photon with more energy than the gap is not absorbed by that transition.
CThe resonance condition is right, but absorption is not a threshold effect: a photon with more energy than the difference between two levels is not absorbed by that transition.
DThe energy-level condition is right, but driving at a frequency far from the natural frequency produces only a small amplitude, so little energy is absorbed.
9IBP2S-011-03-01Solar constant & S/4Paper 1AEasy1 mark
The mean intensity of solar radiation incident on the top of Earth's atmosphere, averaged over the whole of Earth's surface, is S/4, where S is the solar constant. What is the reason for the factor of 4?
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Earth intercepts the Sun's parallel rays over its cross-sectional disc, πR2, so the intercepted power is SπR2. As Earth rotates, this power is spread over the whole surface, 4πR2. The mean intensity is therefore SπR2/4πR2 = S/4, which is option B. Option A is wrong because the value S/4 is defined at the top of the atmosphere, so atmospheric absorption plays no part in it. Option C confuses the geometric factor with albedo, which is a separate factor. Option D misapplies the inverse-square law, which describes how intensity changes with distance from the Sun, not this geometry.
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Answer B
Answer: B
Common error: Do not combine the albedo factor with the geometric factor of 4. The mean absorbed intensity is (1 − α)S/4, and the two factors have different causes.
10IBP2S-011-03-02Solar constant & S/4Paper 1AEasy1 mark
A space probe travels outwards until it is three times as far from the Sun as Earth is. At Earth's distance, the intensity of solar radiation is S. What intensity does the probe receive on a surface facing the Sun?
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The Sun's power spreads over a sphere of area 4πd2, so intensity ∝ 1/d2. Tripling the distance divides the intensity by 32 = 9. Option A assumes intensity ∝ 1/d. Option B uses a square root wrongly. Option D incorrectly applies the factor of 4 used for averaging over a planet's surface.
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Answer C
Answer: C
Exam tip: Measuring distances in units of Earth's orbital radius (AU) makes inverse-square scaling quick: I = S/d², with d in AU.
11IBP2S-011-03-05Solar constant & S/4Paper 1AEasy1 mark
Which statement best describes the solar constant?
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The solar constant is an intensity (W m−2) measured at Earth's mean orbital distance, on a surface normal to the Sun's rays and above the atmosphere, so that there is no absorption or scattering. Option A is the luminosity, measured in W. Option B describes a quantity close to S/4 reduced by the atmosphere. Option C includes atmospheric losses.
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Answer D
Answer: D
Exam tip: A full definition needs three conditions: at Earth's mean distance, perpendicular to the rays, and above the atmosphere.
12IBP2S-011-01-01Solar constant & S/4Paper 1AEasy1 mark
Two rocky planets orbit the same star at the same orbital radius. They have the same albedo and the same emissivity. Planet X has twice the radius of planet Y. What is the ratio \(\frac{\text{equilibrium temperature of X}}{\text{equilibrium temperature of Y}}\)?
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The power absorbed is proportional to the cross-sectional area πR2. The power emitted is proportional to the surface area 4πR2. Both depend on R2, so R cancels in the energy balance. The equilibrium temperature therefore does not depend on the planet's size and the ratio is 1. Option 2 comes from assuming T ∝ R. Option √2 comes from assuming T2 ∝ R. Option 4 comes from using only the absorbed power, which scales with R2.
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Answer C
Answer: C
Exam tip: Write out the full balance (1 − α)SπR² = eσ4πR²T⁴ before reasoning. You will see straight away which quantities cancel.
13IBP2S-011-01-02Solar constant & S/4Paper 1AEasy1 mark
In a simple energy-balance model of the Earth, the mean solar intensity received per square metre of the Earth's surface is taken as S/4, where S is the solar constant. Why does the factor 4 appear?
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The Earth intercepts sunlight over its disc of area πR2. That energy is shared, on average, over the whole spherical surface of area 4πR2. The ratio of these areas gives S/4. Atmospheric transmission and reflection are handled by the albedo, not by this factor. The inverse-square law sets the value of S itself.
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Answer D
Answer: D
Common error: Do not confuse the factor 4 from geometry with the effect of albedo. These are two separate corrections, and both appear in (1 − α)S/4.
14IBP2S-011-01-06Energy balance of a planetPaper 1AEasy1 mark
The mean temperature of a planet stays constant over many years. Which statement must be true?
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A constant temperature means there is no net gain or loss of internal energy. The power absorbed must therefore equal the power emitted. The incident power is larger than the absorbed power, because a fraction α is reflected. The emitted intensity is spread over the whole surface, so it is not equal to S. Albedo and emissivity are independent properties.
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Answer B
Answer: B
Common error: Balance the absorbed power, not the incident power. Reflected radiation never enters the planet's energy budget.
A manufacturer states that the glass door of a kitchen oven has an emissivity of 0.92. What does this value mean?
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Emissivity is the ratio of the power radiated per unit area by a surface to the power radiated per unit area by a black body at the same temperature. An emissivity of 0.92 means the glass radiates 0.92 × σT4 per unit area. Option A describes reflection, which relates to albedo, not emission. Option C confuses emissivity with the spectrum of the emitted radiation. Option D wrongly applies the factor to temperature, when it actually scales the emitted power.
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Answer B
Answer: B
Common error: Emissivity multiplies σT⁴. It does not multiply T. A surface with e = 0.92 behaves like a black body whose emitted power, not temperature, is reduced to 92%.
16IBP2S-011-00-02Greenhouse gasesPaper 1AEasy1 mark
A gas sample taken from a hospital operating theatre contains nitrous oxide (N2O) from anaesthetic equipment, water vapour, carbon dioxide and nitrogen. Which of these components is not a greenhouse gas?
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Greenhouse gases absorb infrared radiation because the frequencies of their molecular vibrations match infrared frequencies. N2O, H2O and CO2 all do this. Nitrogen (N2) is a symmetric diatomic molecule and does not significantly absorb infrared, so it is not a greenhouse gas. Nitrous oxide is easily confused with nitrogen because of its name. Water vapour is often overlooked, yet it is the largest contributor to the natural greenhouse effect.
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Answer D
Answer: D
Exam tip: Learn the main greenhouse gases: H₂O, CO₂, CH₄ and N₂O. The two most abundant atmospheric gases, N₂ and O₂, are not greenhouse gases.
17IBP2S-011-00-06Greenhouse gasesPaper 1AEasy1 mark
Greenhouse gases in the lower atmosphere absorb most strongly which type of radiation?
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The Earth's surface is at about 290 K, so it emits mainly infrared radiation with a peak near 10 μm. Greenhouse gas molecules have vibrational modes at infrared frequencies, so they absorb this radiation strongly. The atmosphere is largely transparent to visible light. Ultraviolet absorption by ozone is a separate process, and microwaves from space make a negligible contribution.
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Answer C
Answer: C
Exam tip: Connect the two halves of the topic. The Sun is hot, so it emits short wavelengths that pass through the atmosphere. The Earth is cool, so it emits long-wavelength infrared, which greenhouse gases absorb.
A fresh snowfield at a ski resort receives solar radiation of intensity 800 W m−2. It reflects radiation of intensity 680 W m−2. What is the albedo of the snow?
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Albedo = reflected power ÷ incident power = 680 ÷ 800 = 0.85. Albedo is a ratio, so it has no unit and lies between 0 and 1. 0.15 is the fraction absorbed, not reflected. 1.2 is the ratio inverted (incident ÷ reflected). 680 W m−2 is the reflected intensity, not the albedo.
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Answer B
Answer: B
Exam tip: Albedo is always reflected ÷ incident, so it is unitless and between 0 and 1. An answer with a unit, or one greater than 1, is wrong.
Which change would increase the albedo of the Earth?
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Low-level clouds reflect a large fraction of incoming sunlight. More cloud cover means a larger fraction of sunlight is reflected back to space, so the albedo increases. Carbon dioxide absorbs infrared radiation emitted by the surface. It does not reflect much sunlight, so it changes the greenhouse effect, not the albedo. Open ocean is much darker than ice, so melting sea ice lowers albedo. Dark panels absorb more than pale sand, so they also lower albedo.
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Answer C
Answer: C
Common error: Do not mix up albedo, which is reflection of incoming short-wavelength radiation, with the greenhouse effect, which is absorption of outgoing infrared.
20IBP2S-011-02-07Greenhouse gasesPaper 1AEasy1 mark
Why do greenhouse gases such as CO2 and H2O absorb strongly in the infrared but let most visible light pass through?
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A molecule absorbs a photon when the photon energy matches the gap between two of its energy levels. For CO2 and H2O, the gaps between vibrational levels correspond to infrared photon energies, so infrared is absorbed through resonance. Infrared photons have less energy than visible photons, not more. The molecules are far smaller than infrared wavelengths. These gases are colourless.
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Answer D
Answer: D
Exam tip: Mention resonance, or matching vibrational energy levels, when you explain why particular gases absorb infrared.
21IBP2S-011-02-10Surface–atmosphere energy balancePaper 1AEasy1 mark
Which of the following is not a way in which energy leaves the Earth's surface?
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Infrared radiation emitted downwards by the atmosphere (back-radiation) is absorbed by the surface. It is an energy input to the surface. The other three options all remove energy from the surface: thermal emission, latent heat carried away by evaporated water, and warm air rising by convection.
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Answer A
Answer: A
Exam tip: In a surface energy budget, sort every flow into 'in' or 'out' before you write the balance equation.
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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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.
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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.
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?
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Step 1At equilibrium the power emitted per unit area equals the power absorbed per unit area: eσT⁴ = I.
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Step 2T = (I/eσ)1/4.
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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.
A flat region is covered by forest of albedo 0.15. After a snowfall, 40 % of the area of the region is covered by snow of albedo 0.80 and the rest is still forest. The solar intensity incident on the region is the same before and after the snowfall.
What is (solar power absorbed by the region after the snowfall) / (solar power absorbed before the snowfall)?
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Step 1The mean albedo after the snowfall is the area-weighted mean: 0.40 × 0.80 + 0.60 × 0.15 = 0.41, so the fraction absorbed is 1 − 0.41 = 0.59.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Before, the fraction absorbed is 1 − 0.15 = 0.85, so the ratio is 0.59/0.85 = 0.69.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe whole region has been treated as snow-covered: 0.20/0.85 = 0.24. Only 40 % of the area is snow.
BThe two albedos have been averaged without weighting by area ((0.80 + 0.15)/2 = 0.475), giving 0.525/0.85 = 0.62.
CCorrect: the absorbed fraction falls from 0.85 to 0.59, a ratio of 0.69.
DThis is the ratio of the mean albedos, 0.41/0.15 = 2.7, which compares the scattered powers, not the absorbed powers.
In the resonance model of the greenhouse effect, a molecule of a greenhouse gas is modelled as atoms joined by springs.
Which statement explains why such a molecule absorbs strongly some of the infrared radiation emitted by the Earth's surface?
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Step 1The oscillating electric field of the radiation is a periodic driving force on the charged atoms. The amplitude, and so the energy absorbed, is large only when the driving frequency is close to a natural frequency of vibration of the molecule, which for CO₂, CH₄, H₂O and N₂O lies in the infrared.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: absorption is strong when the radiation's frequency matches a natural frequency of the molecule — resonance.
BInfrared photons have less energy than visible photons (lower frequency). The absorption depends on matching a natural frequency, not on the photon having a large energy.
CResonance needs the driving frequency to be close to the natural frequency. A driving frequency far from it produces only a small amplitude and little absorption.
DAbsorption is selective: greenhouse gases absorb strongly only in particular bands of frequency, and transmit most visible radiation.
28B-1A-53Equilibrium temperaturePaper 1AMedium1 mark
In a simple energy-balance model the mean surface temperature T of the Earth is given by (1 − α)S/4 = eσT⁴, where α is the albedo, S is the solar constant and e is the effective emissivity of the Earth's surface and atmosphere.
Which row gives the effect on T of an increase in the area covered by ice and of an increase in the concentration of greenhouse gases?
More ice coverMore greenhouse gases
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Step 1Ice has a high albedo, so more ice increases α, reduces the absorbed intensity (1 − α)S/4 and so reduces T.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2More greenhouse gas means that less of the infrared emitted by the surface escapes to space, so the effective emissivity e falls; with the same absorbed intensity, T⁴ = (1 − α)S/(4eσ) increases.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe effect of the ice has been reversed, as if the albedo were the fraction absorbed. Ice scatters more radiation, so less is absorbed and T falls.
BCorrect: a higher albedo lowers the absorbed intensity, and a lower effective emissivity means the surface must be hotter to radiate the same power to space.
CGreenhouse gases have been taken to increase the effective emissivity, as if they helped the Earth radiate to space. They absorb and re-emit infrared back towards the surface, so e falls and T rises.
DBoth effects have been reversed: albedo has been treated as the fraction absorbed, and greenhouse gases as increasing the effective emissivity.
The Sun is a sphere of radius R that radiates as a black body at surface temperature T. The mean distance between the Sun and the Earth is d.
What is the solar constant?
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Step 1Luminosity of the Sun: L = σ(4πR²)T⁴.
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Step 2The solar constant is the intensity at distance d: S = L/4πd² = σT⁴R²/d².
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe luminosity has not been divided by 4πd², only by d²: the power spreads over a sphere of area 4πd².
BThe intensity has been taken as inversely proportional to the distance, and the area of the Sun as proportional to R; both should be squared.
CThis is a quarter of the solar constant. The factor 1/4 gives the mean intensity over the Earth's whole surface, not the intensity of the rays arriving at the Earth.
31B-1A-82Surface–atmosphere energy balancePaper 1AMedium1 mark
Instruments on a flux tower in a desert measure the mean energy flows per unit area at the ground surface during one hour around midday. The values are shown in W m−2. The surface layer has negligible heat capacity, so the energy flows into and out of it balance.
What is the intensity X of the infrared radiation from the atmosphere that is absorbed by the surface?
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Step 1Energy in = energy out: 640 + X = 590 + 280 + 20 + 90.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2X = 980 − 640 = 340 W m−2.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe convection term, 280 W m−2, has been left out of the energy leaving the surface.
BOnly the radiation and convection have been balanced; the evaporation (20) and the conduction into the ground (90) also carry energy away from the surface.
CCorrect: 640 + X = 980, so X = 340 W m−2.
DThe absorbed solar radiation has been left out of the energy entering the surface: 980 is the total leaving, not X.
Which statement about water vapour in the Earth's atmosphere is correct?
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Step 1Water vapour is one of the main greenhouse gases. Its main source is natural evaporation (human sources include irrigation and combustion), and warmer air can hold more of it, so warming increases its concentration.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AWater vapour is one of the main greenhouse gases: its molecules absorb infrared radiation emitted by the surface. Condensation only limits how much the air can hold.
BBurning fuels does release water vapour, but the main source is evaporation from the oceans, a natural process.
CWater vapour is largely transparent to visible radiation; it absorbs strongly in the infrared, where its molecular energy levels match the photon energies.
DCorrect: it absorbs infrared radiation, and because warmer air holds more water vapour, warming increases its concentration and adds to the greenhouse effect.
33B-1A-87Surface–atmosphere energy balancePaper 1AMedium1 mark
Infrared radiation of intensity I emitted by the ground passes into a thin layer of air. The layer absorbs 40 % of this radiation and transmits the rest. In a simple model, the layer re-emits all of the energy it absorbs, half upwards and half downwards. Ignore all other radiation.
What is the intensity of the infrared radiation that leaves the top of the layer?
The layer absorbs 40 % of the infrared from the ground
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Step 1Transmitted upwards: (1 − 0.40)I = 0.60I.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Re-emitted upwards: half of 0.40I = 0.20I; total leaving the top = 0.80I.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AOnly the re-emitted radiation travelling upwards has been counted; the 60 % transmitted also leaves the top.
BThe re-emission has been ignored, as if the absorbed energy stayed in the layer.
34IBP2S-011-03-03Solar constant & S/4Paper 1AMedium1 mark
A planet orbits the Sun at half of Earth's orbital radius and has an albedo of 0.50. The solar constant at Earth is S. What is the mean solar intensity absorbed per unit area of the planet's whole surface?
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At half the distance, the intensity is 22S = 4S. Averaging over the sphere gives 4S/4 = S. Absorbing a fraction (1 − 0.50) gives 0.50S. Option A ignores the change in distance. Option B treats intensity as ∝ 1/d. Option D omits the factor of 1/4 from the averaging geometry.
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Answer C
Answer: C
Exam tip: Apply the three factors in turn: inverse-square for distance, then 1/4 for geometry, then (1 − α) for albedo.
35IBP2S-011-03-06Equilibrium temperaturePaper 1AMedium1 mark
A climate model treats Earth as a black body in radiative equilibrium. If the Sun's luminosity increased by 10%, by what factor would Earth's equilibrium temperature change?
