B.2 Greenhouse Effect Questions | IB Physics HL 2025

1B-1A-08

Resonance model of the greenhouse effect·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In a resonance model of the greenhouse effect, the bonds in a molecule of carbon dioxide are treated as oscillators. One mode of vibration of the molecule has a natural angular frequency of 1.26 × 10¹⁴ rad s⁻¹.

Which row gives the wavelength of the electromagnetic radiation that drives this vibration at resonance, and the direction in which the molecule re-emits the absorbed energy?

WavelengthDirection of re-emission

Show mark scheme

Marking point	Mark	Notes
Step 1Resonance occurs when the frequency of the radiation equals the natural frequency of the vibration: f = ω/2π = 1.26 × 10¹⁴/2π = 2.0 × 10¹³ Hz.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2λ = c/f = 3.00 × 10⁸/2.0 × 10¹³ = 1.5 × 10⁻⁵ m = 15 μm, in the infrared, close to the peak of the radiation emitted by the Earth's surface.	—
Step 3The excited molecule re-emits the energy in a random direction, so the radiation leaves the gas in all directions: part returns to the surface and warms it, part continues upwards.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: f = ω/2π gives 15 μm, and re-emission takes place in all directions.
BThe wavelength is right, but the molecules do not act as mirrors: the absorbed energy is re-emitted in all directions, and only part of it returns to the surface.
CThis uses λ = c/ω, forgetting that ω = 2πf: 3.00 × 10⁸/1.26 × 10¹⁴ = 2.4 μm.
DThis combines both errors: the angular frequency is used as if it were the frequency, and the re-emission is taken to be directed back to the surface only.

Syllabus understandingB.2 — the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; that the greenhouse effect can be explained in terms of both a resonance model and molecular energy levels; C.1 — T = 1/f = 2π/ω Command term: Determine

2B-1A-15

Albedo and the solar constant·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The solar constant at a planet with no atmosphere is S and the albedo of its surface is α.

What is the mean intensity of solar radiation absorbed by the planet, averaged over its whole surface?

Show mark scheme

Marking point	Mark	Notes
Step 1The planet intercepts radiation over its projected disc, area πr², but its total surface area is 4πr², so the mean incoming intensity is S/4.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2A fraction α is scattered, so a fraction (1 − α) is absorbed.	—
Step 3Mean absorbed intensity = (1 − α)S/4.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis uses the albedo as the fraction absorbed; albedo is the fraction scattered.
BThis divides by the area of a hemisphere (2πr²), as if only the day side counted in the average over the whole surface.
CThis is the intensity absorbed by a surface perpendicular to the rays; it ignores the projected-area (S/4) averaging.
DCorrect: projected area πr² shared over 4πr², with a fraction (1 − α) absorbed.

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system; the solar constant S; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4 Command term: Deduce

3B-1A-23

Surface–atmosphere energy balance·B.2 Greenhouse effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The surface of a planet absorbs a mean intensity of 160 W m⁻² of solar radiation and 240 W m⁻² of infrared radiation emitted downwards by its atmosphere. The surface also loses 40 W m⁻² to the atmosphere by convection and evaporation. The emissivity of the surface is 0.90. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴)

What is the equilibrium temperature of the surface?

Show mark scheme

Marking point	Mark	Notes
Step 1In equilibrium the power radiated per unit area by the surface equals the net intensity it receives: eσT⁴ = 160 + 240 − 40 = 360 W m⁻².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2T⁴ = 360/(0.90 × 5.67 × 10⁻⁸) = 7.05 × 10⁹ K⁴.	—
Step 3T = 290 K.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis ignores the 240 W m⁻² returned by the atmosphere: 0.90σT⁴ = 160 − 40 = 120 W m⁻² gives 220 K.
BThis ignores the emissivity, treating the surface as a black body: σT⁴ = 360 W m⁻² gives 282 K.
CCorrect: 0.90σT⁴ = 360 W m⁻².
DThis forgets the 40 W m⁻² lost by convection and evaporation: 0.90σT⁴ = 400 W m⁻² gives 298 K.

Syllabus understandingB.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature as given by emissivity = power radiated per unit area/σT⁴; the conservation of energy (energy balance problems including energy exchanged between the surface and the atmosphere of a body) Command term: Determine

4B-1A-37

Equilibrium temperature·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two planets orbit the same star. Planet Y is at distance d from the star and has an equilibrium surface temperature T. Planet X is at distance 2d and has the same albedo and the same emissivity as Y. Neither planet has an atmosphere.

What is the equilibrium surface temperature of X?

Show mark scheme

Marking point	Mark	Notes
Step 1The intensity of the starlight follows an inverse-square law: at 2d it is ¼ of that at d.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2In equilibrium the power absorbed per unit area equals the power emitted per unit area: (1 − albedo)S/4 = eσT⁴, so T⁴ ∝ S.	—
Step 3T ∝ S^1/4: T_X = T × (¼)^1/4 = T/√2.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis takes the temperature to be proportional to the intensity received, instead of to its fourth root.
BThis takes the temperature to be inversely proportional to the distance.
CCorrect: T⁴ ∝ 1/d², so T ∝ 1/√d.
DThis uses an inverse (not inverse-square) law for the intensity: (½)^1/4.

Syllabus understandingB.2 — the conservation of energy; emissivity and albedo; estimation of the equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity (Guidance) Command term: Deduce

5B-1A-38

Emissivity·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

Two metal panels on the radiator of a spacecraft are at the same temperature. Panel X is coated with a paint of emissivity 0.30 and emits 150 W m⁻². Panel Y is coated with a paint of emissivity 0.90. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

Which row gives the power emitted per unit area by panel Y and the temperature of the panels?

Power emitted per unit area by Y / W m⁻²Temperature / K

Show mark scheme

Marking point	Mark	Notes
Step 1Emissivity = (power radiated per unit area)/σT⁴, so at the same temperature the emission is proportional to the emissivity: 150 × 0.90/0.30 = 450 W m⁻².	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2For X: T⁴ = 150/(0.30 × 5.67 × 10⁻⁸) = 8.82 × 10⁹ K⁴, so T = 306 K.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThis inverts the ratio of emissivities: the higher-emissivity paint emits more, not less.
BThis assumes surfaces at the same temperature emit equally whatever their emissivity.
CThe temperature here comes from σT⁴ = 150 W m⁻², treating X as a black body and ignoring its emissivity.
DCorrect: Y emits three times as much as X, and both are at 306 K.

6B-1A-39

Albedo and latitude·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

At a high-latitude site, sunlight of intensity 950 W m⁻² (measured on a surface perpendicular to the rays) falls on level ground covered in snow. The rays make an angle of 65° with the vertical. The albedo of the snow is 0.80.

What is the power absorbed per unit area of the ground?

Show mark scheme

Marking point	Mark	Notes
Step 1The beam is spread over a larger area of level ground: the intensity on the ground is 950 cos 65° = 401 W m⁻².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The fraction absorbed is 1 − albedo = 0.20.	—
Step 3Absorbed: 0.20 × 401 = 80 W m⁻². Oblique rays and high-albedo snow are why high latitudes absorb so little.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: 950 × cos 65° × (1 − 0.80) = 80 W m⁻².
BThis uses sin 65° instead of cos 65°: 950 × 0.906 × 0.20 = 172 ≈ 170 W m⁻².
CThis ignores the angle of the rays: 950 × 0.20 = 190 W m⁻².
DThis uses the albedo as the fraction absorbed: 401 × 0.80 = 321 ≈ 320 W m⁻².

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; that the incoming radiative power is dependent on the projected surface along the direction of the path of the rays Command term: Determine

7B-1A-40

Equilibrium temperature·B.2 Greenhouse effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A thin, flat sunshade is in space at the same distance from the Sun as the Earth, with one face perpendicular to the Sun's rays. The intensity of sunlight there is 1.36 × 10³ W m⁻². The sunlit face has an albedo of 0.20. The sunshade conducts well, so both faces are at the same temperature, and both faces have an emissivity of 0.80. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

What is the equilibrium temperature of the sunshade?

Show mark scheme

Marking point	Mark	Notes
Step 1Power absorbed per unit area of the sunshade: (1 − 0.20) × 1360 = 1088 W m⁻² (a flat sheet facing the Sun intercepts with its full area, so there is no factor of 1/4).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Power emitted per unit area of the sunshade: two faces, each 0.80σT⁴, total 1.60σT⁴.	—
Step 3T⁴ = 1088/(1.60 × 5.67 × 10⁻⁸) = 1.20 × 10¹⁰ K⁴, so T = 331 K.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses the albedo as the fraction absorbed: 0.20 × 1360 = 272 W m⁻² gives 234 K.
BThis treats both faces as black bodies (emissivity 1): 2σT⁴ = 1088 W m⁻² gives 313 K.
CCorrect: 1088 = 1.60σT⁴.
DThis lets only the sunlit face radiate: 0.80σT⁴ = 1088 W m⁻² gives 394 K.

Syllabus understandingB.2 — the conservation of energy; emissivity; albedo; the estimation of the equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity and the solar constant Command term: Determine

8B-1A-64

Greenhouse gases·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The graph shows the fraction of radiation absorbed by a layer of gas X in the atmosphere, for wavelengths from 0.20 μm to 12 μm.

The surface temperature of the Sun is 5800 K and the mean surface temperature of the Earth is 288 K. Treat both as black bodies. Which row is correct?

Fraction of radiation absorbed by gas X against wavelength (drawn to scale)

X absorbs strongly at the peak wavelength of the Sun's radiationX absorbs strongly at the peak wavelength of the radiation from the Earth's surface

Show mark scheme

Marking point	Mark	Notes
Step 1Wien: λ_max(Sun) = 2.9 × 10⁻³/5800 = 5.0 × 10⁻⁷ m = 0.50 μm, in the visible, where X absorbs only 0.02.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2λ_max(Earth) = 2.9 × 10⁻³/288 = 1.01 × 10⁻⁵ m = 10.1 μm, inside the infrared band (9.3–10.5 μm), where X absorbs 0.80.	—
Step 3So X lets most sunlight through but absorbs strongly near the peak of the Earth's emission: no / yes.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis takes the ultraviolet band to contain the solar peak; 0.50 μm lies in the visible, beyond the end of the UV band at about 0.33 μm.
BThis uses λ_max ∝ T, which gives the hotter Sun the longer peak wavelength; Wien's law gives λ_max ∝ 1/T.
CThis uses the Earth's Celsius temperature: 2.9 × 10⁻³/15 = 1.9 × 10⁻⁴ m (190 μm), far beyond the graph, so the infrared band seems irrelevant.
DCorrect: the solar peak (0.50 μm) is where X is almost transparent; the Earth's peak (10 μm) lies inside the strongly absorbing infrared band, so X acts as a greenhouse gas.

Syllabus understandingB.2 — the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels; B.1 — the emission spectrum of a black body and Wien's displacement law as given by λ_maxT = 2.9 × 10⁻³ m K Command term: Deduce

9B-1A-65

Equilibrium temperature·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A moon with no atmosphere is in thermal equilibrium. When its only energy source is the solar radiation it absorbs, its mean surface temperature is T₀.

Tidal heating then generates thermal energy inside the moon at a rate equal to a fraction f of the solar power it absorbs. This energy also leaves as radiation from the surface. The albedo and the emissivity of the surface do not change.

What is the new mean surface temperature in equilibrium?

Show mark scheme

Marking point	Mark	Notes
Step 1Energy balance before: eσAT₀⁴ = P_abs, where P_abs is the solar power absorbed.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2After: emitted power = P_abs + fP_abs, so eσAT⁴ = (1 + f)P_abs.	—
Step 3Dividing: (T/T₀)⁴ = 1 + f, so T = T₀(1 + f)^1/4. (For f = 0.70 the temperature rises by 14 %.)	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis takes the emitted power to be proportional to T; the Stefan–Boltzmann law makes it proportional to T⁴.
BThis uses a square root, as if emitted power were proportional to T² (or by confusion with T ∝ d^−1/2 for distance from the Sun).
CThis takes the fourth root of the extra term only; the fourth root of (1 + f) is not 1 + f^1/4.
DCorrect: the total emitted power is (1 + f) times the original, and emitted power ∝ T⁴.

Syllabus understandingB.2 — the conservation of energy; emissivity; the estimation of the equilibrium temperature of a body using energy balance between incoming and outgoing radiation (Guidance); B.1 — the Stefan–Boltzmann law Command term: Deduce

10B-1A-86

Greenhouse gases from human activity·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

Some human activities add greenhouse gases to the atmosphere.

Which activities increase the concentration of a greenhouse gas in the atmosphere?

I. Rearing large herds of cattle
II. Clearing forests by burning the trees
III. Spreading nitrogen-based fertiliser on fields

Show mark scheme

Marking point	Mark	Notes
Step 1I: the digestion of food by cattle releases methane, CH₄.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: burning the trees releases carbon dioxide, CO₂, and removes plants that would have absorbed CO₂.	—
Step 3III: bacteria in soil convert part of the nitrogen fertiliser into nitrous oxide, N₂O. All three add a greenhouse gas.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis omits III: nitrous oxide is one of the main greenhouse gases, and agricultural fertiliser is its main human source.
BThis omits II: burning forests releases carbon dioxide just as burning fossil fuels does.
CThis omits I: the digestion of food by cattle releases methane, one of the main greenhouse gases, so cattle farming is a major human source.
DCorrect: cattle (CH₄), burning forests (CO₂) and fertiliser (N₂O) all add greenhouse gases.

11B-1A-87

Albedo of a region·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Sunlight of the same intensity falls on every part of a large region of ocean. 60 % of the area of the region is clear ocean of albedo 0.06. The remaining 40 % is covered by cloud of albedo 0.50.

What is the albedo of the whole region?

Show mark scheme

Marking point	Mark	Notes
Step 1Albedo = total scattered power/total incident power. Per unit incident intensity, the incident power is proportional to the area.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Scattered power ∝ 0.60 × 0.06 + 0.40 × 0.50 = 0.036 + 0.200 = 0.236.	—
Step 3Albedo of the region = 0.236/1.00 = 0.24.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: each albedo is weighted by the fraction of the area (and so of the incident power) it covers.
BThis is the simple mean (0.06 + 0.50)/2 = 0.28, which ignores that the clear ocean covers more of the area.
CThis swaps the area fractions: 0.40 × 0.06 + 0.60 × 0.50 = 0.32.
DThis is the fraction absorbed, 1 − 0.24 = 0.76, not the fraction scattered.

12B-1A-88

Effect of cloud cover·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A thick layer of low cloud stays over a region of land for a whole day and the following night.

Compared with a clear day and a clear night, what is the effect of the cloud on the temperature of the ground during the day and during the night?

DayNight

Show mark scheme

Marking point	Mark	Notes
Step 1Day: the cloud has a high albedo, so it scatters much of the sunlight back to space and less sunlight reaches the ground: the ground is cooler.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Night: there is no sunlight. The water droplets and water vapour of the cloud absorb the infrared radiation emitted by the ground and re-emit it in all directions, part of it back down to the ground.	—
Step 3So the ground loses energy more slowly at night and stays warmer: lower by day, higher at night.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis considers only the absorption of infrared by the cloud and ignores the sunlight it scatters back to space during the day.
BThis considers only the high albedo of the cloud and ignores the infrared that the cloud returns to the ground at night.
CCorrect: by day the cloud's albedo dominates; at night only its absorption and re-emission of infrared acts.
DThis reverses both effects: the cloud reduces the sunlight reaching the ground and reduces the net infrared loss.

Syllabus understandingB.2 — that Earth's albedo varies daily and is dependent on cloud formations and latitude; the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; that water vapour H₂O is one of the main greenhouse gases Command term: Deduce

13B-1A-89

Emissivity from a graph·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows how the power radiated per unit area I by a surface S varies with its kelvin temperature T.

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

What is the emissivity of S?

Power radiated per unit area I by surface S against kelvin temperature T (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
Step 1Read one point from the curve, e.g. I = 1.0 × 10³ W m⁻² at T = 400 K (or 320 W m⁻² at 300 K).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2A black body at 400 K radiates σT⁴ = 5.67 × 10⁻⁸ × 400⁴ = 1450 W m⁻².	—
Step 3Emissivity = 1016/1450 = 0.70.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is 1 − 0.70: it confuses emissivity with the fraction "missing" from the black-body value.
BCorrect: emissivity = (power radiated per unit area)/σT⁴ = 0.70 at every temperature on the curve.
CThis compares temperatures instead of powers: (1016/σ)^1/4/400 = 0.91.
DThis inverts the ratio: 1450/1016 = 1.4; an emissivity cannot exceed 1.

14B-1A-90

Resonance model·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

In the resonance model of the greenhouse effect, a molecule of a greenhouse gas absorbs infrared radiation strongly only at certain wavelengths.

Which property of the molecule determines these wavelengths?

Show mark scheme

Marking point	Mark	Notes
Step 1The bonds of the molecule behave like springs, so the molecule has natural frequencies of vibration.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Infrared radiation whose frequency equals one of these natural frequencies drives the vibration at resonance: the amplitude grows and energy is absorbed strongly.	—
Step 3So the absorbed wavelengths are λ = c/f_natural, fixed by the bonds (masses of the atoms and stiffness of the bonds).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThe mean speed of the molecules depends on the temperature and the mass, but the random motion of the molecule as a whole is not the oscillation that is driven at resonance.
BMolecules are about 10⁻¹⁰ m across, far smaller than infrared wavelengths of about 10⁻⁵ m; no size matching is involved.
CCorrect: absorption is strongest when the radiation frequency equals a natural frequency of vibration of the bonds.
DThe collision rate changes with temperature, but the absorbed wavelengths are fixed by the molecule's own natural frequencies.

Syllabus understandingB.2 — that the greenhouse effect can be explained in terms of both a resonance model and molecular energy levels; C.4 — natural frequency and resonance Command term: Identify

15B-1A-91

Atmospheric absorption bands·B.2 Greenhouse effect

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The graph shows the fraction of radiation transmitted vertically through the whole of the Earth's atmosphere, for wavelengths from 0.3 μm to 20 μm.

An instrument on a satellite measures the temperature of the ground, which is about 290 K, by detecting the radiation that the ground emits. Which wavelength should the instrument detect, and why?

Fraction of radiation transmitted vertically through the atmosphere against wavelength (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
Step 1Wien: the ground at 290 K emits most strongly near λ_max = 2.9 × 10⁻³/290 = 1.0 × 10⁻⁵ m = 10 μm.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From the graph, about 0.75 of the radiation between 8.2 μm and 12.6 μm is transmitted to space.	—
Step 3So 10 μm is both strongly emitted by the ground and able to reach the satellite.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AAt 0.5 μm the ground at 290 K emits almost nothing; radiation detected there is sunlight scattered by the ground, which does not measure its temperature.
BThe ground does emit at 6.5 μm, but the graph shows that almost none of this radiation (about 0.05) passes through the atmosphere.
CCorrect: 10 μm is near the peak of the ground's emission and lies in the band where the atmosphere is transparent.
DRadiation from the ground at 15 μm is absorbed by greenhouse gases and does not reach the satellite; strong absorption is the opposite of what is needed.