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Absorbed power ∝ S ∝ L, and the emitted power ∝ T4. So T ∝ L1/4, and 1.101/4 ≈ 1.024. Option B assumes T ∝ L. Option C uses a square root. Option A squares the factor instead of taking a root.
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Answer D
Answer: D
Exam tip: For small changes, a fractional change x in the absorbed power gives a fractional change of roughly x/4 in temperature.
36IBP2S-011-03-07Solar constant & S/4Paper 1AMedium1 mark
Earth has a radius of 6.37 × 106 m and the solar constant is 1.36 × 103 W m−2. What is the total solar power intercepted by Earth?
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Sunlight reaches Earth as parallel rays, so Earth intercepts the power falling on its cross-sectional disc, of area πR2: P = SπR2 = 1360 × π × (6.37 × 106)2 ≈ 1.7 × 1017 W. Option A wrongly uses S/4 together with the disc area. Option C uses the area of a hemisphere, 2πR2. Option D uses the full surface area, 4πR2.
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Answer B
Answer: B
Common error: Use either S with πR² or S/4 with 4πR². Both give the same power. Never use S/4 with πR².
37IBP2S-011-03-08Solar constant & S/4Paper 1AMedium1 mark
Mars orbits the Sun at 1.5 times Earth's orbital radius. The solar constant at Earth is 1360 W m−2. What is the mean solar intensity incident on the top of the Martian atmosphere, averaged over the whole planet?
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The intensity at Mars is 1360/1.52 ≈ 604 W m−2. Averaging over the sphere gives 604/4 ≈ 150 W m−2. Option D omits the factor of 1/4. Option B uses 1/d instead of 1/d2. Option C is Earth's value, S/4.
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Answer A
Answer: A
Exam tip: Check whether a question asks for the intensity facing the Sun or the intensity averaged over the whole planet. These differ by a factor of 4.
38IBP2S-011-03-10Solar constant & S/4Paper 1AMedium1 mark
Earth's distance from the Sun varies from 1.47 × 1011 m to 1.52 × 1011 m during the year. By approximately what percentage is the solar intensity above the atmosphere greater at the closest point than at the farthest point?
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The intensity ratio is (1.52/1.47)2 ≈ 1.07, an increase of about 7%. Option B is the change in distance, not in intensity. Option D doubles the correct value. Option A wrongly treats the solar constant as fixed at every point in the orbit; it is defined at the mean distance.
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Answer C
Answer: C
Common error: A small fractional change in distance produces roughly twice that fractional change in intensity, because intensity ∝ 1/d².
39IBP2S-011-01-03Equilibrium temperaturePaper 1AMedium1 mark
A space agency models a planet that moves from orbital radius d to orbital radius 2d around its star. Its albedo and emissivity do not change. What is the ratio \(\frac{\text{new equilibrium temperature}}{\text{original equilibrium temperature}}\)?
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Doubling the distance reduces the intensity to one quarter, because I ∝ 1/d2. Since T4 ∝ I, the temperature changes by (1/4)1/4 = 1/√2. Option 1/2 assumes T ∝ 1/d. Option 1/4 assumes T ∝ I. Option 1/16 wrongly raises the intensity factor to the fourth power instead of taking its fourth root.
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Answer B
Answer: B
Exam tip: Combine the two proportionalities, I ∝ d⁻² and T ∝ I¹/⁴, to get T ∝ d⁻¹/². This saves time in ratio questions.
40IBP2S-011-01-04Equilibrium temperaturePaper 1AMedium1 mark
Some climate scientists propose that, in the distant past, ice may have covered the whole of the Earth's surface. Suppose such an event raised the planet's albedo from 0.30 to 0.65, with emissivity unchanged. What is the ratio \(\frac{\text{new equilibrium temperature}}{\text{original equilibrium temperature}}\)?
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The absorbed fraction is (1 − α). It changes from 0.70 to 0.35, which halves the absorbed power. Since T4 ∝ (1 − α), the ratio is (1/2)1/4. Option 1/2 forgets the fourth root. Option (1/2)1/2 uses a square root instead of a fourth root. Option (0.65/0.30)1/4 uses the albedo itself in place of (1 − α). That option would wrongly predict warming.
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Answer A
Answer: A
Common error: The albedo is the fraction reflected. The fraction absorbed is 1 − α, so a higher albedo always means less absorbed power and a lower temperature.
41IBP2S-011-01-07Equilibrium temperaturePaper 1AMedium1 mark
A student models the Earth using \(T = \left(\frac{(1-\alpha)S}{4e\sigma}\right)^{1/4}\), where e is the effective emissivity of the Earth and its atmosphere. In this model, how is an increase in the atmospheric concentration of carbon dioxide best represented?
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Extra CO2 absorbs more of the infrared radiation emitted by the surface. Part of this energy is re-radiated back down, so less infrared escapes to space at a given surface temperature. This lowers the effective emissivity e and so raises T. Increasing e would predict cooling. CO2 has little effect on reflected sunlight, so α is not the right quantity. CO2 has no effect on the solar constant S.
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Answer D
Answer: D
Exam tip: Effective emissivity compares the power escaping to space with the power a black body at the surface temperature would emit. Greenhouse gases reduce this ratio.
42IBP2S-011-01-08Energy balance of a planetPaper 1AMedium1 mark
An exoplanet orbits a star whose luminosity is 16 times that of the Sun. Its orbital radius is 4 times the Earth–Sun distance. The exoplanet has the same albedo and emissivity as the Earth. How does the equilibrium temperature of the exoplanet compare with that of the Earth?
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The intensity at the planet is S = L/(4πd2). Multiplying L by 16 and d2 by 16 leaves S unchanged. With the same albedo and emissivity, the equilibrium temperature is therefore the same as the Earth's. The answer '4 times' uses L/d. The answer '16 times' ignores the distance. The answer '2 times' takes a square root of 16/4.
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Answer D
Answer: D
Exam tip: Always find the intensity at the planet first. Two planets receiving the same intensity, with the same α and e, have the same equilibrium temperature.
43IBP2S-011-01-09Greenhouse gasesPaper 1AMedium1 mark
Carbon dioxide in the atmosphere absorbs strongly the infrared radiation emitted by the Earth's surface. It is almost transparent to visible sunlight. What is the reason for this difference?
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A molecule absorbs a photon only if the photon energy matches a difference between two of its energy levels. For CO2, vibrational transitions correspond to infrared photon energies. Molecules are far smaller than visible wavelengths, so the size argument is wrong. Infrared photons have less energy than visible photons, not more. CO2 does not significantly reflect visible light.
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Answer A
Answer: A
Common error: Infrared photons have lower energy than visible photons. Absorption depends on matching energy levels, not on having 'more' energy.
Two panels on a satellite are at the same temperature. Panel A has emissivity 0.8 and surface area 2 m2. Panel B has emissivity 0.2 and surface area 4 m2. What is the ratio of the power radiated by A to the power radiated by B?
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P = eσAT4. T is the same for both panels, so PA/PB = (0.8 × 2)/(0.2 × 4) = 1.6/0.8 = 2. Option D (4) comes from comparing emissivities only and ignoring area. Option A (0.5) compares areas only. Option B wrongly assumes that equal temperatures mean equal powers.
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Answer C
Answer: C
Exam tip: In ratio questions, write P = eσAT⁴ for each object and cancel common factors. Check which quantities really are the same for both.
The absorber plate of a rooftop solar water heater is designed to reach as high a temperature as possible. The plate operates at about 350 K. Which combination of surface properties is best?
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Sunlight is mostly visible and near-infrared, so the plate should absorb strongly at these wavelengths. A plate at about 350 K emits mainly long-wavelength infrared (around 8 μm). A low emissivity at these wavelengths reduces radiative loss, so the plate gets hotter. Option A describes ordinary black paint, which absorbs well but also radiates well. Option C reflects most of the incoming sunlight. Option D describes a radiative-cooling coating.
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Answer B
Answer: B
Exam tip: Emissivity can depend on wavelength. A surface can absorb strongly in the visible and still emit weakly in the infrared. This 'selective surface' idea also explains why greenhouse gases are transparent to sunlight but absorb infrared.
46IBP2S-011-00-05Greenhouse gasesPaper 1AMedium1 mark
Which statement correctly describes how greenhouse gases raise the mean surface temperature of the Earth?
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The surface emits infrared radiation. Greenhouse gas molecules absorb this radiation and then re-radiate infrared in random directions. Some of this re-emitted radiation returns to the surface, so the surface receives more power and reaches a higher equilibrium temperature. Option A describes reflection, but the process is absorption followed by re-emission. Option C describes ozone absorbing ultraviolet, which is not the greenhouse mechanism. Option D confuses the greenhouse effect with ozone depletion.
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Answer B
Answer: B
Common error: Do not say greenhouse gases 'reflect' or 'trap' heat. Say they absorb infrared from the surface and re-emit it in all directions.
47IBP2S-011-00-07Greenhouse gasesPaper 1AMedium1 mark
Methane strongly absorbs infrared radiation of wavelength 7.7 μm. Which is the best explanation for this absorption?
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Absorption is a resonance effect. When the frequency of the incident infrared matches a natural vibrational frequency of the molecule, energy is transferred efficiently and the molecule absorbs. Infrared photons have far too little energy to ionise molecules. A molecule is about 10−10 m across, much smaller than 7.7 μm. Methane in air is in thermal equilibrium with the other gases, so it is not hotter.
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Answer A
Answer: A
Exam tip: Use the word 'resonance' and link it to matching frequencies of molecular vibration. This is the key idea in any explanation of greenhouse absorption.
48IBP2S-011-00-09Equilibrium temperaturePaper 1AMedium1 mark
A thermal camera is set to emissivity 1.00. It is pointed at a metal handrail whose true emissivity is 0.60. Reflected radiation from the surroundings is negligible. The camera calculates temperature from the measured power per unit area. The true kelvin temperature of the handrail is T. What temperature does the camera display? (0.601/4 ≈ 0.88)
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The handrail emits 0.60σT4 per unit area. The camera assumes e = 1, so it solves σTread4 = 0.60σT4. This gives Tread = 0.601/4 T ≈ 0.88 T. The camera reads too low. Option A applies the emissivity to T rather than T4. Option C ignores the effect of emissivity. Option D inverts the correction.
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Answer B
Answer: B
Exam tip: Any device that infers temperature from emitted radiation must be told the emissivity. If it assumes a value too high, it under-reads the temperature.
49IBP2S-011-02-03Energy balance of a planetPaper 1AMedium1 mark
An exoplanet orbits a star. The intensity of the star's radiation at the planet's orbit is 2400 W m−2. The albedo of the planet is 0.25. What is the mean power absorbed per unit area of the planet's surface?
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The planet intercepts radiation over its cross-section, πR2. It spreads that energy over its whole surface, 4πR2. This gives the factor of 1/4. Only the fraction (1 − 0.25) = 0.75 is absorbed. Mean absorbed intensity = 0.75 × 2400 ÷ 4 = 450 W m−2. 1800 W m−2 leaves out the factor of 4. 600 W m−2 ignores albedo. 150 W m−2 uses the albedo as the absorbed fraction.
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Answer D
Answer: D
Common error: Using the albedo itself as the absorbed fraction is a frequent slip. The absorbed fraction is (1 − albedo).
Two identical flat roofs receive the same solar intensity. One is coated white (albedo 0.80). The other is black (albedo 0.10). What is the ratio \(\frac{\text{power absorbed by black roof}}{\text{power absorbed by white roof}}\)?
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Absorbed power ∝ (1 − albedo). Ratio = (1 − 0.10) ÷ (1 − 0.80) = 0.90 ÷ 0.20 = 4.5. 8.0 is the ratio of the albedos, which compares reflected power. 0.22 is the correct ratio inverted. 1.1 is 0.90 ÷ 0.80, which mixes the absorbed fraction of one roof with the albedo of the other.
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Answer C
Answer: C
Exam tip: Convert each albedo to an absorbed fraction (1 − α) before you compare absorbed powers.
51IBP2S-011-02-05Energy balance of a planetPaper 1AMedium1 mark
A simple model of a planet makes three assumptions: • the atmosphere is transparent to incoming sunlight; • the atmosphere absorbs all the infrared radiation emitted by the surface; • the atmosphere re-emits this energy equally upwards and downwards. In equilibrium, what is the ratio of the power radiated per unit area by the surface to the solar power absorbed per unit area by the surface?
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Let the surface emit Ps. The atmosphere absorbs all of Ps and emits Ps/2 upwards and Ps/2 downwards. At the top of the atmosphere, only Ps/2 leaves to space. This must equal the absorbed solar power P, so Ps = 2P. Check at the surface: input = P + Ps/2 = 2P = output. The ratio is 1 if the atmosphere is ignored. 0.5 inverts the ratio. 4 wrongly doubles the factor.
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Answer A
Answer: A
Exam tip: In a layer model, write a separate energy balance for the top of the atmosphere and for the surface. Each must balance on its own.
52IBP2S-011-02-08Solar constant & S/4Paper 1AMedium1 mark
An icy moon has an albedo of 0.60. Dust from a nearby ring settles on the surface and lowers the albedo to 0.20. The emissivity and the incident intensity stay the same. What is the ratio \(\frac{\text{new equilibrium temperature}}{\text{original equilibrium temperature}}\)?
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At equilibrium, σT4 ∝ (1 − α), so T ∝ (1 − α)1/4. The ratio is (0.80 ÷ 0.40)1/4 = 21/4 ≈ 1.19. 2.0 ignores the fourth root. 1.41 uses a square root. 0.84 is the correct ratio inverted, which would mean that darkening cools the moon.
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Answer A
Answer: A
Common error: Compare the absorbed fractions (1 − α), not the albedos themselves, and then take the fourth root.
53IBP2S-011-02-09Energy balance of a planetPaper 1AMedium1 mark
Suppose the concentration of carbon dioxide in the atmosphere suddenly increases. Before any temperatures have had time to change, which statement is correct?
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Extra CO2 absorbs more of the upward infrared radiation. Less infrared escapes at the top of the atmosphere. Incoming absorbed solar power is unchanged, so there is a net energy gain and temperatures start to rise until balance is restored. The surface's own emission depends on its temperature, which has not yet changed. CO2 barely affects albedo or the transmission of sunlight.
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Answer C
Answer: C
Exam tip: Describe the enhanced greenhouse effect as an imbalance: power in exceeds power out until the temperature rises.
55B-1A-34Equilibrium temperaturePaper 1AHard1 mark
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?
The atmosphere absorbs all infrared from the surface and emits equally upwards and downwards
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Step 1Atmosphere: it absorbs σTs⁴ and emits an intensity D upwards and D downwards, so σTs⁴ = 2D.
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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.
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Step 3Eliminate D: σTs⁴ = I + σTs⁴/2, so σTs⁴ = 2I.
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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).
Planets X and Y have no atmosphere and orbit the same star at the same distance. They have the same radius and the same albedo, and both radiate as black bodies.
X rotates rapidly, so its whole surface is at one uniform temperature TX. Y always keeps the same hemisphere facing the star. In a simple model the sunlit hemisphere of Y is at a uniform temperature TY and its dark hemisphere radiates negligible power.
What is TY/TX?
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Step 1Both planets intercept the same power, set by the projected disc πr²: (1 − α)Sπr².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2X radiates from its whole surface, 4πr²; Y radiates only from its sunlit hemisphere, 2πr².
57B-1A-83Energy balance of a planetPaper 1AHard1 mark
A gas-giant planet of radius R and albedo α is at a distance from the Sun where the intensity of the solar radiation is I. Energy from its slow contraction is also released inside the planet at a total rate P. The planet radiates as a black body at a uniform temperature T and is in equilibrium.
What is σT⁴?
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Step 1Power in: (1 − α)IπR² absorbed over the projected disc, plus P from inside.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Power out: 4πR²σT⁴ from the whole surface.
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Step 3Divide by 4πR²: σT⁴ = (1 − α)I/4 + P/4πR².
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: both inputs are spread over the radiating area 4πR², giving (1 − α)I/4 + P/4πR².
BThe internal power has been divided by the projected disc πR². It leaves through the whole surface, 4πR², just like the absorbed solar power.
CThe factor 1/4 has been left out of the solar term: the planet intercepts over πR² but radiates over 4πR².
DThe internal power has been treated as a loss. It is an extra input, so it makes the planet hotter.
The concentration of carbon dioxide in the atmosphere increases and then stays constant at its new value. The solar constant and the albedo of the Earth do not change. Eventually the Earth reaches a new equilibrium.
Which row compares the new equilibrium with the original equilibrium?
Infrared intensity leaving the top of the atmosphereMean surface temperature
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Step 1At any equilibrium the infrared power leaving the top of the atmosphere equals the solar power absorbed, (1 − α)S/4 per unit area, which has not changed.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2More CO₂ means that more of the surface's infrared is absorbed and re-emitted back down, so the surface must be hotter before the same intensity escapes to space.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis describes the Earth just after the increase, before equilibrium is restored. The resulting energy imbalance warms the Earth until the outgoing intensity again equals the absorbed solar intensity.
BCorrect: the outgoing intensity returns to (1 − α)S/4, but with more absorption by CO₂ the surface must be hotter to achieve it.
CA hotter surface emits more, but more of that emission is absorbed by the atmosphere. If more left than is absorbed from the Sun, the Earth would cool.
DThe outgoing intensity is indeed the same, but the surface temperature cannot be: with more absorbing CO₂, a surface at the old temperature would send less to space than it receives.
59IBP2S-011-03-04Solar constant & S/4Paper 1AHard1 mark
Two black bodies are in space at the same distance from the Sun. Body P is a thin flat plate held with one face perpendicular to the Sun's rays; it radiates from both faces. Body Q is a rapidly rotating sphere whose temperature is uniform. What is the ratio of equilibrium temperatures, TP/TQ?
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Plate: it absorbs SA and emits 2AσT4, so σTP4 = S/2. Sphere: it absorbs Sπr2 and emits 4πr2σT4, so σTQ4 = S/4. The ratio of fourth powers is 2, so TP/TQ = 21/4 ≈ 1.19. Options C and D come from forgetting the fourth root, or taking only a square root. Option A assumes that the shape makes no difference.
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Answer B
Answer: B
Common error: In an energy balance, the absorbing area and the emitting area are usually different. Write each one out explicitly before cancelling.
60IBP2S-011-03-09Energy balance of a planetPaper 1AHard1 mark
An airless, rapidly rotating planet behaves as a black body and has equilibrium temperature T at a distance d from its star. An identical planet orbits the same star at a distance 4d. What is the equilibrium temperature of the second planet?
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σT4 ∝ S ∝ 1/d2, so T ∝ d−1/2. At 4d, T becomes T/41/2 = T/2. Option A is the change in intensity, not in temperature. Option B assumes T ∝ 1/d. Option D takes a fourth root of 4 rather than of 16.
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Answer C
Answer: C
Exam tip: Combining the inverse-square law with the Stefan–Boltzmann law gives T ∝ 1/√d, a useful result for comparing planets.
Two black-body objects are at the same distance from the Sun. • Object P is a thin flat plate that always faces the Sun. Its back is perfectly insulated, so it radiates only from its sunlit face. • Object Q is a small, rapidly rotating sphere with a uniform surface temperature. What is the ratio \(\frac{\text{equilibrium temperature of P}}{\text{equilibrium temperature of Q}}\)?
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The plate absorbs SA and emits σAT4 from one face, so TP4 = S/σ. The sphere absorbs SπR2 and emits σ4πR2T4, so TQ4 = S/(4σ). The ratio of the fourth powers is 4, so the temperature ratio is 41/4 = √2. Option 4 ignores the fourth root. Option 2 takes only a square root. Option 1 assumes that equal intensity means equal temperature, whatever the geometry.
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Answer C
Answer: C
Exam tip: For any shape, compare the area that absorbs with the area that emits. The ratio of these areas sets the factor inside the fourth root.
62IBP2S-011-01-10Energy balance of a planetPaper 1AHard1 mark
A planet is modelled with albedo α, effective emissivity e and orbital radius d around a star of luminosity L. Each change below is made on its own. Which change produces the largest increase in equilibrium temperature?
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T4 ∝ (1 − α)L/(ed2), so compare the factor by which each change multiplies T4. • Albedo: (1 − α) goes from 0.40 to 0.80, a factor of 2. • Emissivity: 1/e increases by a factor of 3. • Distance: 1/d2 increases by a factor of 1/0.64 ≈ 1.56. • Luminosity: a factor of 1.5. The emissivity change gives the largest factor, so it gives the largest temperature rise. The albedo change looks largest only if you compare α values directly rather than (1 − α).
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Answer B
Answer: B
Exam tip: Because the fourth root is the same for every option, compare the factor by which T⁴ changes. There is no need to take roots.
A CubeSat in deep shadow dissipates a constant electrical power from its electronics. It loses energy only by radiation and receives no external radiation. Over time its coating degrades, and its emissivity falls from 0.80 to 0.05. By what factor does its equilibrium kelvin temperature change?
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At equilibrium the radiated power equals the constant dissipated power, so eσAT4 is constant. Therefore T4 ∝ 1/e. T2/T1 = (0.80/0.05)1/4 = 161/4 = 2. Option D forgets the fourth root. Option C takes a square root instead. Option A assumes that the power input alone fixes the temperature, ignoring the fact that a poorer emitter must be hotter to radiate the same power.
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Answer B
Answer: B
Common error: Identify which quantity is fixed. Here the power is fixed, so a lower emissivity forces a higher temperature, with T ∝ e⁻¹/⁴.
64IBP2S-011-00-10Equilibrium temperaturePaper 1AHard1 mark
A small asteroid in sunlight has a surface that behaves as a grey body, so its absorptivity equals its emissivity at all wavelengths. Space weathering darkens the surface, and the emissivity rises from 0.30 to 0.90. The asteroid's distance from the Sun does not change. What happens to its equilibrium temperature?
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Absorbed power = eIAcross and emitted power = eσAsurfaceT4, where I is the solar intensity. At equilibrium the factor e appears on both sides and cancels, so T does not depend on e. Tripling the emissivity triples both the absorption and the emission. Option A considers only the increased absorption. Option B considers only the increased emission. Option D applies the factor directly to T.
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Answer C
Answer: C
Common error: 'Darker means hotter' only holds when absorptivity in sunlight differs from emissivity in the infrared. For a true grey body, the two effects cancel exactly.
65IBP2S-011-02-06Surface–atmosphere energy balancePaper 1AHard1 mark
A rocky planet with no atmosphere has an equilibrium surface temperature of 250 K. Suppose an atmosphere is added that is transparent to sunlight and absorbs all infrared radiation from the surface. It re-emits this energy equally upwards and downwards. The albedo is unchanged. What is the new equilibrium surface temperature?
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With the absorbing layer, the surface must radiate twice the absorbed solar power, so σT4 doubles. The new temperature is T = 21/4 × 250 ≈ 1.19 × 250 ≈ 297 K. 500 K assumes temperature is proportional to emitted power. 354 K uses a square root instead of a fourth root. 250 K ignores the downward re-emission from the atmosphere.
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Answer B
Answer: B
Common error: Radiated power depends on T⁴. Doubling the power raises the temperature by a factor of 2¹/⁴, not 2.
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.
A student uses a satellite dataset for a region of open ocean. For each of eight months the dataset gives the mean cloud fraction f (the fraction of the area of the region covered by cloud) and the mean albedo α of the region. Each value of α has an absolute uncertainty of ±0.02; the uncertainty in f is negligible.
The student models the region as a mixture of cloud-free ocean of albedo αo and cloud of albedo αc, so that α = αo + (αc − αo)f.
f
0.18
0.27
0.35
0.44
0.52
0.61
0.70
0.83
α
0.16
0.18
0.24
0.26
0.31
0.34
0.39
0.44
Graph drawn to scale
(a)
Outline why the albedo of the region increases as the cloud fraction increases.
(1)
(b)
Calculate the percentage uncertainty in α for the month with the smallest cloud fraction.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine the intercept of your line on the α axis, and state what it represents.
(2)
(e)
Determine, using your line, the albedo αc of the cloud.
(2)
(f)
Published values of the albedo of this type of cloud lie between 0.45 and 0.60. Comment on the student’s result.
(1)
(g)
The months in the dataset are spread through the year, so the angle at which the Sun’s rays meet the region at noon is different in each month. Suggest how this could affect the data.
(1)
(h)
In one month the mean solar intensity incident on the region is 280 W m−2. Calculate, using the gradient of your line, the decrease in the mean intensity absorbed by the region if the cloud fraction increased from 0.20 to 0.80.
(2)
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Part (a)
cloud scatters a much larger fraction of the incident sunlight than open ocean, so a larger cloud fraction scatters more of the incident power
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OWTTE.
Part (b)
0.02/0.16 × 100 = 13 %
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Accept 12.5 %.
Part (c)
single straight line through all the error bars, extended to f = 0 and f = 1
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Part (d)
intercept ≈ 0.07
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Accept 0.05–0.09.
the albedo of the cloud-free ocean, αo
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Part (e)
value of α on the line at f = 1 OR αc = gradient + intercept
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αc = 0.52
✓ 1
Accept 0.49–0.55. Allow ECF from (d).
Part (f)
0.52 lies within the published range, so the result is consistent with the published values
✓ 1
Allow ECF from (e).
Part (g)
the albedo depends on the angle of the Sun’s rays «surfaces scatter more when the Sun is low», so part of the variation in α is not caused by cloud: this adds scatter / a systematic effect to the data
✓ 1
OWTTE. Accept: the ocean albedo αo is not constant from month to month.
Part (h)
Δα = gradient × 0.60 = 0.444 × 0.60 = 0.266
✓ 1
Allow ECF from (c).
decrease in absorbed intensity = 280 × 0.266 = 75 W m−2
✓ 1
Accept 69–81 W m−2. Award [2] for CNA.
Answers: (b) 13 % · (d) 0.07 · (e) 0.52 · (h) 75 W m−2(the remaining parts are explanations — see the table above)
A student uses a dataset of global five-year mean values to test whether the burning of fossil fuels is the main cause of the rise in the concentration of carbon dioxide in the atmosphere. For each period the dataset gives the mean mass E of CO₂ released each year by burning fossil fuels, in gigatonnes (1 Gt = 1012 kg), and the mean rise ΔC each year in the CO₂ concentration of the atmosphere, in parts per million (ppm). Each value of ΔC has an uncertainty of ±0.15 ppm per year; the uncertainty in E is negligible.
An increase of 1 ppm in the concentration corresponds to 7.8 Gt of CO₂ added to the atmosphere.
period
E / Gt per year
ΔC / ppm per year
1985–89
22.4
1.52
1990–94
23.6
1.40
1995–99
25.0
1.66
2000–04
27.2
1.67
2005–09
29.8
2.01
2010–14
32.4
2.03
2015–19
34.6
2.31
2020–24
36.0
2.23
Graph drawn to scale
(a)
Draw the line of best fit for the data.
(1)
(b)
Determine the gradient of your line. Include an appropriate unit.
(2)
(c)
Calculate, using your answer to (b), the fraction of the CO₂ released by burning fossil fuels that remains in the atmosphere.
(1)
(d)
Calculate the percentage uncertainty in ΔC for the period 1985–89.
(1)
(e)
Outline whether the graph supports the hypothesis that the rise in CO₂ concentration is caused by burning fossil fuels.
(1)
(f)
Suggest what happens to the CO₂ released by burning fossil fuels that does not remain in the atmosphere.
(1)
(g)
State one natural process that releases CO₂ into the atmosphere and could add to the scatter of the data.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
single straight line through all the error bars
✓ 1
Part (b)
gradient from a large triangle
✓ 1
0.063 ppm Gt−1
✓ 1
Accept 0.057–0.069. Unit needed: ppm per Gt OR ppm Gt−1.
Part (c)
0.063 × 7.8 = 0.49
✓ 1
Allow ECF from (b). Accept 0.45–0.55.
Part (d)
0.15/1.52 × 100 = 10 %
✓ 1
Part (e)
yes: ΔC increases linearly with E AND the line passes through/close to the origin «within the error bars», so ΔC is proportional to the emissions
✓ 1
Both features needed. OWTTE.
Part (f)
it is absorbed by natural sinks: it dissolves in the oceans and is taken up by plants in photosynthesis
✓ 1
Either sink.
Part (g)
volcanic eruptions OR respiration / decay of organic matter OR forest fires OR release from the oceans when they warm
✓ 1
Any one.
Answers: (b) 0.063 ppm Gt−1 · (c) 0.49 · (d) 10 % (the remaining parts are explanations — see the table above)
A student determines the emissivity e of a black paint used on spacecraft radiators. A thin square aluminium plate of side 50.0 mm, painted on both faces, is hung by two fine leads in an evacuated chamber. The walls of the chamber are kept at Tw = 290 K. A heater on the plate is adjusted until the plate reaches a steady temperature T, measured with a thermocouple to ±2 K. The electrical power P supplied to the heater is then recorded, with an uncertainty of ±0.20 W.
In the steady state P = eσA(T⁴ − Tw⁴), where A is the total area of the two faces. The graph shows P against (T⁴ − Tw⁴), with error bars, for five of the six readings.
T / K
P / W
(T⁴ − Tw⁴) / 1010 K⁴
350
2.23
0.793
380
3.69
1.378
410
5.83
2.118
440
8.19
470
11.35
4.172
500
14.89
5.543
Graph drawn to scale
(a)
Suggest why the plate is placed in an evacuated chamber.
(1)
(b)
Calculate the missing value in the table for T = 440 K and plot the point on the graph.
(1)
(c)
Show that the absolute uncertainty in (T⁴ − Tw⁴) when T = 500 K is about 0.1 × 1010 K⁴.
(1)
(d)
Draw the line of best fit for the data.
(1)
(e)
Determine e using the gradient of your line.
(3)
(f)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in e. Ignore the uncertainty in A.
(2)
(g)
The manufacturer states that the emissivity of the paint is 0.92. Comment on the student's result.
(1)
(h)
Some energy is conducted away from the plate along the leads. State and explain the effect of this on the value of e obtained.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
so that no energy is transferred from the plate by conduction or convection through air: all the energy supplied leaves by radiation
✓ 1
OWTTE.
Part (b)
(440⁴ − 290⁴)/1010 = 3.041 AND point plotted correctly
A student determines the emissivity e of black insulating tape. A strip of the tape is stuck to an aluminium block that is warmed on a hotplate. A thermocouple inside the block measures its temperature T with an uncertainty of ±1.0 K. An infrared thermometer, set to an emissivity of 1.00, is aimed at the tape and gives a reading TIR with an uncertainty of ±0.5 K. The room is at Tr = 293 K; treat this value as exact.
The infrared thermometer receives the radiation emitted by the tape and the radiation from the room reflected by the tape, so that TIR⁴ − Tr⁴ = e(T⁴ − Tr⁴).
The graph shows (TIR⁴ − Tr⁴) against (T⁴ − Tr⁴), with error bars, for five of the six readings.
T / K
TIR / K
(T⁴ − Tr⁴) / 109 K⁴
(TIR⁴ − Tr⁴) / 109 K⁴
323.3
321.0
3.55
3.25
342.6
340.4
6.41
6.06
363.2
359.6
10.03
383.5
378.2
14.26
13.09
402.7
398.0
18.93
17.72
423.1
417.1
24.68
22.90
Graph drawn to scale
(a)
Outline why the reading of the infrared thermometer is lower than the temperature of the tape.
(1)
(b)
Calculate the missing value in the table for T = 363.2 K and plot the point on the graph.
(1)
(c)
In a first analysis the student ignored the radiation from the room reflected by the tape and used TIR⁴ = eT⁴. Calculate the value of e that this analysis gives for the reading at T = 423.1 K.
(1)
(d)
Draw the line of best fit for the data.
(1)
(e)
Determine e using the gradient of your line.
(2)
(f)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in e.
(2)
(g)
The manufacturer states that the emissivity of the tape is 0.95. Comment on the student’s result.
(1)
(h)
The outer surface of the tape is slightly cooler than the block. State and explain the effect of this on the value of e obtained.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the tape radiates less power per unit area than a black body at the same temperature «e < 1», but the thermometer assumes a black body, so it displays a lower temperature
✓ 1
OWTTE.
Part (b)
(359.6⁴ − 293⁴)/109 = 9.35 AND point plotted correctly
✓ 1
Accept 9.4. Point within half a small square.
Part (c)
e = (417.1/423.1)⁴ = 0.944
✓ 1
Accept 0.94. The fourth power must be seen. «Larger than the value from the graph: ignoring the reflected room radiation overestimates e.»
Part (d)
single straight line through all the error bars «passing through or very close to the origin»
✓ 1
Part (e)
gradient = Δy/Δx from a large triangle «e = gradient»
✓ 1
e = 0.93
✓ 1
Accept 0.92–0.94. No unit. Award [2] for CNA.
Part (f)
steepest and shallowest lines through all the error bars «gradients ≈ 0.94 and 0.92»
✓ 1
Δe = ½(emax − emin) ≈ ±0.01
✓ 1
Accept ±0.01 to ±0.03 «lines through the corners of the error boxes give about 0.96 and 0.91». Allow ECF.