16B-1A-92

Outgoing infrared spectrum·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksExplain

A satellite measures the spectrum of the infrared radiation leaving the top of the Earth's atmosphere above a region where the surface temperature is 288 K. The solid line on the graph shows the result. The dashed lines are the spectra of black bodies at 288 K and at 220 K.

Which statement explains the shape of the measured spectrum between 13 μm and 17 μm?

Measured spectrum of the infrared radiation leaving the top of the atmosphere (solid) and black-body spectra at 288 K and 220 K (dashed) (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
Step 1The 288 K curve shows that the surface emits strongly between 13 μm and 17 μm, so the dip is caused by the atmosphere.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Carbon dioxide has molecular energy levels whose spacing matches photons of about 15 μm, so it absorbs this radiation from the surface; the excited molecules re-emit in all directions.	—
Step 3The radiation that finally escapes at these wavelengths is emitted by carbon dioxide high in the atmosphere, where it is cold (about 220 K), so the intensity follows the 220 K curve.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThe dashed 288 K curve shows that a surface at 288 K emits strongly at 13–17 μm; the dip is not in the emission of the surface.
BNitrogen and oxygen are not greenhouse gases: they do not absorb infrared at these wavelengths.
CRe-emission takes place in all directions, not only downwards; and the measured intensity is not zero in the band.
DCorrect: CO₂ absorbs the surface radiation near 15 μm, and the radiation reaching space there comes from cold CO₂ high up, matching the 220 K curve.

17B-1A-93

Albedo and equilibrium temperature·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksEstimate

A climate model treats the Earth as a body of constant emissivity at a mean temperature of 288 K, in equilibrium with the solar radiation it absorbs. The albedo of the Earth is 0.30.

A decrease in cloud cover reduces the albedo to 0.29. What is the change in the equilibrium temperature?

Show mark scheme

Marking point	Mark	Notes
Step 1In equilibrium eσT⁴ = (1 − α)S/4, so T⁴ ∝ (1 − α).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The absorbed fraction rises from 0.70 to 0.71, a fractional increase of 0.01/0.70 = 1.4 %.	—
Step 3T ∝ (1 − α)^1/4, so T rises by about ¼ × 1.4 % = 0.36 %: ΔT ≈ 0.0036 × 288 = +1.0 K.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis has the right size but the wrong sign: a lower albedo means more sunlight is absorbed, so the temperature rises.
BCorrect: T ∝ (1 − α)^1/4, giving +1.0 K.
CThis uses the fractional change of the albedo itself, 0.01/0.30, instead of that of the absorbed fraction 1 − α: ¼ × 3.3 % × 288 = 2.4 K.
DThis forgets the fourth root, taking T ∝ (1 − α): 1.4 % × 288 = 4.1 K.

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; Guidance: the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants Command term: Estimate

18B-1A-94

Energy balance and latitude·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The graph shows, for one hemisphere, how the annual mean intensity of solar radiation absorbed by the Earth and its atmosphere, and the annual mean intensity of infrared radiation emitted to space, vary with latitude. The mean temperature at each latitude does not change from year to year.

What can be deduced from the graph?

Annual mean absorbed solar intensity and emitted infrared intensity against latitude (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
Step 1At low latitudes more energy is absorbed than is emitted; near the poles more is emitted than is absorbed.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The temperatures are constant, so by conservation of energy each region must be in overall energy balance.	—
Step 3The surplus at low latitudes must therefore be carried to high latitudes by convection in the atmosphere and by ocean currents.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the radiative surplus at low latitudes balances the deficit at high latitudes only if energy is transported polewards.
BThis ignores the constant temperatures given in the stem: a radiative imbalance at one latitude does not mean warming when energy is carried away.
CThis reverses the effect of latitude: snow, ice and low Sun angles make the albedo larger near the poles; the low absorbed intensity there is consistent with that.
DThe emitted intensity falls towards the poles because the temperature falls there; the graph gives no information about the emissivity.

Syllabus understandingB.2 — the conservation of energy; that Earth's albedo varies daily and is dependent on cloud formations and latitude; B.1 — qualitative description of thermal energy transferred by convection due to fluid density differences Command term: Deduce

19B-1A-95

Albedo and emissivity of snow·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Fresh snow at a temperature of 263 K has an albedo of 0.85 for sunlight and an emissivity of 0.98. Sunlight of intensity 500 W m⁻² falls on the snow at right angles to its surface. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

What is (power emitted per unit area by the snow)/(power per unit area of sunlight absorbed by the snow)?

Show mark scheme

Marking point	Mark	Notes
Step 1Absorbed from sunlight: (1 − 0.85) × 500 = 75 W m⁻².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Emitted: 0.98 × 5.67 × 10⁻⁸ × 263⁴ = 266 W m⁻².	—
Step 3Ratio = 266/75 = 3.5. The albedo refers to sunlight and the emissivity to the snow's own infrared emission; they are independent quantities.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis makes both errors of B and C: emissivity taken as 1 − albedo and albedo taken as the fraction absorbed: 40.7/425 = 0.096.
BThis takes the emissivity to be 1 − albedo = 0.15: 0.15σ(263)⁴/75 = 40.7/75 = 0.54.
CThis uses the albedo as the fraction absorbed: 266/(0.85 × 500) = 266/425 = 0.63.
DCorrect: 266 W m⁻² emitted against 75 W m⁻² absorbed from sunlight.

20B-1A-96

Power absorbed by a planet·B.2 Greenhouse effect

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A planet of radius R orbits a star of luminosity L at a distance d. The albedo of the planet is α.

What is the total power absorbed by the planet from the star?

Show mark scheme

Marking point	Mark	Notes
Step 1Intensity at the planet: S = L/4πd².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The planet intercepts the starlight over its projected disc, area πR²: incident power = L/4πd² × πR² = LR²/4d².	—
Step 3A fraction (1 − α) is absorbed: P = (1 − α)LR²/4d².	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses the mean intensity S/4 and then also multiplies by the projected area πR², counting the factor of ¼ twice.
BThis uses the albedo as the fraction absorbed; the albedo is the fraction scattered.
CCorrect: (1 − α) × (L/4πd²) × πR².
DThis multiplies the intensity by the whole surface area 4πR², but only the projected disc πR² intercepts the starlight.

21B-1A-97

Equilibrium temperature of a tidally locked planet·B.2 Greenhouse effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Planets X and Y have no atmosphere. They are identical, have the same albedo and emissivity, and receive starlight of the same intensity.

X rotates rapidly, so its whole surface is at the same temperature T.

Y always keeps the same face towards its star. Assume that the sunlit hemisphere of Y has a uniform temperature, and that its dark hemisphere radiates negligible power.

What is the temperature of the sunlit hemisphere of Y?

Show mark scheme

Marking point	Mark	Notes
Step 1Both planets intercept and absorb the same power P_abs over the same projected disc.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2X radiates it from its whole surface 4πR²; Y radiates it from one hemisphere only, area 2πR².	—
Step 3eσ(2πR²)T_Y⁴ = eσ(4πR²)T⁴, so T_Y⁴ = 2T⁴ and T_Y = 2^1/4T ≈ 1.19T.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes that equal absorbed power gives equal temperatures, ignoring that Y radiates from only half the area.
BCorrect: half the radiating area must radiate twice the power per unit area, and the power per unit area is proportional to T⁴.
CThis uses the ratio of the areas inside a square root, as if the power per unit area were proportional to T².
DThis takes the temperature to be proportional to the power radiated per unit area, ignoring the fourth power in the Stefan–Boltzmann law.

Syllabus understandingB.2 — the solar constant S; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4; the conservation of energy; Guidance: the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants Command term: Deduce

22B-1A-98

Enhanced greenhouse effect and energy balance·B.2 Greenhouse effect

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The concentration of carbon dioxide in a planet's atmosphere increases and then stays constant. The planet reaches a new equilibrium in which its surface temperature is higher than before. The albedo of the planet and the energy output of its star do not change.

Which statements about the new equilibrium, compared with the original equilibrium, are correct?

I. The mean power per unit area radiated into space by the planet is the same.
II. The power per unit area radiated by the surface of the planet is greater.
III. The planet absorbs more power from its star than it radiates into space.

Show mark scheme

Marking point	Mark	Notes
Step 1In any equilibrium, power radiated into space = power absorbed from the star = (1 − α)S/4 per unit area. Neither α nor S has changed, so I is correct.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The surface is hotter, so it radiates more power per unit area (∝ T⁴): II is correct. More of this radiation is absorbed by the atmosphere and returned to the surface.	—
Step 3III describes the period of warming, not the equilibrium: once equilibrium is reached the net energy gain is zero, so III is wrong.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the outgoing power to space returns to its original value, while the warmer surface radiates more.
BThis accepts I and III together, which contradict each other: if the outgoing power equals the old value, it equals the absorbed power, so there is no net gain.
CThis is the "trapped heat" misconception: in equilibrium the planet must lose to space as much power as it absorbs, so the outgoing power is not reduced.
DIII is true only while the planet is warming up; at the new equilibrium absorbed and emitted powers are equal.