Part (g)
comparison of 0.95 with the candidate’s range and a consistent conclusion: e.g. 0.95 lies outside the range 0.92–0.94, so the result is not consistent with the manufacturer’s value «the student’s value is too small»
✓ 1
Allow ECF from (e) and (f): with a range of 0.90–0.96, ‘consistent’ is accepted. A reference to the uncertainty is needed.
Part (h)
the thermocouple value of T is too large, so (T⁴ − Tr⁴) is too large and the gradient, and so e, is too small «underestimated»
✓ 1
Direction and reason both needed.
Answers: (b) 9.35 × 109 K⁴ · (c) 0.944 · (e) e = 0.93 · (f) ±0.01 (the remaining parts are explanations — see the table above)
71B-1B-38Equilibrium temperaturePaper 1BMedium11 marks
Engineers test white and grey coatings for a small satellite. Each coating is applied to a thin flat coupon whose back face is insulated. The coupons are placed in a vacuum chamber whose walls are cooled by liquid nitrogen, so that the radiation they receive from the walls is negligible. A solar simulator lamp, set to give an intensity of 1361 W m−2, shines perpendicularly onto the front face of each coupon in turn.
Every coating has an emissivity e = 0.85 for infrared radiation. The albedo α of each coating for the simulator’s radiation was measured separately with an uncertainty of ±0.01. The steady temperature T of each coupon is measured with an uncertainty of ±1 K. The graph shows T⁴ against (1 − α), with error bars, for six of the seven coupons.
α
(1 − α)
T / K
T⁴ / 109 K⁴
0.10
0.90
394
24.1
0.22
0.78
379
20.6
0.35
0.65
363
17.4
0.47
0.53
344
0.58
0.42
326
11.3
0.70
0.30
298
7.9
0.82
0.18
263
4.8
Graph drawn to scale
(a)
Show that, in the steady state, T⁴ = I(1 − α)/eσ, where I is the intensity incident on the coupon.
(1)
(b)
Calculate the missing value in the table for α = 0.47 and plot the point on the graph.
(1)
(c)
Calculate the percentage uncertainty in (1 − α) for the most reflective coating.
(1)
(d)
Draw the line of best fit for the data.
(1)
(e)
Determine, using the gradient of your line, the intensity I incident on the coupons.
(3)
(f)
The uncertainty in the value of I found from the graph is ±4 %. Deduce whether the result is consistent with the intensity set on the simulator.
(1)
(g)
Suggest one reason for the difference between the two values of the intensity.
(1)
(h)
Determine, using the graph, the albedo of a coating that would keep a coupon at 293 K in this chamber.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
power absorbed per unit area (1 − α)I = power radiated per unit area eσT⁴ «back face insulated, no radiation from the walls», so T⁴ = I(1 − α)/eσ
✓ 1
Must see the energy balance stated.
Part (b)
344⁴ = 14.0 × 109 K⁴ AND point plotted correctly
✓ 1
Accept 14.00.
Part (c)
0.01/0.18 × 100 = 5.6 %
✓ 1
Accept 6 %.
Part (d)
single straight line through all the error bars «passing close to the origin»
✓ 1
Part (e)
gradient ≈ 26.7 × 109 K⁴ from a large triangle
✓ 1
Accept 25.9–27.5 × 109 K⁴.
gradient = I/eσ
✓ 1
I = 26.7 × 109 × 0.85 × 5.67 × 10−8 = 1290 W m−2
✓ 1
Allow ECF. Accept 1250–1330 W m−2. Award [3] for CNA.
Part (f)
range 1240 to 1340 W m−2 does not include 1361 W m−2, so the result is not consistent «the intensity reaching the coupons is lower than the set value»
✓ 1
Allow ECF from (e). A conclusion consistent with the candidate’s range is needed.
Part (g)
any one: energy is conducted from the coupon through its mounts / the insulation, so T is lower than radiation alone predicts; the lamp intensity at the coupon is less than the set value «e.g. absorption by the chamber window»; the albedo for the simulator’s spectrum differs from the value measured
✓ 1
OWTTE.
Part (h)
293⁴ = 7.37 × 109 K⁴, read from the line: (1 − α) ≈ 0.28
✓ 1
Allow use of the gradient: (1 − α) = 7.37/gradient. Allow ECF from (e).
α ≈ 0.72
✓ 1
Accept 0.69–0.75.
Answers: (b) 14.0 × 109 K⁴ · (c) 5.6 % · (e) I = 1290 W m−2 · (h) α ≈ 0.72 (the remaining parts are explanations — see the table above)
72B-1B-04Equilibrium temperaturePaper 1BHard11 marks
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)
Show mark scheme
Marking point
Mark
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)
73B-1B-25Equilibrium temperaturePaper 1BHard11 marks
A student uses a database of dark, airless asteroids to test a simple energy-balance model. For each asteroid the database gives its mean distance d from the Sun, in astronomical units (AU), and its mean surface temperature T, found from the infrared radiation it emits. The uncertainty in each value of T is ±3 %.
In the model, each asteroid has albedo α and emissivity e, is in radiative equilibrium, and absorbs a mean intensity equal to (1 − α)/4 times the solar intensity incident on it. The solar intensity at a distance d is S0/d², where S0 = 1361 W m−2 is the solar constant at 1 AU. The student plots lg T against lg d. The graph shows the data for six of the seven asteroids.
Asteroid
d / AU
T / K
lg(d / AU)
lg(T / K)
P
1.12
266
0.049
2.425
Q
1.45
228
0.161
2.358
R
1.93
202
0.286
2.305
S
2.36
184
T
2.81
164
0.449
2.215
U
3.30
152
0.519
2.182
V
4.05
140
0.607
2.146
Graph drawn to scale
(a)
Show that the model predicts T ∝ d−1/2.
(2)
(b)
Calculate the missing values for asteroid S and plot its point on the graph.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine the gradient of your line.
(2)
(e)
The uncertainty in the gradient is ±0.03. Comment on whether the data support the model.
(1)
(f)
Using your line, determine the mean surface temperature that the model predicts at d = 1 AU, and hence determine the albedo of these asteroids. Assume e = 0.90.
(3)
(g)
Outline one reason why the temperature of an individual asteroid may not fit the model exactly.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
energy balance: (1 − α)(S0/d²)/4 = eσT⁴
✓ 1
Accept without the factor ¼ «it does not affect the proportionality».
α, e, S0 constant so T⁴ ∝ d−2 and T ∝ d−1/2
✓ 1
Part (b)
0.373 AND 2.265 AND point plotted correctly
✓ 1
Accept 2 to 4 decimal places.
Part (c)
single straight line through all the error bars
✓ 1
Part (d)
gradient from a large triangle «negative sign shown»
✓ 1
gradient = −0.50
✓ 1
Accept −0.53 to −0.47. No unit.
Part (e)
the model predicts −0.50, which lies within −0.53 to −0.47, so the data support the model
✓ 1
Allow ECF from (d). Reference to the uncertainty is needed.
Part (f)
at lg d = 0 the line gives lg T ≈ 2.446, so T ≈ 279 K
✓ 1
Accept 274–284 K. Accept the line extended to lg d = 0 or a point on the line combined with the gradient.
Accept 0.02–0.16 «the result is very sensitive to T». Allow ECF.
Part (g)
any one: its albedo or emissivity differs from the assumed values; its orbit is elliptical so d varies; it rotates so the day and night sides differ in temperature; T is not a uniform surface temperature
✓ 1
OWTTE.
Answers: (b) 0.373, 2.265 · (d) −0.50 · (f) 279 K, α ≈ 0.09 (the remaining parts are explanations — see the table above)
A student at a mountain observatory estimates the solar constant S. On a cloudless morning, when the Earth is at its mean distance from the Sun, she measures the intensity I of the direct solar radiation with a radiometer pointed at the Sun. Each value of I has an uncertainty of ±10 W m−2. For each reading she records the zenith angle z, the angle between the direction of the Sun and the vertical.
The radiation crosses a thickness of atmosphere proportional to the air mass m = 1/cos z. A model of the absorption and scattering in the atmosphere gives I = S e−km, where k is a constant. The graph shows ln(I / W m−2) against m, with error bars, for seven of the eight readings.
z / °
m
I / W m−2
ln(I / W m−2)
72
3.236
942
6.848
68
2.669
996
6.904
64
2.281
1050
6.957
60
2.000
1079
6.984
55
1119
50
1.556
1134
7.034
45
1.414
1159
7.055
40
1.305
1168
7.063
Graph drawn to scale
(a)
Show that the model predicts a straight-line graph of ln I against m.
(1)
(b)
Calculate the two missing values in the table for z = 55° and plot the point on the graph.
(2)
(c)
Show that the absolute uncertainty in ln I for I = 942 W m−2 is about 0.01.
(1)
(d)
Draw the line of best fit for the data.
(1)
(e)
Determine k using the gradient of your line.
(2)
(f)
Determine S using your line.
(2)
(g)
Lines of maximum and minimum gradient show that the uncertainty in the intercept on the ln I axis is ±0.017. Determine whether the student’s value of S is consistent with the accepted value of the solar constant, 1361 W m−2.
(2)
(h)
Suggest why the measurements are made on a cloudless morning at a high-altitude site.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ln I = ln S − km, which has the form y = c + mx with gradient −k and intercept ln S
✓ 1
Must see the logarithm of both sides taken.
Part (b)
m = 1/cos 55° = 1.743
✓ 1
Accept 1.74.
ln 1119 = 7.020 AND point plotted correctly
✓ 1
Accept 3 or 4 d.p.
Part (c)
Δ(ln I) = ΔI/I = 10/942 = 0.011 «≈ 0.01»
✓ 1
Must see ΔI/I with the values substituted.
Part (d)
single straight line through all the error bars, extended to m = 0
✓ 1
Part (e)
gradient from a large triangle «negative» = −k
✓ 1
k = 0.114
✓ 1
Accept 0.104–0.124. No unit «m has no unit».
Part (f)
intercept at m = 0: ln S ≈ 7.214
✓ 1
Accept 7.204–7.224.
S = e7.214 = 1360 W m−2
✓ 1
Allow ECF. Accept 1340–1380 W m−2.
Part (g)
range of S: e7.197 to e7.231 = 1335 to 1381 W m−2 OR ΔS = S × 0.017 ≈ ±23 W m−2
✓ 1
Allow ECF from (f).
1361 W m−2 lies within this range, so the value is consistent with the accepted value
✓ 1
MP2 needs a comparison and a conclusion consistent with the range found.
Part (h)
k must stay constant during the measurements: cloud or haze would change the absorption; at high altitude there is less atmosphere «air, water vapour, dust» above the site, so the absorption is smaller and more stable
✓ 1
Any one valid point. OWTTE.
Answers: (b) 1.743, 7.020 · (e) k = 0.114 · (f) S = 1360 W m−2(the remaining parts are explanations — see the table above)
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
Marking point
Mark
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)
A classroom air-quality monitor measures the concentration of carbon dioxide. Infrared radiation from a small lamp passes along a tube containing air from the room and falls on a detector. A filter in front of the detector transmits only radiation of wavelength 4.26 μm, which carbon dioxide molecules absorb. The higher the concentration of carbon dioxide, the smaller the power reaching the detector.
The diagram shows four of the energy levels of a carbon dioxide molecule.
Energy levels drawn to scale
(a)
Show that the energy of a photon of wavelength 4.26 μm is about 0.29 eV.
(1)
(b)
Draw arrows on the diagram to show the transition when a carbon dioxide molecule in its lowest energy level absorbs a photon of wavelength 4.26 μm (label this arrow A) and when it absorbs a photon of wavelength 15.0 μm emitted by the Earth’s surface (label this arrow B).
(2)
(c)
An excited molecule in the tube soon emits a photon of the same wavelength. Explain why, nevertheless, the power reaching the detector decreases as the concentration of carbon dioxide increases.
(2)
(d)
The lamp in the monitor is a small filament at a temperature of 1100 K. Determine the wavelength at which the filament emits most strongly, and outline why the filament is nevertheless a suitable source for the monitor.
(3)
(e)
In one lesson, 28 people are in a closed classroom of volume 240 m³. Each person breathes out carbon dioxide at a rate of 5.0 × 10−6 m³ s−1, measured at room temperature and pressure. At the start of the 50-minute lesson the air contains 420 parts per million (ppm) of carbon dioxide by volume. Assume that no air enters or leaves the room. Determine whether the concentration rises above 1000 ppm before the end of the lesson.
(2)
(f)
Water vapour absorbs infrared radiation of wavelength 2.7 μm. Suggest why the monitor would give readings that are too high if its filter also transmitted radiation of this wavelength.
The Earth moves around the Sun in a slightly elliptical orbit. Its distance from the Sun is smallest, 1.471 × 1011 m, in early January and greatest, 1.521 × 1011 m, in early July. The solar constant S is the intensity of the Sun’s radiation at the Earth’s mean distance from the Sun.
Luminosity of the Sun = 3.83 × 1026 W.
Axes for (b)
(a)
(i)
Show that the intensity of the Sun’s radiation at the Earth in early January is about 1410 W m−2.
(1)
(ii)
Calculate the intensity of the Sun’s radiation at the Earth in early July.
(1)
(b)
Sketch, on the axes, a graph to show how the intensity of the Sun’s radiation at the Earth varies over one year.
(2)
(c)
State Kepler’s second law and hence outline why the Earth spends less time near its closest point to the Sun than near its furthest point.
(2)
(d)
Treat the Earth as a black body with an albedo of 0.30 that reaches equilibrium at each point of its orbit. Determine the difference between its equilibrium temperatures in early January and in early July.
(3)
(e)
The global mean surface temperature is in fact higher in July than in January. Suggest one reason for this.
(1)
Show mark scheme
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Mark
Notes
Part (a)(i)
3.83 × 1026/(4π × (1.471 × 1011)²) = 1409 W m−2
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
3.83 × 1026/(4π × (1.521 × 1011)²) = 1317 W m−2
✓ 1
Accept 1320 W m−2. ALTERNATIVE: 1409 × (1.471/1.521)².
Part (b)
smooth curve with a single maximum of about 1410 W m−2 in early January «at the start and end» and a minimum of about 1320 W m−2 in early July
✓ 1
Allow ECF from (a).
curve varies smoothly and continuously «not straight segments», with the same value on 1 January at both ends
✓ 1
Part (c)
the line joining the Earth to the Sun sweeps out equal areas in equal times
✓ 1
when the Earth is closer to the Sun it must move faster «to sweep the same area», so it passes the closest part of the orbit more quickly
✓ 1
OWTTE.
Part (d)
σT⁴ = (1 − 0.30)I/4 for each date
✓ 1
Accept a ratio method: T ∝ I1/4 ∝ d−1/2.
TJan = 256.8 K AND TJul = 252.5 K
✓ 1
Allow ECF from (a).
difference = 4.3 K
✓ 1
Accept 4.2–4.4 K. Award [3] for CNA.
Part (e)
most of the Earth’s land is in the northern hemisphere: land has a smaller heat capacity than ocean, so it warms more during the northern summer OR snow cover on northern land in January raises the Earth’s albedo, so less radiation is absorbed
✓ 1
Any one reason. OWTTE.
Answers: (a)(ii) 1317 W m−2 · (d) 4.3 K (the remaining parts are explanations — see the table above)
A horticultural glasshouse is a building with walls and roof made of glass. The glass transmits most electromagnetic radiation with wavelengths between 0.3 μm and 2.5 μm but absorbs almost all radiation with wavelengths greater than 5 μm.
The surface temperature of the Sun is 5800 K. The soil and plants inside the glasshouse are at about 300 K.
Diagram not to scale
(a)
Calculate the wavelength at which the emission is most intense for the Sun and for the soil.
(2)
(b)
Hence explain why the glass lets in most of the energy radiated by the Sun but absorbs most of the energy radiated by the soil.
(2)
(c)
Solar radiation of intensity 650 W m−2 is incident perpendicularly on a horizontal glass roof of area 40 m2. The glass transmits 88 % of this radiation to the floor below, which has an albedo of 0.25. Calculate the power absorbed by the floor.
(2)
(d)
The glass absorbs infrared radiation from the soil and then emits infrared radiation. Outline why only part of this emitted radiation returns to the soil.
(1)
(e)
When a roof vent is opened, the air temperature inside the glasshouse falls quickly, although the glass still absorbs the infrared radiation from the soil. Explain what this shows about the main reason why a closed glasshouse stays warm.
(2)
(f)
Discuss whether a glasshouse is a good model for the warming of the Earth’s surface by its atmosphere.