Syllabus understandingB.2 — the conservation of energy; that the augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect (Guidance: the burning of fossil fuels is a primary cause of the enhanced greenhouse effect); Guidance: energy balance problems will include energy exchanged between the surface and the atmosphere of a body Command term: Deduce

23B-1B-03

Energy balance of a planet·B.2 Greenhouse effect

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

Three planets b, c and d orbit a faint star of mass 4.0 × 10²⁹ kg and luminosity 2.0 × 10²⁴ W in circular orbits. Their orbital periods are found from the regular dips in the star's brightness as each planet passes in front of it. Infrared observations of planet c suggest a mean surface temperature of 341 K. Astronomers want to know whether planet c could have an atmosphere. They model each planet as a black body with an albedo of 0.30 that radiates from its whole surface.

Planet	b	c	d
Orbital period / days	4.0	9.2	20.5
Orbital radius / 10¹⁰ m	0.432		1.28

(a)

(i)

Show that the orbital radius of planet c is about 7.5 × 10⁹ m.

(2)

(b)

(i)

Determine the equilibrium surface temperature of planet c predicted by the model.

(3)

(c)

(i)

Discuss, with reference to your answer to (b), whether the infrared observation supports the suggestion that planet c has an atmosphere.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
r³ = GMT²/4π² with T = 9.2 × 86 400 = 7.95 × 10⁵ s	✓ 1	Kepler's third law from the data booklet; the conversion of days to seconds is required.
r = (6.67 × 10⁻¹¹ × 4.0 × 10²⁹ × (7.95 × 10⁵)²/4π²)^1/3 = 7.53 × 10⁹ m	✓ 1	Answer to at least 3 s.f. Alternative: scale from planet b or d with r³ ∝ T².
Part (b)(i)
Intensity at c: S = L/4πr² = 2.0 × 10²⁴/(4π × (7.53 × 10⁹)²) = 2810 W m⁻²	✓ 1	Allow ECF from (a). Using 7.5 × 10⁹ m gives 2830 W m⁻².
Absorbed mean intensity = (1 − 0.30) × S/4 = 491 W m⁻²	✓ 1	The factor 1/4 (projected disc versus whole sphere) is required.
T = (491/5.67 × 10⁻⁸)^1/4 = 305 K	✓ 1	Accept 300–310 K.
Part (c)(i)
341 K is about 36 K higher than the model value, more than the model's uncertainty would suggest	✓ 1	Allow ECF from (b). The comparison must use the candidate's value.
An atmosphere containing greenhouse gases absorbs part of the outgoing infrared and re-emits some back to the surface, raising the surface temperature, so the observation supports the suggestion	✓ 1	Accept also a lower true albedo, or internal/tidal heating, as an alternative explanation that weakens the conclusion.

Answers: (a)(i) 7.53 × 10⁹ m · (b)(i) 305 K (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the solar constant; that the incoming radiative power is dependent on the projected surface of a planet … resulting in a mean value of the incoming intensity being S/4; equilibrium temperature from energy balance including albedo; D.1 — Kepler's third law Command term: Determine

24B-1B-07

Solar constant·B.2 Greenhouse effect

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

To estimate the solar constant, a student places a thin aluminium disc of diameter 4.00 cm and mass 12.0 g, blackened on its upper face, so that it faces the Sun directly at midday. The disc rests on polystyrene. Its temperature θ is read every 30 s with a digital thermometer of resolution 0.1 °C. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. The table and the graph show the results with the line of best fit.

t / s	0	30	60	90	120	150
θ / °C	18.0	21.2	23.8	27.1	30.2	32.9

Temperature θ of the disc against time t with the line of best fit (drawn to scale)

(a)

(i)

A classmate calculates the rate of rise of temperature from the first two readings only. Determine the rate of rise of temperature in a way that is more reliable, and explain your choice.

(2)

(b)

(i)

Determine the intensity of the solar radiation absorbed by the disc.

(2)

(c)

(i)

At the time of the experiment the atmosphere transmits 72 % of the solar radiation arriving at its top. Estimate the solar constant from the student's data.

(1)

(ii)

The accepted value of the solar constant is 1360 W m⁻². Suggest, with a reason, why the student's estimate differs from this value.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient of the line of best fit over the full range, e.g. (34.0 − 18.0)/160 = 0.10 K s⁻¹	✓ 1	Accept 0.097–0.103 K s⁻¹. The first two readings give 0.107 K s⁻¹: award [0] for this value.
The line uses all six readings over the whole range, so the random scatter (±0.2 °C, larger than the 0.1 °C resolution) has much less effect than in a 30 s interval	✓ 1	OWTTE.
Part (b)(i)
Power absorbed = mc × gradient = 0.0120 × 900 × 0.10 = 1.08 W	✓ 1	Assumes all the energy absorbed raises the temperature of the disc.
Area = π × 0.0200² = 1.26 × 10⁻³ m²; intensity = 859 W m⁻²	✓ 1	Allow ECF from (a). Accept 830–880 W m⁻².
Part (c)(i)
S = 859/0.72 = 1190 W m⁻²	✓ 1	Allow ECF from (b).
Part (c)(ii)
The estimate is about 12 % too low	✓ 1	Allow ECF from (c)(i). The direction of the difference must agree with the candidate's value.
The disc loses energy to the air (convection) and to its support while it warms, and its surface is not a perfect absorber, so less power raises its temperature than is incident on it	✓ 1	Accept also: the disc is not exactly perpendicular to the rays. "Human error" scores [0].

Answers: (a)(i) 0.10 K s⁻¹ · (b)(i) 859 W m⁻² · (c)(i) 1190 W m⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the solar constant S; the conservation of energy; B.1 — quantitative analysis of thermal energy transfers with Q = mcΔT; Tools — using the full data range, comparing an experimental value with an accepted value Command term: Determine

25B-1B-14

Albedo·B.2 Greenhouse effect

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A satellite instrument measures the monthly mean albedo α of several regions of tropical ocean, together with the fraction f of each region covered by cloud. The graph shows α against f. Each albedo has an uncertainty of ±0.02, shown by the error bars. The solid line is the line of best fit; the dashed lines are the steepest and the shallowest lines that pass through all the error bars. The mean solar intensity arriving at the top of the atmosphere above these regions is 410 W m⁻².

Cloud fraction f	0.10	0.25	0.40	0.55	0.70	0.85
Albedo α	0.12	0.20	0.29	0.35	0.44	0.50

Albedo against cloud fraction with error bars, line of best fit (solid) and steepest/shallowest lines (dashed) (drawn to scale)

(a)

(i)

Determine the gradient of the line of best fit.

(1)

(ii)

Determine the absolute uncertainty in the gradient.

(2)

(b)

(i)

Determine the difference between the mean intensities absorbed by a region with f = 0.20 and a region with f = 0.70, with its uncertainty.

(2)

(c)

(i)

A climate model assumes that the albedo of tropical ocean completely covered by cloud is 0.65. Deduce whether the data are consistent with this assumption.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient = (0.58 − 0.074)/1.0 = 0.51	✓ 1	Accept 0.49–0.53.
Part (a)(ii)
Steepest gradient ≈ 0.55; shallowest ≈ 0.47	✓ 1	Accept ±0.01 on each.
Uncertainty = (0.55 − 0.47)/2 = ±0.04, so gradient = 0.51 ± 0.04	✓ 1	Allow ECF from (a)(i). Accept ±0.03 to ±0.05.
Part (b)(i)
Difference in albedo = 0.51 × 0.50 = 0.255; difference in absorbed intensity = 0.255 × 410 = 105 W m⁻²	✓ 1	Allow ECF from (a)(i). Accept 100–109 W m⁻².
Same fractional uncertainty as the gradient: ±0.04 × 0.50 × 410 = ±8 W m⁻²	✓ 1	Allow ECF from (a)(ii).
Part (c)(i)
At f = 1 the best-fit line gives 0.58 and even the steepest line gives only 0.60	✓ 1	Allow ECF from (a)(ii). Values read from the candidate's lines.
0.65 lies outside the range allowed by the data (even allowing for extrapolation from f = 0.85), so the assumption is not consistent	✓ 1	The conclusion must follow from the comparison.

Answers: (a)(i) 0.51 · (a)(ii) ±0.04 · (b)(i) 105 ± 8 W m⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; Tools — error bars, steepest and shallowest lines, uncertainty in a gradient Command term: Determine

26B-1B-15

Emissivity·B.2 Greenhouse effect

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student measures the emissivity e of a matt paint. A metal plate painted on one face is heated electrically, and its temperature is measured with a thermometer embedded in the metal. A calibrated detector gives the net power radiated per unit area, I, from the painted face. The surroundings are at an unknown kelvin temperature T_r. The net intensity is modelled by

I = eσ(T⁴ − T_r⁴)

where T is the kelvin temperature of the plate. The graph shows I against T⁴ with the line of best fit, extended to lower values of T⁴.

θ / °C	40	60	80	100	120	140
T / K	313	333	353	373	393	413
T⁴ / 10⁹ K⁴	9.60	12.30	15.53	19.36	23.85	29.09
I / W m⁻²	87	203	333	489	692	924

I against T⁴ with the line of best fit; the grey line marks I = 0 (drawn to scale)

(a)

(i)

Explain why the net intensity depends on T⁴ − T_r⁴ rather than on T⁴ alone.

(1)

(b)

(i)

Determine the emissivity of the paint.

(3)

(c)

(i)

Determine T_r in °C.