(3)
Show mark scheme
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Notes
Part (a)
Sun: 2.9 × 10−3/5800 = 5.0 × 10−7 m «0.50 μm»
✓ 1
soil: 2.9 × 10−3/300 = 9.7 × 10−6 m «9.7 μm»
✓ 1
Part (b)
the Sun’s radiation is mostly at wavelengths around 0.5 μm, inside the range 0.3–2.5 μm that the glass transmits
✓ 1
Allow ECF from (a).
the soil’s radiation is mostly at wavelengths around 10 μm, greater than 5 μm, which the glass absorbs
✓ 1
Part (c)
650 × 0.88 × (1 − 0.25) × 40
✓ 1
= 1.7 × 104 W «17 kW»
✓ 1
Award [2] for CNA. Award [1 max] for 2.3 × 104 W «albedo ignored».
Part (d)
the glass emits from both of its faces / in all directions, so part of the radiation goes outwards to the surroundings
✓ 1
OWTTE.
Part (e)
warm air escapes through the vent and is replaced by cooler air from outside «convection»
✓ 1
so a closed glasshouse stays warm mainly because it stops warm air being carried away by convection, not mainly because the glass absorbs infrared radiation
✓ 1
OWTTE.
Part (f)
similarity: both are largely transparent to incoming solar radiation and absorb outgoing infrared, re-emitting part of it back towards the surface
✓ 1
difference: the atmosphere has no lid and convection still carries energy upwards; its warming is due to absorption and re-emission by greenhouse-gas molecules at particular wavelengths, so the glasshouse is only a partial model
✓ 1
conclusion: the glasshouse is a poor model of the mechanism, because a glasshouse warms mainly by stopping convection, while the atmosphere warms the surface by absorbing and re-emitting infrared
✓ 1
Award [1] for a similarity, [1] for a difference and [1] for a conclusion consistent with them.
Answers: (a) 5.0 × 10−7 m and 9.7 × 10−6 m · (c) 1.7 × 104 W (the remaining parts are explanations — see the table above)
A family compares the carbon dioxide emissions of two cars.
The petrol car uses 6.0 litres of petrol for every 100 km travelled. Petrol has a density of 740 kg m−3 and an energy density of 46 MJ kg−1. Burning 1.0 kg of petrol releases 3.1 kg of carbon dioxide.
The electric car takes 0.18 kW h of electrical energy from its charger for every kilometre travelled. The electricity is generated in a gas-fired power station of efficiency 0.50, and 92 % of the electrical energy generated reaches the charger. Burning natural gas releases 0.056 kg of carbon dioxide for every 1.0 MJ of energy released. 1 kW h = 3.6 MJ.
(a)
State what is meant by the enhanced greenhouse effect.
(1)
(b)
(i)
Show that the mass of petrol used by the petrol car per kilometre is about 0.044 kg.
(1)
(ii)
Calculate the mass of carbon dioxide released per kilometre by the petrol car.
(1)
(iii)
The engine of the petrol car does 0.50 MJ of useful work per kilometre. Calculate its efficiency.
(1)
(c)
Determine whether the electric car is responsible for less carbon dioxide per kilometre than the petrol car.
(3)
(d)
Natural gas is mainly methane. Some methane leaks from gas pipelines. Methane molecules absorb infrared radiation strongly at a wavelength of 7.7 μm.
(i)
Calculate the frequency of radiation of wavelength 7.7 μm.
(1)
(ii)
Outline, using the resonance model, why methane absorbs radiation of this frequency strongly.
(2)
Show mark scheme
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Mark
Notes
Part (a)
the additional warming of the Earth caused by the increase in the concentration of greenhouse gases due to human activities «e.g. burning fossil fuels»
✓ 1
OWTTE.
Part (b)(i)
0.060 × 10−3 m3 × 740 kg m−3 = 0.0444 kg
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
0.0444 × 3.1 = 0.138 kg «138 g»
✓ 1
Accept 0.14 kg.
Part (b)(iii)
energy from fuel = 0.0444 × 46 = 2.04 MJ; efficiency = 0.50/2.04 = 0.24
✓ 1
Accept 24 %.
Part (c)
electrical energy at charger = 0.18 × 3.6 = 0.648 MJ per km
✓ 1
energy from gas = 0.648/(0.50 × 0.92) = 1.41 MJ; CO₂ = 1.41 × 0.056 = 0.079 kg per km
✓ 1
Allow ECF. Award [1 max] if one efficiency is omitted.
0.079 kg < 0.138 kg, so yes, the electric car is responsible for less CO₂ per kilometre
✓ 1
MP3 needs the comparison and a consistent conclusion.
Part (d)(i)
f = 3.00 × 108/7.7 × 10−6 = 3.9 × 1013 Hz
✓ 1
Part (d)(ii)
the molecule is modelled as atoms joined by springs, with a natural frequency of vibration equal to this frequency
✓ 1
the oscillating electric field of the radiation drives the vibration at resonance, so the amplitude is large and energy is absorbed strongly
✓ 1
OWTTE.
Answers: (b)(ii) 0.138 kg · (b)(iii) 0.24 · (c) 0.079 kg per km · (d)(i) 3.9 × 1013 Hz (the remaining parts are explanations — see the table above)
A solar-powered spacecraft orbits Jupiter, at a mean distance of 5.2 AU from the Sun. Its solar arrays have a total area of 60 m2 and an efficiency of 0.28, and they are kept perpendicular to the Sun's rays. The solar constant at Earth (1.0 AU) is 1360 W m−2.
(a)
State what is meant by the solar constant.
(1)
(b)
Calculate the intensity of solar radiation at the spacecraft.
(2)
(c)
Calculate the electrical power output of the arrays.
(2)
Show mark scheme
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Notes
Part (a)
The power per unit area received from the Sun on a surface perpendicular to the rays, at Earth's mean distance from the Sun and outside the atmosphere
✓ 1
Part (b)
Uses the inverse-square law: I = 1360/5.22
✓ 1
I = 50.3 W m−2 (accept 50 W m−2)
✓ 1
Part (c)
P = 0.28 × 50.3 × 60
✓ 1
P = 845 W (accept 840–850 W from rounded intermediate values; ECF from (b))
✓ 1
Exam tip: A Sun-tracking panel receives the full intensity, not a quarter of it. The factor of 1/4 applies only when averaging over a rotating sphere.
81IBP2S-011-03-12Solar constant & S/4Paper 2Easy6 marks
The Sun has a luminosity of 3.83 × 1026 W. Earth's mean distance from the Sun is 1.50 × 1011 m and Earth's albedo is 0.30.
(a)
Show that the solar constant S is about 1.4 × 103 W m−2.
(2)
(b)
Outline why the mean intensity incident on Earth's whole surface is S/4.
(2)
(c)
Calculate the mean intensity absorbed by the Earth–atmosphere system per unit area of Earth's surface.
(2)
Show mark scheme
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Notes
Part (a)
S = L/(4πd2)
✓ 1
= 3.83 × 1026/(4π × (1.50 × 1011)2) = 1.35 × 103 W m−2, which is about 1.4 × 103 W m−2. The answer must be given to at least 3 s.f.
✓ 1
Part (b)
Earth intercepts solar power over its cross-sectional area πR2
✓ 1
Averaged over time, this power is spread over the total surface area 4πR2, so the mean intensity is SπR2/4πR2 = S/4
✓ 1
Part (c)
Mean incident intensity S/4 = 339 W m−2
✓ 1
Absorbed = (1 − 0.30) × 339 ≈ 237 W m−2
✓ 1
Accept answers from 236 to 245 W m−2, for example from using S = 1.35 × 103 or 1.4 × 103 W m−2. Allow follow-through from the value of S found in the first part.
—
Common error: Albedo is the fraction reflected, so the fraction absorbed is (1 − α). Multiplying by α instead is a frequent slip.
82IBP2S-011-01-11Equilibrium temperaturePaper 2Easy6 marks
A spherical space probe of radius 0.80 m is sent to study the inner Solar System. It orbits the Sun at 0.72 AU. The probe spins slowly, so its surface temperature can be taken as uniform.
Data: • Solar intensity at 1.00 AU = 1.36×103 W m−2 • Albedo of probe surface = 0.60 • Emissivity of probe surface = 0.85 • σ = 5.67×10−8 W m−2 K−4
(a)
Calculate the intensity of solar radiation at the probe.
(2)
(b)
Calculate the solar power absorbed by the probe.
(2)
(c)
Determine the equilibrium temperature of the probe.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Use the inverse-square law: I = 1.36×103 / 0.722
✓ 1
I = 2.6×103 W m−2
✓ 1
Part (b)
Cross-sectional area = π × 0.802 = 2.0 m2
✓ 1
Fraction absorbed = 1 − albedo = 0.40, so P = 0.40 × 2.6×103 × 2.0 = 2.1×103 W
✓ 1
Allow ECF from (a).
—
Part (c)
Power emitted = eσ(4πr2)T4, with 4πr2 = 8.0 m2, set equal to the power absorbed
✓ 1
T4 = 2.1×103 / (0.85 × 5.67×10−8 × 8.0), giving T ≈ 272 K
✓ 1
Allow ECF from (b).
—
Common error: The probe absorbs over its cross-section πr² but emits over its whole surface 4πr². Using the same area for both overestimates T by a factor of √2.
83IBP2S-011-01-15Energy balance of a planetPaper 2Easy6 marks
Instruments on a weather satellite measure the Earth's radiation budget. Averaged over the whole surface of the Earth and over a year, they record: • Solar constant: 1361 W m−2 • Mean solar intensity reflected back to space: 102 W m−2 • Mean surface temperature of the Earth: 288 K
σ = 5.67×10−8 W m−2 K−4
(a)
Show that the mean incoming solar intensity per square metre of the Earth's surface is about 340 W m−2.
(2)
(b)
Calculate the albedo of the Earth.
(1)
(c)
Calculate the mean intensity that the Earth must emit for its temperature to remain constant.
(1)
(d)
Determine the effective emissivity of the Earth.
(2)
Show mark scheme
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Mark
Notes
Part (a)
The Earth intercepts SπR2, which is spread over a surface area of 4πR2
✓ 1
1361/4 = 340.25 ≈ 340 W m−2
✓ 1
Part (b)
α = 102/340 = 0.30
✓ 1
Part (c)
340 − 102 = 238 W m−2
✓ 1
Part (d)
σT4 = 5.67×10−8 × 2884 = 390 W m−2
✓ 1
e = 238/390 = 0.61
✓ 1
Allow error carried forward from the emitted intensity found in the previous part.
—
Common error: Using the full solar constant, 1361 W m⁻², in place of S/4 gives an emissivity greater than 1. An emissivity above 1 shows that a step has been missed.
84IBP2S-011-00-11Energy balance of a planetPaper 2Easy4 marks
During heavy braking in a motorsport race, a carbon brake disc reaches a temperature of 1100 K. The disc has emissivity 0.75 and an exposed radiating surface area of 0.090 m2. The surroundings are at about 300 K.
(a)
State what is meant by a black body.
(1)
(b)
Calculate the power radiated by the disc at 1100 K.
(2)
(c)
Outline why radiation absorbed by the disc from the surroundings can be neglected when calculating the power radiated by the disc.
(1)
Show mark scheme
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Notes
Part (a)
A body that absorbs all radiation incident on it, at all wavelengths. It is also a perfect emitter, with emissivity 1.
✓ 1
Part (b)
P = eσAT4 = 0.75 × 5.67×10−8 × 0.090 × 11004
✓ 1
P ≈ 5.6×103 W
✓ 1
Part (c)
The absorbed power is proportional to (300)4, so it is only about (300/1100)4 ≈ 0.55% of the emitted power (about 30 W compared with 5600 W), which is negligible.
✓ 1
Exam tip: Raise the kelvin temperature to the fourth power before multiplying. Never use °C in the Stefan–Boltzmann law.
85IBP2S-011-00-12Greenhouse gasesPaper 2Easy6 marks
Agriculture contributes significantly to the atmospheric concentration of several greenhouse gases. Water vapour, however, makes the largest contribution to the natural greenhouse effect.
(a)
State two greenhouse gases, other than carbon dioxide and water vapour. For each gas, state one human activity that releases it.
(2)
(b)
Explain why a rise in CO2 concentration can lead to an increase in atmospheric water vapour, and why this amplifies the warming.
(2)
(c)
Distinguish between the greenhouse effect and the enhanced greenhouse effect.
(2)
Show mark scheme
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Mark
Notes
Part (a)
First gas named, with a matching human activity
✓ 1
Second gas named, with a matching human activity
✓ 1
Acceptable pairs:
—
• Methane: cattle/livestock digestion, rice paddies, landfill or natural gas leaks
—
• Nitrous oxide: nitrogen fertiliser use or industrial processes
—
• CFCs/HFCs: refrigerants, aerosols or air conditioning
—
• Tropospheric ozone: vehicle or industrial emissions
—
Each mark needs both the gas and a correct activity.
—
Part (b)
Warming from CO2 increases evaporation, and warmer air can hold more water vapour
✓ 1
Water vapour is itself a greenhouse gas. It absorbs more infrared and causes further warming (positive feedback)
✓ 1
Part (c)
Greenhouse effect: a natural process. Atmospheric gases absorb infrared emitted by the surface and re-emit it in all directions, some back towards the surface. This keeps the surface warmer than it would otherwise be
✓ 1
Enhanced greenhouse effect: additional warming caused by increased concentrations of greenhouse gases from human activity
✓ 1
Exam tip: For 'distinguish', make a clear contrast. The key difference is natural greenhouse gas levels versus human-caused increases in their concentrations.
A ski resort covers a slope of area 2.0 × 104 m2 with machine-made snow. On a clear day, the intensity of sunlight on the slope is 650 W m−2. Fresh snow has an albedo of 0.80.
(a)
State what is meant by albedo.
(1)
(b)
Calculate the total solar power absorbed by the fresh snow.
(2)
(c)
Over several days, soot and dust lower the albedo of the snow to 0.50. Calculate the increase in the solar power absorbed by the slope.
(1)
(d)
Outline why melting of the snow tends to speed up once it has started.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Albedo is the ratio of the reflected (scattered) power to the incident power on a surface.
Allow ECF from (b), e.g. 6.5 × 106 W − 2.6 × 106 W.
—
Part (d)
Melting exposes darker ground or dirty, wet snow, which has a lower albedo.
✓ 1
More solar power is absorbed, causing more melting. This is a positive feedback.
✓ 1
Common error: In (c), it is the absorbed fraction (1 − α) that rises from 0.20 to 0.50. Multiplying by the albedos instead gives the change in reflected power.
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)
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)
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)
Solar radiation of intensity S = 1360 W m−2 arrives at the Earth as parallel rays. The diagram shows the Earth at an equinox, when the Sun is directly overhead at the equator at local noon. At latitude 60° N the rays at noon make an angle of 60° with the vertical. Ignore the atmosphere in (a) and (b).
latitude / °
mean incident intensity / W m−2
albedo
mean absorbed intensity / W m−2
0
433
0.24
329
30
375
0.25
281
60
216
0.40
80
75
0.60
Diagram NOT accurately drawnAxes for (c)(ii)
(a)
Explain why the intensity of the solar radiation on horizontal ground at noon is smaller at latitude 60° N than at the equator.
(3)
(b)
Calculate the intensity of the solar radiation on horizontal ground at latitude 60° N at noon.
(1)
(c)
The table shows, for four latitudes at the equinox, the mean intensity of the solar radiation over 24 hours on horizontal ground at the top of the atmosphere, and the mean albedo of the Earth–atmosphere system.
(i)
Complete the table.
(2)
(ii)
Draw, on the axes, a graph to show the variation of the mean absorbed intensity with latitude.
(2)
(d)
Determine the equilibrium temperature at latitude 80° predicted by this model, assuming that the Earth–atmosphere system there radiates as a black body and that no energy is transferred between latitudes.
(2)
(e)
The measured mean temperature near the surface at latitude 80° at the equinox is about 245 K. Explain why this is much higher than your answer to (d).
(3)
Show mark scheme
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Notes
Part (a)
at 60° N the same beam of radiation «the same power» is spread over a larger area of ground
✓ 1
the area of ground is larger by a factor 1/cos 60° = 2 / the ground has a smaller projected area perpendicular to the rays
✓ 1
intensity is power per unit area, so the intensity is smaller «halved»
✓ 1
Part (b)
1360 × cos 60° = 680 W m−2
✓ 1
Part (c)(i)
60°: 216 × 0.60 = 130 W m−2
✓ 1
Accept 129.6.
80°: 75 × 0.40 = 30 W m−2
✓ 1
Part (c)(ii)
all four points plotted correctly to within half a small square
✓ 1
Allow ECF from (c)(i).
smooth curve through the points, decreasing with latitude
✓ 1
Do not accept straight segments joining the points dot to dot.
Part (d)
σT4 = 30 W m−2
✓ 1
T = (30 / 5.67 × 10−8)1/4 = 152 K
✓ 1
Award [2] for CNA. Allow ECF from (c)(i).