(2)

(d)

(i)

The manufacturer gives the emissivity of the paint as 0.90. Air flowing past the plate cools the painted surface slightly below the temperature of the metal. Explain how this accounts for the difference from your answer to (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The plate emits eσT⁴ per unit area but also absorbs radiation from the surroundings that depends on T_r⁴; I is the difference	✓ 1
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (924 − 87)/((29.09 − 9.60) × 10⁹) = 4.3 × 10⁻⁸ W m⁻² K⁻⁴	✓ 1	Accept 4.1–4.5 × 10⁻⁸. The factor 10⁹ is required.
Gradient = eσ, so e = gradient/σ	✓ 1
e = 4.3 × 10⁻⁸/5.67 × 10⁻⁸ = 0.76	✓ 1	Accept 0.72–0.79.
Part (c)(i)
I = 0 when T⁴ = T_r⁴: the extended line crosses I = 0 at T⁴ ≈ 7.7 × 10⁹ K⁴	✓ 1	Allow ECF from (b). Accept 7.4–7.9 × 10⁹ K⁴, or \|intercept\|/gradient = 328/4.3 × 10⁻⁸.
T_r = (7.7 × 10⁹)^1/4 = 296 K = 23 °C	✓ 1	Accept 20–26 °C.
Part (d)(i)
The surface radiates at a lower temperature than the recorded T, so each measured I is too small for the plotted T⁴: the gradient, and hence e, is underestimated (0.76 < 0.90)	✓ 1	Allow ECF from (b). The direction must be explained, not just stated.

Answers: (b)(i) 0.76 · (c)(i) 23 °C (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature as given by emissivity = power radiated per unit area/σT⁴; B.1 — the Stefan–Boltzmann law; the Kelvin and Celsius scales; Tools — linearising, extrapolation, systematic error Command term: Determine

27B-1B-29

Laboratory model of planetary equilibrium·B.2 Greenhouse effect

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student models a planet in a laboratory. A small copper plate, blackened on its front face and insulated at the back, is held inside an evacuated glass chamber whose walls are at T_r = 293.0 K. A lamp outside the chamber acts as a point source that emits radiation of total power P. The front face of the plate faces the lamp at a distance d. When the temperature T of the plate is steady it is recorded with a thermocouple.

The front face has an emissivity of 0.95 and an albedo of 0.05 for the radiation of the lamp. The plate also absorbs radiation from the walls of the chamber, so the energy balance of the plate is modelled as

(1 − 0.05)P/4πd² = 0.95σ(T⁴ − T_r⁴)

The table and the graph show the results.

d / m	0.20	0.25	0.30	0.35	0.40	0.50
T / K	323.3	312.7	307.7	303.6	301.7	298.1
(1/d²) / m⁻²	25.00	16.00	11.11	8.16	6.25	4.00
(T⁴ − T_r⁴) / 10⁹ K⁴	3.55	2.19		1.13	0.92	0.53

Graph of (T⁴ − T_r⁴) against 1/d² with the line of best fit (drawn to scale)

(a)

(i)

Explain why the plate is placed in an evacuated chamber.

(1)

(b)

(i)

Calculate the missing value in the table.

(1)

(c)

(i)

Determine the gradient of the line of best fit. State an appropriate unit for your answer.

(2)

(d)

(i)

Determine P.

(2)

(e)

(i)

The glass wall of the chamber absorbs and reflects part of the radiation from the lamp. State and explain how this affects the value of P found in (d).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Without air there is no convection (or conduction through air) from the plate, so the plate loses (and gains) energy only by radiation, as the model assumes	✓ 1	Reference to the model / radiation as the only transfer is required. "To keep it clean" or "to avoid draughts" alone scores [0].
Part (b)(i)
(307.7⁴ − 293.0⁴) = 1.59 × 10⁹ K⁴, so the entry is 1.59	✓ 1	Accept 1.59 or 1.594. Award [0] for (307.7 − 293.0)⁴.
Part (c)(i)
Gradient from two well-separated points on the line, e.g. (3.67 − (−0.02))/(26.0 − 0) × 10⁹ = 1.42 × 10⁸	✓ 1	Accept 1.36–1.48 × 10⁸. The factor 10⁹ must be included. Using two table points that are not on the line scores [0] for this mark unless they lie on the line.
Unit: K⁴ m²	✓ 1	Allow the unit mark independently of the value.
Part (d)(i)
From the model, gradient = (1 − 0.05)P/(4π × 0.95σ), so P = 4π × 0.95σ × gradient/0.95	✓ 1	The relation between the gradient and P must be seen or used.
P = 4π × 0.95 × 5.67 × 10⁻⁸ × 1.42 × 10⁸/0.95 = 1.0 × 10² W	✓ 1	Allow ECF from (c). Accept 97–106 W.
Part (e)(i)
The intensity reaching the plate is less than P/4πd², so each T is lower than the model predicts and the gradient is too small: the value found in (d) is smaller than the true power of the lamp (it is the power transmitted by the glass)	✓ 1	Allow ECF from (d). The direction (underestimate) and a reason are both required.

Answers: (b)(i) 1.59 · (c)(i) 1.42 × 10⁸ K⁴ m² · (d)(i) 1.0 × 10² W (the remaining parts are explanations — see the table above)

28B-1B-30

Measuring the albedo of sand·B.2 Greenhouse effect

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

To measure the albedo of dry desert sand, a student places a sample at the bottom of a white dome. A beam of light enters the dome through a small hole and falls on the sand. A detector inside the dome is calibrated so that its reading P_s gives the total power scattered by the sand, and a second detector measures the incident power P_i of the beam. The student changes P_i by changing the current in the lamp. Each reading of P_s is uncertain by ±1.0 mW.

The table and the graph show the results with the line of best fit.

P_i / mW	20	40	60	80	100	120
P_s / mW	10.1	16.2	23.6	29.9	37.4	43.6

Scattered-power reading P_s against incident power P_i with error bars and the line of best fit (drawn to scale)

(a)

(i)

Determine the albedo of the sand from the graph.

(2)

(b)

(i)

State the value of the intercept on the P_s axis and suggest a cause.

(1)

(c)

(i)

Another student calculates the albedo from the single reading at P_i = 40 mW. Calculate the value obtained, and explain why it is not reliable.

(2)

(d)

(i)

When the sand is wet, its albedo falls to 0.18. Sunlight of intensity 850 W m⁻² falls on the sand at right angles. Determine the increase in the intensity absorbed by the sand when it is wetted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Albedo = gradient of the line (scattered power per unit incident power)	✓ 1	Using a single ratio P_s/P_i scores [0] for this mark.
Gradient = (47.1 − 3.1)/(130 − 0) = 0.34	✓ 1	Accept 0.32–0.36.
Part (b)(i)
Intercept ≈ 3 mW; light from the room leaking into the dome (or a zero error of the detector) gives a reading even when no light from the beam is scattered	✓ 1	Accept 2.5–3.5 mW. Both the value and a plausible cause are needed.
Part (c)(i)
16.2/40 = 0.41	✓ 1
The reading includes the offset of about 3 mW, a systematic error, so the ratio is too large; the gradient method removes the offset (and uses all the data)	✓ 1	Allow ECF from (a) and (b). The comparison must agree with the candidate's values.
Part (d)(i)
Increase = (albedo_dry − albedo_wet) × 850 = (0.34 − 0.18) × 850	✓ 1	Allow ECF from (a).
= 1.4 × 10² W m⁻²	✓ 1	Accept 1.2–1.5 × 10² W m⁻².

Answers: (a)(i) 0.34 · (b)(i) 3 mW · (c)(i) 0.41 · (d)(i) 1.4 × 10² W m⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; Tools — gradient and intercept of a linear graph, systematic error (zero offset) Command term: Determine

29B-1B-31

Absorption of infrared by carbon dioxide·B.2 Greenhouse effect

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A beam of infrared radiation with wavelengths close to 15 μm passes through a sealed tube of length 0.50 m and is then measured by a detector, whose reading is I. The tube is filled with carbon dioxide at a pressure p and a constant temperature of 295 K. It is suggested that

I = I₀e^−kp

where I₀ and k are constants. The table and the graph show the results.

p / Pa	100	250	400	550	700	850
I / mV	72.2	55.7	44.8	34.6	27.7	21.3
ln(I / mV)	4.28	4.02	3.80	3.54	3.32	3.06

Graph of ln(I / mV) against p with the line of best fit (drawn to scale)

(a)

(i)

Outline why a graph of ln I against p is used to test the suggestion.

(1)

(b)

(i)

Determine k, including an appropriate unit.

(2)

(c)

(i)

Calculate the pressure at which the gas absorbs half of the radiation that enters the tube.

(1)

(d)

(i)

Determine the number of carbon dioxide molecules per unit volume in the tube at the pressure found in (c).

(2)

(e)

(i)

The detector also responds to some radiation at wavelengths that carbon dioxide does not absorb. Suggest how this would affect the graph at high pressures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
ln I = ln I₀ − kp, so if the suggestion is correct the graph is a straight line of gradient −k (and intercept ln I₀)	✓ 1	The linear form must be seen.
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (2.82 − 4.44)/(1000 − 0) = −1.62 × 10⁻³	✓ 1	Accept 1.55–1.70 × 10⁻³ in magnitude.
k = 1.6 × 10⁻³ Pa⁻¹	✓ 1	The positive value and the unit Pa⁻¹ are both needed for this mark.
Part (c)(i)
p_½ = ln 2/k = 0.693/1.6 × 10⁻³ = 4.3 × 10² Pa	✓ 1	Allow ECF from (b). Accept 4.1–4.5 × 10² Pa.
Part (d)(i)
N/V = p/k_BT	✓ 1	From pV = Nk_BT.
= 433/(1.38 × 10⁻²³ × 295) = 1.1 × 10²³ m⁻³	✓ 1	Allow ECF from (c). Accept 1.0–1.1 × 10²³ m⁻³.
Part (e)(i)
The reading cannot fall below the value due to the unabsorbed radiation, so at high pressures ln I falls less steeply and the graph curves towards a horizontal line (so a gradient taken there underestimates k)	✓ 1	OWTTE. A reference to the graph levelling off (curving) is required.