Part (e)
energy is transported from low latitudes to high latitudes
✓ 1
by convection in the atmosphere «winds» and by ocean currents carrying warm water polewards
✓ 1
the greenhouse effect: the atmosphere absorbs infrared radiation from the surface and re-emits some of it back to the surface, so the surface is warmer than the black-body estimate
✓ 1
OWTTE
Answers: (b) 680 W m−2 · (c)(i) 130 and 30 W m−2 · (d) 152 K (the remaining parts are explanations — see the table above)
91B-2-65Energy balance of a planetPaper 2Medium13 marks
Instruments on satellites measure the Earth’s energy budget at the top of the atmosphere. Averaged over the whole Earth and over one year, the mean intensity of the incoming solar radiation is 340.0 W m−2, the mean intensity of the solar radiation scattered back to space is 100.0 W m−2 and the mean intensity of the infrared radiation emitted to space is 239.2 W m−2.
Radius of the Earth = 6.37 × 106 m; 1 year = 3.15 × 107 s.
Mean intensities over the whole Earth and one year; diagram not to scale
(a)
(i)
Show that these data correspond to a solar constant of about 1360 W m−2.
(1)
(ii)
Calculate the albedo of the Earth.
(1)
(b)
(i)
Calculate the net intensity absorbed by the Earth.
(1)
(ii)
Explain how the enhanced greenhouse effect causes the infrared intensity emitted to space to be less than the solar intensity absorbed.
(2)
(c)
Determine the energy gained by the whole Earth in one year.
(2)
(d)
About 90 % of this energy is stored in the upper 700 m of the oceans, which cover an area of 3.6 × 1014 m2. The density of sea water is 1030 kg m−3 and its specific heat capacity is 3990 J kg−1 K−1. Estimate the rise in the mean temperature of this layer of ocean in one year.
(3)
(e)
Each of the two intensities leaving the Earth is measured with an uncertainty of ±2.0 W m−2; the uncertainty in the incoming intensity is negligible. Determine whether these satellite measurements alone show that the Earth is gaining energy.
(2)
(f)
Suggest one way in which the warming of the oceans could further increase the greenhouse effect.
(1)
Show mark scheme
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Mark
Notes
Part (a)(i)
S = 4 × 340.0 = 1360 W m−2 «the mean incoming intensity is S/4»
✓ 1
Must see the factor 4.
Part (a)(ii)
100.0/340.0 = 0.294
✓ 1
Accept 0.29.
Part (b)(i)
340.0 − 100.0 − 239.2 = 0.8 W m−2
✓ 1
Part (b)(ii)
human activity «burning fossil fuels» has increased the concentration of greenhouse gases, which absorb more of the infrared radiation emitted by the surface
✓ 1
and re-emit it in all directions, part of it back towards the surface, so less of it escapes to space «until the surface warms enough to restore the balance»
Allow ECF from (b)(i). Award [2] for CNA. Award [1 max] for πR² used «3.2 × 1021 J».
Part (d)
mass of water = 3.6 × 1014 × 700 × 1030 = 2.6 × 1020 kg
✓ 1
energy stored = 0.90 × 1.3 × 1022 = 1.2 × 1022 J
✓ 1
Allow ECF from (c).
ΔT = 1.2 × 1022/(2.6 × 1020 × 3990) = 0.011 K
✓ 1
Accept 0.011–0.012 K. Award [3] for CNA.
Part (e)
uncertainty in the net intensity = ±(2.0 + 2.0) = ±4.0 W m−2
✓ 1
this is much larger than 0.8 W m−2, so the net intensity could be negative: the satellite measurements alone do not show that the Earth is gaining energy
✓ 1
MP2 needs the comparison and a consistent conclusion. Accept ±2.8 W m−2 from adding in quadrature.
Part (f)
more water evaporates, and water vapour is a greenhouse gas OR warmer oceans dissolve less CO₂ / release CO₂ OR melting sea ice lowers the albedo
✓ 1
Any one.
Answers: (a)(ii) 0.294 · (b)(i) 0.8 W m−2 · (c) 1.3 × 1022 J · (d) 0.011 K (the remaining parts are explanations — see the table above)
92B-2-66Surface–atmosphere energy balancePaper 2Medium12 marks
A fruit grower uses a weather station to forecast frost. The surface of the ground has an emissivity of 0.95 and absorbs 95 % of the infrared radiation incident on it. At dusk the surface is at 279 K.
On a clear, dry night the infrared intensity incident on the ground from the atmosphere is 250 W m−2. On a cloudy night a layer of low cloud covers the sky; its base is at 276 K and radiates as a black body.
Diagram NOT accurately drawn
(a)
During the day the same cloud would reduce the solar intensity absorbed by the ground. Outline why.
(1)
(b)
(i)
Calculate the power per unit area radiated by the ground at 279 K.
(1)
(ii)
Show that, at dusk on the clear night, the net rate of energy loss by radiation from the ground is about 90 W m−2.
(1)
(iii)
Calculate the net rate of energy loss by radiation from the ground at dusk on the cloudy night.
(2)
(c)
Draw arrows on the diagram to show what happens to the infrared radiation emitted by the ground on the cloudy night.
(2)
(d)
On the clear night, energy is lost from a surface layer of soil 4.0 cm deep. The density of the soil is 1600 kg m−3 and its specific heat capacity is 900 J kg−1 K−1. Estimate the time taken for this layer to cool by 5.0 K, assuming the net rate of loss stays constant.
(2)
(e)
Frost is more likely on a clear night when the air is dry than on a clear night when the air is humid. Explain why.
(3)
Show mark scheme
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Notes
Part (a)
cloud has a high albedo: it scatters a large fraction of the solar radiation back to space, so less reaches the ground
✓ 1
OWTTE.
Part (b)(i)
0.95 × 5.67 × 10−8 × 279⁴ = 326 W m−2
✓ 1
Accept 330 W m−2.
Part (b)(ii)
326 − 0.95 × 250 = 326 − 237.5 = 88.9 W m−2
✓ 1
Must see full substitution OR answer to at least 3 s.f. The factor 0.95 for absorption must be seen.
Part (b)(iii)
intensity from cloud = 5.67 × 10−8 × 276⁴ = 329 W m−2
✓ 1
net loss = 326 − 0.95 × 329 = 14 W m−2
✓ 1
Accept 13–15 W m−2. Award [1 max] for −2.6 W m−2 «0.95 omitted». Award [2] for CNA.
Part (c)
arrow(s) upwards from the ground to the cloud, ending in/at the cloud «absorbed»
✓ 1
arrows from the cloud in several directions, including at least one downwards reaching the ground «re-emission in all directions»
✓ 1
Do not accept arrows that bounce off the cloud base as if reflected.
Part (d)
energy to be removed per square metre = 1600 × 0.040 × 900 × 5.0 = 2.9 × 105 J
✓ 1
t = 2.9 × 105/89 = 3.2 × 103 s «≈ 0.9 h»
✓ 1
Allow ECF from (b)(ii). Accept 3200–3300 s; accept 3.2 × 103 s using 90 W m−2.
Part (e)
water vapour is a greenhouse gas: it absorbs infrared radiation emitted by the ground
✓ 1
and re-emits it in all directions, so part of it returns to the ground «the downward infrared intensity is greater in humid air»
✓ 1
so the net rate of energy loss from the ground is smaller and its temperature falls more slowly, making it less likely to fall below 273 K
✓ 1
OWTTE.
Answers: (b)(i) 326 W m−2 · (b)(iii) 14 W m−2 · (d) 3.2 × 103 s (the remaining parts are explanations — see the table above)
Humans use energy at a total rate of about 1.9 × 1013 W, and 80 % of this energy is obtained by burning fossil fuels. Eventually all of the energy used becomes thermal energy in the environment. On average, 1.0 kg of carbon dioxide is released for every 14 MJ of energy released by burning fossil fuels.
Radius of the Earth = 6.37 × 106 m; 1 year = 3.15 × 107 s; solar constant S = 1360 W m−2; albedo of the Earth = 0.30.
(a)
(i)
Calculate the mean intensity, over the Earth’s surface, of the thermal energy released by human energy use.
(2)
(ii)
Calculate the ratio of your answer to (a)(i) to the mean intensity of the solar radiation absorbed by the Earth.
(1)
(b)
Show that the mass of carbon dioxide released each year by burning fossil fuels is about 3.4 × 1013 kg.
(1)
(c)
The carbon dioxide added to the atmosphere by human activity has reduced the intensity of the infrared radiation leaving the top of the atmosphere by about 2.2 W m−2, before the surface has warmed in response.
(i)
Explain how the added carbon dioxide reduces the infrared intensity leaving the top of the atmosphere.
(2)
(ii)
Determine whether the direct release of thermal energy or the added carbon dioxide has the larger effect on the energy balance of the Earth.
(2)
(d)
A student claims that replacing fossil-fuel power stations by nuclear power stations would not slow global warming, because a nuclear power station releases as much thermal energy to the environment as a fossil-fuel power station of the same output. Evaluate this claim.
(3)
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Notes
Part (a)(i)
area = 4πR² = 5.10 × 1014 m2
✓ 1
Do not accept πR².
1.9 × 1013/5.10 × 1014 = 0.037 W m−2
✓ 1
Award [2] for CNA.
Part (a)(ii)
absorbed = 0.70 × 1360/4 = 238 W m−2; ratio = 1.6 × 10−4
✓ 1
Allow ECF from (a)(i). Accept the ratio inverted «6.4 × 103» if clearly stated.
Must see full substitution OR answer to at least 3 s.f.
Part (c)(i)
CO₂ molecules absorb infrared radiation emitted by the surface «photon energies matching differences between molecular energy levels»
✓ 1
and re-emit it in all directions, so part returns to the surface and less escapes to space
✓ 1
Do not accept ‘reflects’ or ‘traps heat’.
Part (c)(ii)
2.2/0.037 ≈ 59
✓ 1
Allow ECF from (a)(i).
the added CO₂ has the larger effect, about 60 times greater than the direct heating
✓ 1
MP2 needs a comparison based on a calculation and a consistent conclusion.
Part (d)
the direct heating is very small «about 0.04 W m−2» compared with the effect of CO₂ «2.2 W m−2», so equal waste heat makes little difference
✓ 1
a nuclear power station releases no CO₂ when it operates, so it does not add to the enhanced greenhouse effect
✓ 1
CO₂ remains in the atmosphere for a very long time, so its effect accumulates year after year, whereas the waste heat stops when generation stops; the claim is therefore not correct
✓ 1
Award [3] only for a conclusion supported by at least two arguments. OWTTE.
Answers: (a)(i) 0.037 W m−2 · (a)(ii) 1.6 × 10−4 · (b) 3.42 × 1013 kg · (c)(ii) ratio ≈ 59 (the remaining parts are explanations — see the table above)
94B-2-69Equilibrium temperaturePaper 2Medium10 marks
Models of the Sun’s evolution suggest that 3.8 billion years ago its luminosity was 75 % of its present value. Rocks formed at that time show that liquid water then covered much of the Earth.
A simple energy-balance model gives the mean surface temperature T of the Earth from (1 − α)S/4 = eσT⁴, where S is the solar constant, α the albedo and e the effective emissivity of the Earth–atmosphere system. Today S = 1360 W m−2, α = 0.30 and e = 0.61, which give T = 288 K.
(a)
(i)
Calculate the solar constant 3.8 billion years ago.
(1)
(ii)
Show that, if α and e were the same as today, the model gives a mean surface temperature at that time of about 270 K.
(1)
(b)
Outline why this result is difficult to reconcile with the evidence from the rocks.
(1)
(c)
Determine the effective emissivity that would have given a mean surface temperature of 288 K, with α = 0.30.
(2)
(d)
Life on the early Earth had not yet produced an atmosphere containing oxygen. Suggest one natural source of carbon dioxide and one natural source of methane in the early atmosphere.
(2)
(e)
Another suggestion is that the early Earth had little land and few clouds, so that its albedo was lower. Determine whether a lower albedo alone, with e = 0.61, could give a mean surface temperature of 288 K.
(3)
Show mark scheme
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Notes
Part (a)(i)
0.75 × 1360 = 1020 W m−2
✓ 1
Part (a)(ii)
T = (0.70 × 1020/4/(0.61 × 5.67 × 10−8))1/4 = 268.0 K
✓ 1
Must see full substitution OR answer to at least 3 s.f. ALTERNATIVE: 288 × 0.751/4 = 268.0 K.
Part (b)
the temperature is below 273 K, the melting point of ice, so the oceans would have frozen
this is about the albedo of open ocean with no cloud or ice at all, which is not realistic; a lower albedo alone cannot explain it «extra greenhouse gases are needed»
✓ 1
MP3 needs a conclusion consistent with the albedo found.
Answers: (a)(i) 1020 W m−2 · (a)(ii) 268 K · (c) 0.46 · (e) α = 0.07 (the remaining parts are explanations — see the table above)
96IBP2S-011-03-14Equilibrium temperaturePaper 2Medium8 marks
One geoengineering proposal is to place a large sunshade in space between the Sun and Earth. The shade would block a fixed fraction of the sunlight that would otherwise reach Earth. Earth's radius is 6.4 × 106 m and its effective radiating temperature is 255 K.
(a)
Outline why blocking 1.0% of Earth's cross-sectional area reduces the mean intensity S/4 incident on Earth by 1.0%.
(2)
(b)
Calculate the area that the shade must block to reduce the incident power by 1.0%.
(2)
(c)
Assuming Earth's albedo is unchanged and Earth remains in radiative equilibrium, determine the new effective radiating temperature of Earth when the incident power is reduced by 1.0%.
(2)
(d)
Explain why the actual change in Earth's mean surface temperature might differ from the change in effective radiating temperature calculated in the previous part.
(2)
Show mark scheme
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Notes
Part (a)
The incident power is proportional to the unshaded cross-sectional (disc) area, πR2
✓ 1
This power is still spread over the same surface area, 4πR2, so the mean intensity falls in the same proportion
✓ 1
Part (b)
A = 0.010 × π × (6.4 × 106)2
✓ 1
A ≈ 1.3 × 1012 m2
✓ 1
Part (c)
Emitted power = absorbed power and emitted power ∝ T4, so T = 255 × 0.991/4
✓ 1
T ≈ 254.4 K, a decrease of about 0.6 K
✓ 1
Part (d)
One valid reason, e.g. feedback mechanisms: cooling increases ice and snow cover, raising the albedo, or reduces the water vapour content of the atmosphere
✓ 1
A second, different valid reason, e.g. the surface temperature (≈ 288 K) depends on the greenhouse effect, which the simple model ignores; or greenhouse gas concentrations may change at the same time
✓ 1
Common error: A 1% drop in power does not give a 1% drop in temperature. Because T ∝ P¹/⁴, the temperature falls by only about 0.25%.
97IBP2S-011-03-15Equilibrium temperaturePaper 2Medium8 marks
An orbiter is sent to study Venus, which orbits the Sun at 0.72 AU. The albedo of Venus is 0.76. The solar constant at Earth is 1360 W m−2. Treat Venus as a black-body emitter of uniform temperature.
(a)
Calculate the intensity of solar radiation at the distance of Venus.
(2)
(b)
Determine the mean intensity absorbed per unit area, averaged over the whole surface of Venus.
(2)
(c)
Calculate the equilibrium temperature of Venus predicted by this model.
(2)
(d)
The measured surface temperature of Venus is about 740 K. Explain the difference between this value and the temperature predicted by the black-body model.
(2)
Show mark scheme
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Mark
Notes
Part (a)
I = 1360/0.722
✓ 1
I ≈ 2.6 × 103 W m−2 (2620 W m−2)
✓ 1
Part (b)
Uses I/4 (intercepting disc area πR2 divided by surface area 4πR2) to obtain about 656 W m−2
✓ 1
Multiplies by (1 − 0.76) to obtain about 157 W m−2
✓ 1
Allow error carried forward from the intensity calculated in the previous part.
—
Part (c)
Uses σT4 = 157, so T4 = 157/(5.67 × 10−8)
✓ 1
T ≈ 230 K (accept 229–230 K)
✓ 1
Allow error carried forward from the absorbed intensity found in the previous part.
—
Part (d)
Venus has a very dense, CO2-rich atmosphere that absorbs the infrared radiation emitted by the surface
✓ 1
The atmosphere re-radiates part of this energy back to the surface (an extreme greenhouse effect), so the surface must be much hotter to emit enough power for energy balance
✓ 1
Exam tip: Venus is closer to the Sun but absorbs less per square metre than Earth, because its very high albedo outweighs the larger intensity. Check albedo before drawing conclusions about distance.
98IBP2S-011-01-12Energy balance of a planetPaper 2Medium8 marks
A space telescope collects the following data for a rocky exoplanet and its star.
• Luminosity of star: 2.4×1026 W • Orbital radius of planet: 1.1×1011 m • Albedo of planet: 0.25 • Mean surface temperature (inferred from infrared spectrum): 310 K
σ = 5.67×10−8 W m−2 K−4
(a)
Calculate the intensity of the star's radiation at the planet.
(2)
(b)
Show that the equilibrium temperature of the planet would be about 270 K if it radiated as a black body.
(2)
(c)
Determine the effective emissivity of the planet that is consistent with the inferred surface temperature.
(2)
(d)
Outline why an atmosphere containing greenhouse gases gives the planet an effective emissivity less than 1.
T = (296 / 5.67×10−8)1/4 = 269 K, which rounds to 270 K
✓ 1
Allow error carried forward from the intensity calculated in the first part.
—
Part (c)
Uses e = (1 − α)I / (4σT4) with T = 310 K, or e = (Tblack body/310)4
✓ 1
e = 0.57 (accept any value from 0.56 to 0.58, depending on the rounding of earlier values)
✓ 1
Allow error carried forward from the values found in the first two parts.
—
Part (d)
Greenhouse gases absorb some of the infrared radiation emitted by the surface
✓ 1
They re-radiate part of it back towards the surface, so less power escapes to space than a black body at the surface temperature would emit
✓ 1
Exam tip: In a 'show that' part, give the unrounded value (here 269 K) and then state that it rounds to the given value. You need a full working line to earn the marks.
Engineers designing a lunar habitat test a flat roof panel. The panel faces the Sun directly. Its underside is perfectly insulated, so it radiates only from its upper surface. The Moon has no atmosphere, and the solar intensity at the panel is 1.36×103 W m−2.
The panel is first coated in white paint, which has albedo 0.70 and emissivity 0.90. The albedo refers to the incoming solar radiation. The emissivity refers to the infrared radiation the panel emits. σ = 5.67×10−8 W m−2 K−4
(a)
Explain why the panel's temperature eventually becomes constant.
(2)
(b)
Calculate the equilibrium temperature of the white panel.
(3)
(c)
The white paint is replaced by black paint with albedo 0.05 and emissivity 0.95. Determine the new equilibrium temperature.
(2)
(d)
Suggest why white paint is preferred for the habitat roof.
(1)
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Notes
Part (a)
At first, the power absorbed is greater than the power emitted, so the temperature rises
✓ 1
Emitted power increases as T4 until it equals the absorbed power; the net power is then zero and T stays constant
✓ 1
Part (b)
Power absorbed per m2 = (1 − 0.70) × 1.36×103 = 408 W m−2
✓ 1
Set this equal to 0.90σT4, emitted from one face only
White paint reflects most of the solar radiation but still emits strongly in the infrared, so the roof stays cooler (about 299 K compared with 394 K for black paint)
✓ 1
Common error: The panel radiates from one face only, so there is no factor of 4 or 2 here. Check which surfaces absorb and which emit before writing the balance.
100IBP2S-011-01-16Equilibrium temperaturePaper 2Medium7 marks
A Mars rover mission team models how global dust storms affect Mars. Mars orbits the Sun at 1.52 AU. At 1.00 AU, the solar intensity is 1361 W m−2. In this model, treat Mars as a black-body emitter (e = 1). Ignore any greenhouse effect of its thin atmosphere.
During a global dust storm, suspended dust reflects some incoming sunlight back into space. It also absorbs much of the remaining sunlight high in the atmosphere.
σ = 5.67×10−8 W m−2 K−4
(a)
Calculate the solar intensity at Mars.
(1)
(b)
Before a storm, the albedo of Mars is 0.25. Calculate the equilibrium temperature of Mars.
(2)
(c)
During the storm, the albedo of Mars rises to 0.35. Determine the change in the equilibrium temperature.
(2)
(d)
Outline why, during the storm, the Martian atmosphere warms while the surface cools.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = 1361/1.522 = 589 W m−2
✓ 1
Part (b)
σT4 = (1 − 0.25) × 589/4 = 110 W m−2
✓ 1
T = 210 K
✓ 1
Allow ECF from (a).
—
Part (c)
Tnew = 210 × (0.65/0.75)1/4 = 203 K, or recalculated directly from σT4 = 0.65 × 589/4
✓ 1
The temperature decreases by 7.4 K; accept 7 K
✓ 1
Allow ECF from (b).
—
Part (d)
Dust absorbs incoming solar radiation high in the atmosphere, so the atmosphere gains energy and warms
✓ 1
Less solar radiation reaches the surface because of reflection and absorption by the dust, so the surface absorbs less power than it emits and cools
✓ 1
Exam tip: For a change in temperature, use a ratio such as Tnew = Told(ratio)¹/⁴. This avoids recalculating from scratch and reduces rounding errors.
A patient recovering from surgery lies uncovered in a room. The patient's skin has temperature 34 °C and emissivity 0.98. The walls of the room are at 22 °C. The area of skin exposed to the room is 1.5 m2.
(a)
Show that the power emitted per unit area by the skin is approximately 490 W m−2.
(2)
(b)
Calculate the net rate of energy loss by radiation from the exposed skin. Assume that the fraction of the walls' radiation absorbed by the skin is equal to the skin's emissivity.
(2)
(c)
The patient is then wrapped in a 'space blanket'. Both surfaces of the blanket are metallised, so each surface has a very low emissivity and is a good reflector of infrared radiation. Explain why the blanket reduces the patient's radiative energy loss.
(2)
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Notes
Part (a)
T = 34 + 273 = 307 K, and I = eσT4 = 0.98 × 5.67×10−8 × 3074
Radiated power is proportional to emissivity (P = eσAT4), so the low-emissivity outer surface radiates very little power to the room
✓ 1
The inner surface reflects most of the infrared emitted by the skin back towards the body instead of absorbing it, so little energy reaches the blanket to be lost
✓ 1
Common error: Find the net loss using T₁⁴ − T₂⁴. It is wrong to use (T₁ − T₂)⁴, and both temperatures must be in kelvin.
Engineers test coatings for a satellite thermal blanket. Each coated sample plate is held at 353 K in a vacuum chamber. The power emitted per unit area by each sample is measured.
Coating — emitted intensity / W m−2 black anodised — 860 white ceramic paint — 810 bare aluminium — 44 polished gold — 26
(a)
Show that a black body at 353 K emits about 880 W m−2.
(2)
(b)
Determine the emissivity of bare aluminium.
(2)
(c)
The white ceramic paint looks white in sunlight, yet the data show it is an excellent emitter. Explain this observation.
(2)
(d)
In a later test, the black anodised sample emits 600 W m−2. Assuming its emissivity is the same as in the original test, determine its temperature.
(3)
Show mark scheme
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Notes
Part (a)
I = σT4 = 5.67×10−8 × 3534
✓ 1
I = 880.4 W m−2 ≈ 880 W m−2
✓ 1
Part (b)
e = measured intensity / black-body intensity = 44/880
✓ 1
e = 0.050
✓ 1
Part (c)
Emissivity/absorptivity depends on wavelength: the paint reflects visible light, so it looks white
✓ 1
At 353 K the emission is in the infrared, where the paint has high emissivity (810/880 ≈ 0.92)
✓ 1
Part (d)
e = 860/880 = 0.977
✓ 1
T4 = 600/(0.977 × 5.67×10−8) = 1.08×1010 K4
✓ 1
T ≈ 323 K
✓ 1
Allow the ratio method T = 353 × (600/860)1/4 for full marks; allow error carried forward from the black-body intensity in the first part.
—
Exam tip: If temperature is the unknown, rearrange first and take the fourth root last. A quick check: T = 353 × (600/860)¹/⁴ gives the same answer.
103IBP2S-011-00-15Energy balance of a planetPaper 2Medium6 marks
Astronomers model a rocky exoplanet that has an atmosphere. The mean surface temperature is 265 K. Measurements from space show that the planet emits 170 W m−2 averaged over its surface. Treat the surface as a black body.
(a)
Calculate the intensity emitted by the planet's surface.
(1)
(b)
Determine the effective emissivity of the planet–atmosphere system.
(1)
(c)
Explain, with reference to greenhouse gases, why the intensity emitted to space is less than the intensity emitted by the surface.
(2)
(d)
An increase in greenhouse gases reduces the effective emissivity by 0.028. The intensity emitted to space remains 170 W m−2. Determine the new mean surface temperature.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = σT4 = 5.67×10−8 × 2654 ≈ 280 W m−2
✓ 1
Part (b)
e = 170/280 ≈ 0.61 (0.608) ECF from (a)
✓ 1
Part (c)
Greenhouse gases absorb infrared emitted by the surface
✓ 1
They re-emit it in all directions, so part returns to the surface and less escapes to space
✓ 1
Part (d)
New e = 0.608 − 0.028 = 0.580, so T4 = 170/(0.580 × 5.67×10−8)
✓ 1
T ≈ 268 K
✓ 1
ECF from (b)
—
Exam tip: The power emitted to space stays fixed because it must balance the absorbed stellar power. A smaller emissivity therefore requires a higher surface temperature.
104IBP2S-011-02-12Equilibrium temperaturePaper 2Medium6 marks
A team planning a crewed mission is modelling the surface of Mars. The intensity of solar radiation at the orbit of Mars is 590 W m−2. The mean albedo of Mars is 0.25. Treat the Martian surface as a black body. Stefan–Boltzmann constant σ = 5.67 × 10−8 W m−2 K−4.
(a)
Show that the mean solar power absorbed per unit area of the Martian surface is about 110 W m−2.
(2)
(b)
Calculate the equilibrium surface temperature of Mars predicted by this model.
(2)
(c)
The measured mean surface temperature of Mars is only a few kelvin above the value in (b). On Earth, the measured value is about 33 K above the equivalent prediction. Explain this difference.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses the factor 1/4: area intercepted πR2 ÷ total surface area 4πR2.
The Martian atmosphere is very thin, with a low total amount of greenhouse gas above each square metre.
✓ 1
Little of the infrared emitted by the surface is absorbed and re-radiated back down, so the greenhouse effect on Mars is very small.
✓ 1
Exam tip: In a 'show that' part, give your answer to one more significant figure than the value quoted (here 111 W m⁻²). This shows you calculated it rather than copied it.
105IBP2S-011-02-13Energy balance of a planetPaper 2Medium6 marks
A climate model of a fictional Earth-like planet gives the mean energy flows below. All values are per square metre of the planet's surface.
Top of atmosphere Incoming solar radiation: 400 W m−2 Solar radiation reflected to space (atmosphere and surface combined): 120 W m−2 Solar radiation absorbed by the atmosphere: 70 W m−2
Surface Infrared radiation emitted by the surface: 460 W m−2 Energy lost by convection (sensible heat): 25 W m−2 Energy lost by evaporation (latent heat): 85 W m−2 Infrared back-radiation from the atmosphere absorbed by the surface: B
Stefan–Boltzmann constant σ = 5.67 × 10−8 W m−2 K−4.
All values in W m⁻² per square metre of the planet's surface
(a)
Calculate the albedo of the planet.
(1)
(b)
Determine the solar power per unit area absorbed by the surface.
(1)
(c)
The surface is in equilibrium. Determine B.
(2)
(d)
The emissivity of the surface is 0.95. Calculate the mean surface temperature.
Common error: In (c), include the convection and evaporation losses as well as the radiation. Leaving them out gives a back-radiation that is too small.
A hospital wants to reduce overheating in an operating theatre that sits directly under a flat roof of area 150 m2. At midday, the solar intensity on the roof is 900 W m−2. The existing grey roof has an albedo of 0.15.
(a)
Calculate the solar power absorbed by the existing roof at midday.
(1)
(b)
The engineers want the absorbed power to be no more than 3.0 × 104 W. Determine the minimum albedo of the new roof coating.
(2)
(c)
The coating chosen reflects visible light strongly but has a high emissivity in the infrared. Explain why this combination keeps the roof cool.
(2)
(d)
Outline one reason why covering many roofs in a city with such coatings could lower local air temperatures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
P = (1 − 0.15) × 900 × 150 = 1.1 × 105 W
✓ 1
Part (b)
1 − α = 3.0 × 104 ÷ (900 × 150) = 0.22
✓ 1
α = 0.78
✓ 1
Part (c)
High albedo for short-wavelength sunlight means little solar power is absorbed.
✓ 1
High emissivity at the long (infrared) wavelengths the roof emits means it radiates its thermal energy away efficiently.
✓ 1
Part (d)
The city's mean albedo increases, so less solar energy is absorbed and then transferred to the air by convection. This reduces the urban heat island effect.
✓ 1
Common error: Albedo and emissivity can both be high for the same surface because they apply at different wavelengths. They are not simply 1 minus each other.
107B-2-05Energy balance of a planetPaper 2Hard14 marks
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)
Show mark scheme
Marking point
Mark
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)
108B-2-24Equilibrium temperaturePaper 2Hard15 marks
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)
109B-2-43Equilibrium temperaturePaper 2Hard15 marks
In a simple energy-balance model, the Earth’s surface and atmosphere are treated as a single body at a uniform temperature T. The body absorbs a mean intensity (1 − α)S/4 and emits a mean intensity eσT4, where S = 1360 W m−2 is the solar constant, α is the albedo and e is the effective emissivity of the Earth–atmosphere system.
At present, α = 0.30 and e = 0.610, which give T = 288 K.
(a)
Outline why an increase in the concentration of carbon dioxide in the atmosphere decreases the effective emissivity e.
(2)
(b)
An increase in the concentration of carbon dioxide reduces e to 0.600. Calculate the new equilibrium temperature predicted by the model.
(2)
(c)
It has been proposed that the temperature could be returned to 288 K by spraying reflective particles into the upper atmosphere, which would increase the albedo of the Earth.
(i)
Determine the albedo needed for the temperature to be 288 K when e = 0.600.
(3)
(ii)
Calculate the additional solar power that the whole Earth must reflect to produce this change in albedo. Radius of the Earth = 6.37 × 106 m.
(2)
(d)
Discuss whether increasing the albedo in this way would be an adequate response to the enhanced greenhouse effect.
(4)
(e)
Suggest why the actual rise in temperature caused by the increase in carbon dioxide is likely to be greater than your answer to (b).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
more of the infrared radiation emitted by the surface is absorbed by carbon dioxide molecules
✓ 1
these re-emit radiation in all directions, some back towards the surface, so less power escapes to space for the same surface temperature «emitted intensity is a smaller fraction of σT4»
increase in albedo Δα = 0.3116 − 0.30 = 0.0116 and the power intercepted is SπR2 = 1.73 × 1017 W
✓ 1
Allow ECF from (c)(i).
additional power reflected = 0.0116 × 1.73 × 1017 = 2.0 × 1015 W
✓ 1
Award [2] for CNA. Accept 1.9–2.1 × 1015 W «Δα = 0.012 gives 2.1 × 1015 W». Award [1] max for using 4πR2 (gives 8.1 × 1015 W).
Part (d)
for: only a small change in albedo «about 0.01» is needed and the effect on the global mean temperature would be rapid
✓ 1
Award [1] for each relevant point, up to [4].
against: the particles fall out of the atmosphere within a year or two, so they must be replaced continually; if spraying stopped, rapid warming would follow
✓ 1
against: the carbon dioxide concentration is not reduced, so its other effects «e.g. ocean acidification» continue and the cause «burning fossil fuels» is not removed
✓ 1
against: the cooling would not match the warming everywhere — less sunlight acts only by day and mostly at low latitudes, whereas the greenhouse warming acts day and night at all latitudes / less sunlight reaches crops and solar panels / a single-temperature model cannot predict regional effects
✓ 1
Award [3] max if all points are on one side of the argument.
Part (e)
a warmer atmosphere holds more water vapour, which is a greenhouse gas, so e falls further
✓ 1
melting ice and snow reduces the albedo, so more radiation is absorbed «positive feedbacks»
✓ 1
Accept other valid feedbacks, e.g. release of methane from thawing permafrost.
Answers: (b) 289.2 K · (c)(i) α = 0.312 · (c)(ii) 2.0 × 1015 W (the remaining parts are explanations — see the table above)
A daytime radiative-cooling panel lies horizontally on the flat roof of a building on a clear, sunny day. The upper surface of the panel has an albedo of 0.95 for solar radiation and an emissivity of 0.90 for infrared radiation. Its lower surface is insulated from the roof.
At midday the solar intensity incident on the panel is 900 W m−2 and the air temperature is 303 K. The panel also absorbs infrared radiation of intensity 290 W m−2 emitted downwards by the atmosphere.
Diagram not to scale
(a)
(i)
Calculate the intensity of the solar radiation absorbed by the panel.
(1)
(ii)
Calculate the power per unit area radiated by the panel when its temperature is 303 K.
(1)
(b)
Determine whether the panel, when it is at the air temperature of 303 K, gains or loses energy overall by radiation.