Answers: (b)(i) 1.6 × 10⁻³ Pa⁻¹ · (c)(i) 4.3 × 10² Pa · (d)(i) 1.1 × 10²³ m⁻³ (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; B.3 — the equation of state for an ideal gas pV = Nk_BT; Tools — linearising an exponential relationship with logarithms, gradient with units, systematic error Command term: Determine

30B-2-13

Energy balance of a planet·B.2 Greenhouse effect

Paper 2Medium9 marks

Short answer & extended response9 steps to full marksDetermine

Mars orbits the Sun at a mean distance of 2.28 × 10¹¹ m. Luminosity of the Sun = 3.83 × 10²⁶ W; radius of the Earth's orbit = 1.50 × 10¹¹ m.

(a)

(i)

Show that the intensity of sunlight at the mean distance of Mars is about 590 W m⁻².

(1)

(b)

(i)

Mars has an albedo of 0.25. Determine the equilibrium temperature of Mars, assuming that the surface radiates as a black body and that the atmosphere has no effect.

(3)

(c)

(i)

Determine, in Earth years, the orbital period of Mars.

(1)

(d)

(i)

The distance of Mars from the Sun varies between 2.07 × 10¹¹ m and 2.49 × 10¹¹ m. Show that the equilibrium temperature is proportional to d^−1/2, and determine the difference between the highest and the lowest equilibrium temperatures.

(2)

(e)

(i)

The measured mean surface temperature of Mars is about 210 K. Its atmosphere is mainly carbon dioxide at a very low pressure. Deduce what this suggests about the greenhouse effect on Mars.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
S = L/4πd² = 3.83 × 10²⁶/(4π × (2.28 × 10¹¹)²) = 586 W m⁻²	✓ 1	An answer to at least 3 s.f. is required.
Part (b)(i)
Mean incoming intensity = S/4 = 147 W m⁻² (sunlight is intercepted by the cross-section πR² but the energy is radiated from the whole surface 4πR²)	✓ 1	Allow ECF from (a).
Absorbed intensity = 0.75 × 147 = 110 W m⁻²	✓ 1
T = (110/5.67 × 10⁻⁸)^1/4 = 210 K	✓ 1	Award [2 max] for 297 K (factor 4 omitted).
Part (c)(i)
Kepler's third law: T² ∝ r³, so T_Mars = (2.28/1.50)^3/2 × 1 year = 1.87 years	✓ 1
Part (d)(i)
T⁴ ∝ S ∝ 1/d², so T ∝ d^−1/2	✓ 1
210 × (2.28/2.07)^1/2 = 220 K and 210 × (2.28/2.49)^1/2 = 201 K: difference ≈ 19 K	✓ 1	Allow ECF from (b). Accept 18–20 K.
Part (e)(i)
The measured temperature is about equal to the value from (b), so the greenhouse effect on Mars is very small	✓ 1	Allow ECF from (b).
The thin atmosphere contains few CO₂ molecules above each square metre, so it absorbs, and re-emits back towards the surface, only a small fraction of the infrared emitted by the surface	✓ 1

Answers: (a)(i) 586 W m⁻² · (b)(i) 210 K · (c)(i) 1.87 years · (d)(i) ≈ 19 K (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the solar constant and the mean incoming intensity S/4; albedo; the estimation of the equilibrium temperature of a body using energy balance; the absorption of infrared radiation by greenhouse gases; B.1 — apparent brightness b = L/4πd²; Stefan–Boltzmann law; D.1 — Kepler's three laws of orbital motion Command term: Determine

31B-2-21

Greenhouse gases·B.2 Greenhouse effect

Paper 2Medium8 marks

Short answer & extended response8 steps to full marksDetermine

Carbon dioxide molecules absorb infrared radiation of wavelength 15.0 μm, which makes the molecules vibrate (bend). In a simple resonance model, this vibration is treated as an oxygen atom of mass 2.66 × 10⁻²⁶ kg oscillating on a spring of spring constant k.

(a)

(i)

Calculate the frequency of the radiation.

(1)

(b)

(i)

Calculate, in eV, the energy of a photon of this radiation.

(1)

(c)

(i)

Determine the value of k for which the natural frequency of the model is equal to the frequency of the radiation.

(1)

(d)

(i)

Explain, using the resonance model, how carbon dioxide in the atmosphere reduces the energy radiated from the Earth's surface to space at this wavelength.

(2)

(e)

(i)

In a molecule containing the heavier isotope oxygen-18, the vibrating mass is 18/16 times larger and k is unchanged. Deduce the wavelength absorbed by this molecule.

(2)

(f)

(i)

State one limitation of the resonance model of the greenhouse effect.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
f = c/λ = 3.00 × 10⁸/15.0 × 10⁻⁶ = 2.00 × 10¹³ Hz	✓ 1
Part (b)(i)
E = hf = 6.63 × 10⁻³⁴ × 2.00 × 10¹³ = 1.33 × 10⁻²⁰ J = 0.083 eV	✓ 1	Allow ECF from (a). The answer in eV is required.
Part (c)(i)
T = 2π√(m/k), so k = 4π²f²m; k = 4π² × (2.00 × 10¹³)² × 2.66 × 10⁻²⁶ = 420 N m⁻¹	✓ 1	Allow ECF from (a).
Part (d)(i)
Radiation at the natural frequency of the bond drives the vibration at resonance, so the amplitude becomes large and energy is absorbed from the radiation	✓ 1
The absorbed energy is re-emitted in all directions (or shared with other molecules by collisions), so part of it returns towards the surface and less escapes to space	✓ 1
Part (e)(i)
f ∝ 1/√m, so λ ∝ √m	✓ 1
λ = 15.0 × √(18/16) = 15.9 μm	✓ 1	Award [1 max] for 16.9 μm (λ ∝ m).
Part (f)(i)
It is a classical model: it does not explain why only particular (discrete) photon energies are absorbed, which requires molecular energy levels	✓ 1	Accept any valid limitation, e.g. it cannot predict which gases absorb.

Answers: (a)(i) 2.00 × 10¹³ Hz · (b)(i) 0.083 eV · (c)(i) 420 N m⁻¹ · (e)(i) 15.9 μm (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the absorption of infrared radiation by the main greenhouse gases and the subsequent emission of radiation in all directions; that the greenhouse effect can be explained in terms of both a resonance model and molecular energy levels; C.1 — the time period of a mass–spring system T = 2π√(m/k); C.4 — resonance and natural frequency; E.1 — E = hf Command term: Determine

32B-2-22

Heat shield of a solar probe·B.2 Greenhouse effect

Paper 2Medium13 marks

Short answer & extended response9 steps to full marksDetermine

A spacecraft studying the Sun carries a flat heat shield that always faces the Sun. At its closest approach the spacecraft is 6.9 × 10⁹ m from the centre of the Sun.

Luminosity of the Sun L_☉ = 3.83 × 10²⁶ W; solar constant at the Earth S = 1.36 × 10³ W m⁻²; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

(a)

Intensity.

(i)

Show that the intensity of sunlight at closest approach is about 6.4 × 10⁵ W m⁻².

(2)

(ii)

Use your answer to (a)(i) and the solar constant to determine the distance of the Earth from the Sun.

(2)

(b)

The sunlit face of the shield has an albedo of 0.60 and an emissivity of 0.90. Assume that all the energy absorbed by the shield is re-radiated from its sunlit face.

(i)

Determine the equilibrium temperature of the sunlit face.

(3)

(ii)

State this temperature in °C.

(1)

(iii)

Calculate the wavelength at which the radiation emitted by the shield is most intense.

(1)

(c)

Behind the sunlit face is a layer of carbon foam of thickness 0.11 m and thermal conductivity 0.050 W m⁻¹ K⁻¹. The back of the foam is at 330 K.

(i)

Calculate the rate of conduction of energy per unit area through the foam.

(1)

(ii)

Hence comment on the assumption made in (b).

(1)

(d)

The coating.

(i)

Explain why the sunlit face is given a white, high-albedo coating, and determine the temperature it would reach if its albedo were 0.10 with the same emissivity.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
b = L_☉/4πd² = 3.83 × 10²⁶/(4π × (6.9 × 10⁹)²)	✓ 1
b = 6.40 × 10⁵ W m⁻²	✓ 1	Answer to more figures than the target, e.g. 6.40 × 10⁵ W m⁻².
Part (a)(ii)
b ∝ 1/d², so d_E = 6.9 × 10⁹ × √(6.40 × 10⁵/1360) = 6.9 × 10⁹ × 21.7	✓ 1
d_E = 1.5 × 10¹¹ m	✓ 1	Or from d_E = √(L_☉/4πS).
Part (b)(i)
Absorbed intensity = (1 − 0.60) × 6.40 × 10⁵ = 2.56 × 10⁵ W m⁻²	✓ 1
0.90σT⁴ = 2.56 × 10⁵, so T⁴ = 5.0 × 10¹² K⁴	✓ 1
T = 1.5 × 10³ K	✓ 1	Accept 1490–1500 K.
Part (b)(ii)
1497 − 273 ≈ 1.2 × 10³ °C	✓ 1	Accept 1220–1230 °C; ECF.
Part (b)(iii)
λ_max = 2.9 × 10⁻³/1497 = 1.9 × 10⁻⁶ m	✓ 1	ECF.
Part (c)(i)
ΔQ/Δt = kΔT/Δx = 0.050 × (1497 − 330)/0.11 = 5.3 × 10² W m⁻²	✓ 1	Accept 520–540 W m⁻²; ECF.
Part (c)(ii)
This is about 0.2 % of the absorbed 2.56 × 10⁵ W m⁻², so almost all the absorbed energy is indeed re-radiated from the sunlit face: the assumption is justified	✓ 1
Part (d)(i)
A higher albedo reflects more of the incident radiation, so less is absorbed and the equilibrium temperature is lower	✓ 1
T ∝ (1 − albedo)^1/4: T = 1497 × (0.90/0.40)^1/4 = 1.8 × 10³ K	✓ 1	Accept 1830–1840 K.

Answers: (a)(i) 6.40 × 10⁵ W m⁻² · (a)(ii) 1.5 × 10¹¹ m · (b)(i) 1.5 × 10³ K · (b)(ii) 1.2 × 10³ °C · (b)(iii) 1.9 × 10⁻⁶ m · (c)(i) 5.3 × 10² W m⁻² · (d)(i) 1.8 × 10³ K (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the solar constant S; emissivity; albedo; the estimation of the equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity and solar or other constants; B.1 — apparent brightness b = L/4πd²; Wien's displacement law; conduction ΔQ/Δt = kAΔT/Δx Command term: Determine

33B-2-23

Surface–atmosphere energy balance·B.2 Greenhouse effect

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

A rocky planet with an atmosphere orbits a distant star. The intensity of the starlight at the distance of the planet is S = 2.00 × 10³ W m⁻².

The diagram shows the mean energy flows per unit area of the planet's surface, in W m⁻². Of the 520 W m⁻² emitted by the surface, 40 W m⁻² passes directly through the atmosphere to space and the rest is absorbed by the atmosphere. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

Mean energy flows per unit area of the planet's surface, in W m⁻². Gold: starlight; red: infrared radiation; grey: convection and evaporation. P, Q and R are unknown.

(a)

(i)

Outline why the mean intensity of the incoming starlight shown in the diagram is 500 W m⁻².

(1)

(ii)

Determine the albedo of the planet and the value of P.

(2)

(b)

(i)

By considering the energy balance of the surface, determine Q.

(2)

(ii)

By considering the energy balance of the atmosphere, determine R, and verify that the planet as a whole is in energy balance.

(2)

(c)

(i)

The emissivity of the surface is 0.95. Determine the mean surface temperature.

(1)

(d)

(i)

Explain, with reference to the energy flows in the diagram, why an increase in the concentration of greenhouse gases in the atmosphere would raise the surface temperature.

(3)

(e)

(i)

The star has a luminosity of 2.7 × 10²⁶ W and a mass of 1.8 × 10³⁰ kg. Determine, in Earth days, the orbital period of the planet.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The planet intercepts starlight over its cross-section πR² but the energy is shared over its whole surface 4πR², so the mean is S/4 = 2000/4 = 500 W m⁻²	✓ 1	The ratio of the two areas must be seen.
Part (a)(ii)
Albedo = 160/500 = 0.32	✓ 1
P = 500 − 160 − 90 = 250 W m⁻²	✓ 1
Part (b)(i)
Surface: energy in = energy out, so P + Q = 520 + 110	✓ 1
Q = 630 − 250 = 380 W m⁻²	✓ 1	Allow ECF from (a)(ii).
Part (b)(ii)
Atmosphere: 90 + (520 − 40) + 110 = 680 W m⁻² = Q + R, so R = 680 − 380 = 300 W m⁻²	✓ 1	Allow ECF from (b)(i).
Outgoing at the top: 160 + 40 + 300 = 500 W m⁻², equal to the incoming 500 W m⁻²	✓ 1
Part (c)(i)
0.95σT⁴ = 520, so T⁴ = 520/(0.95 × 5.67 × 10⁻⁸) = 9.65 × 10⁹ K⁴; T = 313 K	✓ 1	Accept 312–314 K. 309 K (emissivity ignored) scores 0.
Part (d)(i)
More greenhouse gas absorbs a larger fraction of the infrared emitted by the surface, so the 40 W m⁻² escaping directly to space decreases	✓ 1
The atmosphere re-emits the absorbed energy in all directions, so the downward flow Q to the surface increases	✓ 1
The surface then gains more than it loses, so its temperature rises until it emits enough (the 520 W m⁻² increases) for the outgoing radiation at the top of the atmosphere to balance the absorbed starlight again	✓ 1	"The greenhouse effect traps heat" alone scores [0].
Part (e)(i)
d = √(L/4πS) = √(2.7 × 10²⁶/(4π × 2000)) = 1.04 × 10¹¹ m	✓ 1
The gravitational force provides the centripetal force: GMm/d² = m(2π/T)²d, so T = 2π√(d³/GM)	✓ 1	Kepler's third law in the data-booklet form is equivalent.
T = 2π√((1.04 × 10¹¹)³/(6.67 × 10⁻¹¹ × 1.8 × 10³⁰)) = 1.9 × 10⁷ s = 221 days	✓ 1	Accept 215–225 days. Allow ECF from the distance.

Answers: (a)(ii) 0.32; 250 W m⁻² · (b)(i) 380 W m⁻² · (b)(ii) 300 W m⁻² · (c)(i) 313 K · (e)(i) 221 days (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the conservation of energy; the solar constant and the mean incoming intensity S/4; albedo; emissivity; energy balance problems including energy exchanged between the surface and the atmosphere of a body; the absorption of infrared radiation by greenhouse gases and the subsequent emission in all directions; the enhanced greenhouse effect; that albedo depends on cloud formations; B.1 — apparent brightness b = L/4πd²; D.1 — Newton's law of gravitation and circular orbits (Kepler's third law) Command term: Determine

34B-2-35

Albedo and melting sea ice·B.2 Greenhouse effect

Paper 2Easy12 marks

Short answer & extended response12 steps to full marksDetermine

In summer the mean intensity of sunlight reaching the surface of a region of the Arctic Ocean, averaged over day and night, is 300 W m⁻². Part of the region is covered by sea ice of albedo 0.60; the rest is open sea water of albedo 0.06.

Density of ice = 917 kg m⁻³; specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹.

(a)

(i)

Calculate the intensity of the sunlight absorbed by the ice.

(1)

(ii)

When the ice melts, open water is exposed. Determine the additional energy absorbed by each square metre of the surface in one day.

(2)

(b)

(i)

A sheet of sea ice 1.2 m thick is at 0 °C. Estimate the number of days for which the additional energy in (a)(ii) would have to be absorbed to melt 1.0 m² of this ice.

(3)

(c)

(i)

Explain, with reference to your answers to (a), why the loss of sea ice tends to cause further warming of the Arctic Ocean.

(2)

(d)

(i)

Sea ice floats. Show that when a block of floating ice melts, the level of the water does not change. Assume that the density of the sea water is the same as the density ρ_w of the melt water.

(2)

(e)

(i)

Thawing of frozen ground releases methane into the atmosphere. State one other natural source of atmospheric methane and one source that is due to human activity.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
(1 − 0.60) × 300 = 120 W m⁻²	✓ 1
Part (a)(ii)
Open water absorbs 0.94 × 300 = 282 W m⁻²; the increase is 282 − 120 = 162 W m⁻²	✓ 1	Allow ECF from (a)(i).
Energy = 162 × 86 400 = 1.40 × 10⁷ J	✓ 1	Accept 1.4 × 10⁷ J. The time must be in seconds.
Part (b)(i)
Mass of ice = 917 × 1.2 × 1.0 = 1100 kg	✓ 1
Energy needed = 1100 × 3.34 × 10⁵ = 3.7 × 10⁸ J	✓ 1
Number of days = 3.7 × 10⁸/1.4 × 10⁷ ≈ 26	✓ 1	Allow ECF from (a)(ii). Accept 25–27 days. Award [2 max] for 15 days (total absorption of the water used instead of the additional absorption).
Part (c)(i)
Open water has a much lower albedo, so it absorbs a larger fraction of the sunlight (282 instead of 120 W m⁻²) and the water warms	✓ 1	Allow ECF from the candidate's values in (a).
The warmer water melts more ice, which exposes more water, so the absorption increases further (the effect reinforces itself)	✓ 1	The self-reinforcing link is required for this mark.
Part (d)(i)
The floating ice is in equilibrium, so the buoyancy force equals its weight: ρ_wV_dispg = mg, giving V_disp = m/ρ_w	✓ 1	Use of F_b = ρVg is required.
The melt water has the same mass m, so its volume is m/ρ_w = V_disp: it exactly fills the volume the ice displaced, and the level is unchanged	✓ 1	Mass is conserved on melting; this must be stated or used.
Part (e)(i)
Natural: any one of wetlands (marshes, swamps); termites; wild animals (ruminants); natural gas seeping from the ground or sea floor; wildfires	✓ 1	Thawing frozen ground (permafrost) is given in the stem and is not accepted.
Human activity: any one of livestock farming (cattle); rice paddies; landfill sites; leaks from natural-gas wells, pipelines or coal mines	✓ 1	"Burning fossil fuels" alone is not accepted (it mainly releases CO₂).