(2)
(c)
Show that, if radiation were the only way in which energy could reach or leave the panel, its steady temperature would be about 285 K.
(1)
(d)
In practice the steady temperature of the panel is 296 K.
(i)
Explain why the steady temperature is greater than 285 K.
(2)
(ii)
Calculate the rate per unit area at which energy reaches the panel from the air when the panel is at 296 K.
(1)
(e)
The surface of the panel is designed to emit strongly at wavelengths between 8 μm and 13 μm, a range in which the clear atmosphere absorbs very little infrared radiation.
(i)
Show that a black body at 296 K emits most strongly at a wavelength within this range.
(1)
(ii)
Explain, with reference to the greenhouse gases in the atmosphere, why emitting strongly at these wavelengths helps the panel to cool.
(3)
(f)
Suggest why the panel cools less effectively on a humid or cloudy day.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
(1 − 0.95) × 900 = 45 W m−2
✓ 1
Part (a)(ii)
0.90 × 5.67 × 10−8 × 303⁴ = 430 W m−2
✓ 1
Accept 4.3 × 102 W m−2 to 2 or 3 s.f.
Part (b)
energy absorbed per unit area = 45 + 290 = 335 W m−2
✓ 1
Allow ECF from (a).
335 < 430, so the panel loses energy overall «at about 95 W m−2» and cools below the air temperature
✓ 1
MP2 needs a comparison and a consistent conclusion.
Part (c)
eσT⁴ = 335: T = (335/(0.90 × 5.67 × 10−8))1/4 = 284.6 K
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (d)(i)
when the panel is colder than the air, energy is transferred to it from the air «by conduction and convection»
✓ 1
so the panel must radiate more, and be at a higher temperature, for the energy it loses to equal the energy it gains
✓ 1
OWTTE.
Part (d)(ii)
0.90 × 5.67 × 10−8 × 296⁴ − 335 = 57 W m−2
✓ 1
Accept 55–59 W m−2. Allow ECF from (b).
Part (e)(i)
λmax = 2.9 × 10−3/296 = 9.8 × 10−6 m «9.8 μm», which is between 8 μm and 13 μm
✓ 1
Must see the substitution. Accept 285 K or 303 K used «10.2 or 9.6 μm».
Part (e)(ii)
most of the panel’s radiation is emitted in the 8–13 μm range «near the peak»
✓ 1
greenhouse-gas molecules «CO₂, H₂O» have few energy-level differences matching photons of these wavelengths, so they absorb little of this radiation
✓ 1
Accept a resonance-model argument: no natural frequencies in this range.
so this radiation passes through the atmosphere to space; at other wavelengths the radiation is absorbed and re-emitted in all directions, partly back to the panel, so it removes less energy
✓ 1
OWTTE. Do not accept ‘reflected back’.
Part (f)
water vapour / cloud droplets absorb infrared in the 8–13 μm range and re-emit part of it back down, so the panel absorbs more and loses less energy
✓ 1
OWTTE.
Answers: (a)(i) 45 W m−2 · (a)(ii) 430 W m−2 · (b) net loss 95 W m−2 · (c) 285 K · (d)(ii) 57 W m−2 · (e)(i) 9.8 μm (the remaining parts are explanations — see the table above)
In a simple climate model the Earth is a single body at a uniform mean temperature T. It absorbs a mean solar intensity (1 − α)S/4, where S = 1360 W m−2, and emits a mean intensity eσT⁴ with an effective emissivity e = 0.61. The albedo α depends on T: below 250 K the Earth is covered by ice and α = 0.62; above 280 K there is little ice and α = 0.30; between these temperatures α changes linearly.
The graph shows how the absorbed intensity varies with T. The table gives the emitted intensity at some temperatures.
T / K
emitted intensity eσT⁴ / W m−2
220
81
240
260
158
280
300
280
Graph drawn to scale
(a)
Outline why the albedo is greater when T is below 250 K.
(1)
(b)
(i)
Show that the absorbed intensity when T is above 280 K is about 240 W m−2.
(1)
(ii)
Complete the table.
(2)
(c)
Draw, on the graph, a curve to show how the emitted intensity varies with T.
(2)
(d)
The Earth is in equilibrium when the absorbed and emitted intensities are equal. Use the graph to state the three temperatures at which this model is in equilibrium.
(2)
(e)
Explain why the equilibrium at the highest of these temperatures is stable.
(2)
(f)
When the Earth is covered in ice, carbon dioxide released by volcanoes is no longer removed from the atmosphere, so its concentration rises slowly. Explain how this could end the ice-covered state.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ice and snow cover a larger area, and ice and snow scatter a much larger fraction of the incident solar radiation than ocean or land
✓ 1
OWTTE.
Part (b)(i)
(1 − 0.30) × 1360/4 = 238 W m−2
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
240 K: 115 W m−2
✓ 1
Accept 114.8.
280 K: 213 W m−2
✓ 1
Accept 212.6.
Part (c)
all five values plotted correctly «within half a small square»
✓ 1
Allow ECF from (b)(ii).
smooth curve through the points, increasing ever more steeply «not straight segments»
✓ 1
Part (d)
about 247 K and about 288 K
✓ 1
Accept 245–249 K and 286–290 K. Allow ECF from the candidate’s curve.
about 254 K
✓ 1
Accept 252–256 K.
Part (e)
if T rises slightly above this value, the emitted intensity becomes greater than the absorbed intensity, so the Earth loses energy and cools back
✓ 1
if T falls slightly, the absorbed intensity exceeds the emitted intensity, so the Earth gains energy and warms back «the change is reversed»
✓ 1
Award [1 max] for ‘the curves cross’ without the energy argument.
Part (f)
more CO₂ absorbs more of the infrared emitted by the surface and re-emits part of it back, so the effective emissivity e decreases «the emitted curve on the graph is lowered»
✓ 1
eventually the emitted intensity is less than the absorbed intensity at all temperatures below the warm state «the cold equilibrium disappears», so the Earth gains energy and T rises
✓ 1
as T rises the ice melts and the albedo falls, so more radiation is absorbed and the Earth warms on to the ice-free state «positive feedback»
✓ 1
OWTTE.
Answers: (b)(i) 238 W m−2 · (b)(ii) 115 W m−2, 213 W m−2 · (d) 247 K, 254 K, 288 K (the remaining parts are explanations — see the table above)
The atmosphere of Mars is 95 % carbon dioxide by mass, but the mean pressure at the surface is only 610 Pa. The gravitational field strength at the surface of Mars is 3.71 N kg−1. At the Earth’s surface the pressure is 1.01 × 105 Pa and carbon dioxide makes up 0.064 % of the mass of the atmosphere.
Without an atmosphere, the mean surface temperatures of Mars and the Earth would be 210 K and 255 K. The measured mean surface temperatures are 215 K and 288 K.
Axes for (c)(i)
(a)
(i)
Explain why the mass of the atmosphere above each square metre of a planet’s surface is equal to p/g, where p is the pressure at the surface and g is the gravitational field strength.
(2)
(ii)
Show that the mass of carbon dioxide above each square metre of the surface of Mars is about 160 kg.
(1)
(iii)
Calculate the ratio (mass of carbon dioxide above each square metre of Mars) / (mass of carbon dioxide above each square metre of the Earth).
(2)
(b)
A student claims that Mars must have a stronger greenhouse effect than the Earth. State one piece of evidence from the data that contradicts this claim.
(1)
(c)
Carbon dioxide molecules absorb infrared radiation strongly at a wavelength of 15 μm, which corresponds to a natural frequency of vibration of 2.0 × 1013 Hz. In the resonance model, collisions with other molecules act as damping on this vibration. The density of the gas at the surface of Mars is about 1 % of that at the surface of the Earth, so collisions are much less frequent.
(i)
Sketch, on the axes, graphs to show how the amplitude of the vibration varies with the driving frequency for a carbon dioxide molecule near the surface of the Earth (label it E) and near the surface of Mars (label it M).
(3)
(ii)
Hence explain why the carbon dioxide on Mars absorbs a smaller fraction of the infrared radiation emitted by the surface than might be expected from its mass.
(2)
(d)
Suggest one other reason why the greenhouse effect on Mars is much weaker than on the Earth.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the weight of the column of air above an area A is supported by the surface, so p = F/A = weight/A
✓ 1
for A = 1 m², weight = p and mass = weight/g = p/g
✓ 1
OWTTE.
Part (a)(ii)
610/3.71 × 0.95 = 156 kg
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(iii)
Earth: 1.01 × 105/9.81 × 6.4 × 10−4 = 6.59 kg
✓ 1
ratio = 156/6.59 = 24
✓ 1
Accept 23–25. Allow ECF from (a)(ii). Award [2] for CNA.
Part (b)
the atmosphere raises the temperature of Mars by only about 5 K, but that of the Earth by about 33 K
✓ 1
Both values needed «or a clear comparison».
Part (c)(i)
both curves have a single peak at about 2.0 × 1013 Hz
✓ 1
Accept the peak of E at a very slightly lower frequency.
M has a higher peak than E
✓ 1
M is much narrower than E «E is broader, with larger amplitude far from the peak»
✓ 1
Labels E and M needed for MP2 and MP3.
Part (c)(ii)
with little damping, CO₂ on Mars absorbs strongly only over a very narrow range of frequencies «wavelengths» close to the natural frequency
✓ 1
radiation at this narrow range is already almost fully absorbed, so the extra CO₂ makes little difference, while the surface’s radiation at nearby frequencies escapes to space «on the Earth the broader curve absorbs over a wider range»
✓ 1
OWTTE.
Part (d)
there is almost no water vapour «the main greenhouse gas on the Earth» and there are few clouds, so infrared at wavelengths away from the CO₂ band is hardly absorbed
✓ 1
Accept other sensible reasons, e.g. no other greenhouse gases such as methane in significant amounts.
Answers: (a)(ii) 156 kg · (a)(iii) 24 (the remaining parts are explanations — see the table above)
113IBP2S-011-03-16Solar constant & S/4Paper 2Hard9 marks
Engineers designing a lunar base need to estimate surface temperatures on the Moon. The Moon has no atmosphere, an albedo of 0.12, and rotates once every 29.5 days relative to the Sun. Take the solar constant at the Moon to be 1360 W m−2. Treat the lunar surface as a black-body emitter.
(a)
Explain why the S/4 averaging model is a poor guide to daytime temperatures on the Moon.
(2)
(b)
Calculate the equilibrium temperature at a point where the Sun is directly overhead, assuming the energy absorbed there is all re-radiated locally.
(3)
(c)
Show that the S/4 model predicts a temperature smaller than your answer to the previous part by a factor of √2.
(2)
(d)
The proposed base is at a site where, at local noon, the Sun's rays make an angle of 60° with the normal to the ground. Determine the equilibrium temperature of the ground there at local noon.
(2)
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Notes
Part (a)
The Moon rotates slowly, so each region is lit continuously for about two weeks rather than receiving a time-averaged intensity
✓ 1
There is no atmosphere or ocean to transport energy, so each region reaches equilibrium with its local intensity
The S/4 model gives σT′4 = (1 − α)S/4, so T′ = T × (1/4)1/4
✓ 1
(1/4)1/4 = 1/√2, so T′ = 381/√2 ≈ 270 K
✓ 1
Allow follow-through from the temperature found for the overhead Sun.
—
Part (d)
Intensity per unit ground area = 1360 cos 60° = 680 W m−2, so absorbed = 0.88 × 680 ≈ 598 W m−2
✓ 1
T = (598/5.67 × 10−8)1/4 ≈ 321 K (accept 320–321 K)
✓ 1
Alternative method: T = 381 × (cos 60°)1/4 ≈ 320 K earns full credit.
—
Common error: When the rays are not perpendicular to a surface, the intensity on it is reduced by cos θ, where θ is measured from the normal. Using sin θ by mistake gives the wrong value here.
114IBP2S-011-01-14Equilibrium temperaturePaper 2Hard9 marks
Astronomers searching for habitable planets study a red dwarf star of luminosity L = 5.8×1025 W. For an Earth-like planet of radius R orbiting this star, assume: • albedo α = 0.30 • effective emissivity e = 0.62 • a uniform surface temperature T • a surface atmospheric pressure similar to Earth's, so water freezes at 273 K and boils at 373 K
σ = 5.67×10−8 W m−2 K−4
(a)
Show that the orbital radius d at which the planet has surface temperature T is given by \(d = \sqrt{\frac{(1-\alpha)L}{16\pi e\sigma T^4}}\).
(3)
(b)
Calculate the orbital radius at which the surface temperature is 273 K.
(2)
(c)
Determine the orbital radius at which the surface temperature is 373 K. Hence state the range of orbital radii for which liquid water could exist on the surface.
(2)
(d)
Explain why the inner limit of this range may, in reality, lie further from the star than calculated.
(2)
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Notes
Part (a)
Intensity at the planet = L/(4πd2)
✓ 1
Power absorbed = (1 − α)(L/4πd2)πR2, set equal to power emitted = eσ4πR2T4
115IBP2S-011-00-16Greenhouse gasesPaper 2Hard8 marks
An indoor air-quality monitor measures CO2 concentration. It passes infrared radiation through a sample of air and measures the transmitted intensity at 4.26 μm, a wavelength strongly absorbed by CO2. The infrared source is a small heated filament with emissivity 0.70 and surface area 2.0 mm2. Wien's displacement constant = 2.90×10−3 m K.
(a)
Calculate the frequency of radiation of wavelength 4.26 μm.
(1)
(b)
Air contains far more nitrogen and oxygen than CO2. Outline why the monitor can still measure the CO2 concentration.
(2)
(c)
The filament must radiate 0.15 W. Determine the filament temperature required.
(3)
(d)
Determine the peak emission wavelength of the filament, and comment on whether the filament is a suitable source.
(2)
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Notes
Part (a)
f = c/λ = 3.00×108/4.26×10−6 = 7.04×1013 Hz
✓ 1
Part (b)
N2 and O2 do not absorb infrared at this wavelength. Their molecular vibrations produce no change in dipole moment, so they do not interact with the radiation; accept equivalent statements that they are not greenhouse gases or have no IR absorption here
✓ 1
The reduction in transmitted intensity is therefore due only to CO2, and a greater concentration gives greater absorption, so a lower transmitted intensity
λmax = 2.90×10−3/1172 ≈ 2.5×10−6 m; allow error carried forward from the temperature found in the previous part
✓ 1
The peak is shorter than 4.26 μm, but the black-body spectrum is broad, so the filament still emits a significant fraction of its power at 4.26 μm and is suitable. Accept a point that only a narrow band, e.g. selected by a filter, is used
✓ 1
Common error: Convert mm² to m² by multiplying by 10⁻⁶, not 10⁻³. Because of the fourth root, a wrong area makes the temperature wrong by a factor of about 5.6.
116IBP2S-011-02-14Energy balance of a planetPaper 2Hard7 marks
A space telescope observes a rocky exoplanet. The intensity of starlight at the planet's orbit is 1800 W m−2, and the planet's albedo is 0.35. In a simple model, the planet has an atmosphere with these properties: • it is transparent to starlight; • it absorbs a fraction f of the infrared power Ps emitted per unit area by the surface; • it re-emits all the power it absorbs, half upwards and half downwards. Treat the surface as a black body. σ = 5.67 × 10−8 W m−2 K−4.
The atmosphere absorbs a fraction f of Pₛ
(a)
Show that the mean stellar power absorbed per unit area of the planet's surface, I, is about 290 W m−2.
(2)
(b)
By considering the energy balance at the top of the atmosphere, show that \(P_s = \frac{I}{1 - f/2}\).
(2)
(c)
Spectroscopic data suggest that f = 0.80. Determine the surface temperature of the planet.
(2)
(d)
Explain, using the model, how an increase in f affects the surface temperature.
(1)
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Notes
Part (a)
Uses (1 − α) and the factor 1/4.
✓ 1
I = 0.65 × 1800 ÷ 4 = 292.5 W m−2 ≈ 290 W m−2
✓ 1
Part (b)
Power leaving to space = transmitted surface emission (1 − f)Ps + upward atmospheric emission fPs/2.
✓ 1
Setting this equal to I gives Ps(1 − f + f/2) = I, so Ps = I ÷ (1 − f/2).
✓ 1
Part (c)
Ps = 292.5 ÷ (1 − 0.40) = 488 W m−2
✓ 1
T = (488 ÷ 5.67 × 10−8)1/4 ≈ 305 K
✓ 1
Allow ECF from (a).
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Part (d)
A larger f makes (1 − f/2) smaller, so Ps = σT4 must be larger and the surface temperature rises.
✓ 1
Exam tip: Balancing energy at the top of the atmosphere is usually the quickest route. It avoids having to find the atmosphere's own temperature.
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