Answers: (a)(i) 120 W m⁻² · (a)(ii) 1.40 × 10⁷ J · (b)(i) 26 days (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; that methane, water vapour, carbon dioxide and nitrous oxide are the main greenhouse gases, with origins that are both natural and created by human activity; B.1 — Q = mL; A.2 — buoyancy F_b = ρVg Command term: Determine

35B-2-45

Molecular energy levels of nitrous oxide·B.2 Greenhouse effect

Paper 2Easy8 marks

Short answer & extended response8 steps to full marksCalculate

Nitrous oxide, N₂O, is one of the main greenhouse gases. The diagram shows four of the vibrational energy levels of the N₂O molecule. E₀ is the lowest level.

Four vibrational energy levels of the N₂O molecule (energies drawn to scale)

(a)

(i)

Calculate the wavelength of the photon absorbed when a molecule makes a transition from E₀ to E₂.

(1)

(b)

(i)

The mean temperature of the Earth's surface is 288 K. Show that the surface emits most intensely at a wavelength of about 10 μm.

(1)

(c)

(i)

The Sun emits most intensely at about 0.5 μm. Explain, with reference to the diagram and your answers to (a) and (b), why N₂O absorbs strongly some of the radiation emitted by the Earth's surface but very little of the radiation from the Sun.

(2)

(d)

(i)

Outline how the absorption of infrared radiation by N₂O molecules in the atmosphere leads to a higher surface temperature.

(2)

(e)

(i)

The N₂O in a column of the atmosphere absorbs 1.5 W m⁻² of the radiation emitted by the surface, at the wavelength found in (a). Determine the number of photons absorbed per second per square metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
λ = hc/E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/(0.159 × 1.60 × 10⁻¹⁹) = 7.8 × 10⁻⁶ m	✓ 1	The conversion from eV to J is required. Accept 7.8–7.9 μm.
Part (b)(i)
λ_max = 2.9 × 10⁻³/288 = 1.01 × 10⁻⁵ m	✓ 1	An answer to at least 3 s.f. or the full substitution is required.
Part (c)(i)
The transitions from E₀ correspond to wavelengths of about 4.5 μm to 17 μm (7.8 μm in (a)), which lie in the infrared range emitted strongly by the surface, around 10 μm	✓ 1	Allow ECF from (a) and (b). At least one other transition wavelength or the range must be seen for this mark.
Photons near 0.5 μm have energies of about 2.5 eV, which do not match the spacing of these vibrational levels, so they are not absorbed	✓ 1	The comparison of photon energy with level spacing is required. "Sunlight is not infrared" alone scores [0].
Part (d)(i)
The excited molecules re-emit infrared photons in all directions (or share the energy with other molecules by collisions, and the warmed atmosphere radiates in all directions)	✓ 1
So part of the absorbed energy returns to the surface, which receives more power than from sunlight alone and reaches equilibrium at a higher temperature	✓ 1	"The gas traps heat" alone scores [0].
Part (e)(i)
Energy of one photon = 0.159 × 1.60 × 10⁻¹⁹ = 2.54 × 10⁻²⁰ J	✓ 1	Or hc/λ using the answer to (a). Allow ECF from (a).
Number = 1.5/2.54 × 10⁻²⁰ = 5.9 × 10¹⁹ s⁻¹ m⁻²	✓ 1	Accept 5.8–6.0 × 10¹⁹.

Answers: (a)(i) 7.8 × 10⁻⁶ m · (b)(i) 1.01 × 10⁻⁵ m · (e)(i) 5.9 × 10¹⁹ s⁻¹ m⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; that methane CH₄, water vapour H₂O, carbon dioxide CO₂, and nitrous oxide N₂O, are the main greenhouse gases and each of these has origins that are both natural and created by human activity; E.1 — photon energy E = hf; B.1 — Wien's displacement law λ_maxT = 2.9 × 10⁻³ m K Command term: Calculate

36B-2-46

Daily variation of albedo·B.2 Greenhouse effect

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

A satellite measures, for a region of tropical ocean, the intensity of sunlight arriving at the top of the atmosphere and the intensity scattered back into space by the region (the ocean, the atmosphere and any cloud). The graph shows both intensities during one day. On this day thick clouds formed over the region during the early afternoon.

Intensity of sunlight arriving at the top of the atmosphere (incident) and intensity scattered into space (scattered) above the region against local time (drawn to scale)

(a)

(i)

Determine the albedo of the region at 09:00 and at 16:00.

(2)

(b)

(i)

Suggest why the albedo at 16:00 is different from the albedo at 09:00.

(1)

(c)

(i)

Estimate the energy absorbed per square metre by the region during the day.

(3)

(d)

(i)

Over the 24 hours, the region emits infrared radiation into space at a mean intensity of 240 W m⁻². Deduce whether the region gains or loses energy by radiation over the 24 hours, and suggest what happens to the difference.

(2)

(e)

(i)

Determine the albedo of the region for the whole day.

(2)

(f)

(i)

Assume that 25 % of the energy in (c) is used to evaporate water from the sea surface. Specific latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹. Determine the mass of water evaporated per square metre during the day.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
09:00: 110/919 = 0.12	✓ 1	Accept 0.11–0.13. Readings within half a small square.
16:00: 312/650 = 0.48	✓ 1	Accept 0.45–0.51.
Part (b)(i)
The clouds that formed in the afternoon scatter (reflect) a much larger fraction of the sunlight than the clear sea and air, so the albedo is greater	✓ 1	Allow ECF from (a); the direction of the change must agree with the candidate's values.
Part (c)(i)
Energy absorbed = area between the incident curve and the scattered curve	✓ 1	Identifying the area between the curves (not under the incident curve) is required.
Area estimated by counting squares or by trapezia/rectangles, e.g. ≈ 7.4 × 10³ W m⁻² h	✓ 1	Accept 6.7–8.1 × 10³ W m⁻² h.
× 3600 s: ≈ 2.7 × 10⁷ J m⁻²	✓ 1	Accept 2.4–2.9 × 10⁷ J m⁻². Award [2 max] for 3.6 × 10⁷ J m⁻² (area under the incident curve).
Part (d)(i)
Emitted = 240 × 86 400 = 2.1 × 10⁷ J m⁻², less than the energy absorbed, so the region gains energy by radiation	✓ 1	Allow ECF from (c). The comparison must use the candidate's value.
The surplus is carried away from the region (towards higher latitudes) by winds and ocean currents, so the temperature need not rise	✓ 1	The transport of the surplus away from the region is required for this mark; "the region warms up" alone scores [0].
Part (e)(i)
Total incident energy per square metre = area under the incident curve ≈ 9.9 × 10³ W m⁻² h = 3.6 × 10⁷ J m⁻²	✓ 1	Accept 3.4–3.8 × 10⁷ J m⁻² (or the area in W m⁻² h, consistently with (c)).
Albedo = scattered energy/incident energy = (3.6 × 10⁷ − 2.7 × 10⁷)/3.6 × 10⁷ = 0.25	✓ 1	Allow ECF from (c). Accept 0.22–0.29 (0.26 from unrounded areas). 0.30, the simple mean of the two values in (a), scores [0]: the albedo is a ratio of total powers, and the low-albedo hours before 11:00 receive more of the day’s sunlight than the high-albedo hours after 15:00.
Part (f)(i)
Energy used = 0.25 × 2.7 × 10⁷ = 6.7 × 10⁶ J	✓ 1	Allow ECF from (c).
m = Q/L = 6.7 × 10⁶/2.26 × 10⁶ = 3.0 kg	✓ 1	Accept 2.6–3.3 kg.

Answers: (a)(i) 0.12; 0.48 · (c)(i) 2.7 × 10⁷ J m⁻² · (e)(i) 0.25 · (f)(i) 3.0 kg (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; that Earth's albedo varies daily and is dependent on cloud formations and latitude; the conservation of energy; B.1 — specific latent heat Q = mL Command term: Determine

37B-2-47

Fossil fuels and the enhanced greenhouse effect·B.2 Greenhouse effect

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

The graph shows the annual mean concentration C of carbon dioxide in the atmosphere, in parts per million (ppm), from 1960 to 2020.

An increase of 1 ppm in C corresponds to an extra 7.8 × 10¹² kg of carbon dioxide in the atmosphere. In 2015, human activity released 4.0 × 10¹³ kg of carbon dioxide. Molar mass of carbon dioxide = 0.044 kg mol⁻¹.

Annual mean concentration C of carbon dioxide in the atmosphere against year (drawn to scale)

(a)

(i)

Determine the rate of increase of C in 2015, in ppm per year.

(2)

(b)

(i)

Determine the fraction of the carbon dioxide released by human activity in 2015 that remained in the atmosphere.

(1)

(c)

(i)

Calculate the number of carbon dioxide molecules added to the atmosphere in 2015.

(2)

(d)

(i)

The reduction in the infrared power per unit area leaving the Earth that is caused by the increase in carbon dioxide is ΔF = 5.35 ln(C/280 ppm) W m⁻². Determine ΔF in 2015.

(2)

(e)

(i)

In 2015 the total energy released by the burning of fuels by human activity was 5.6 × 10²⁰ J. All of this energy ends up as thermal energy. Radius of the Earth = 6.37 × 10⁶ m; 1 year = 3.16 × 10⁷ s. Compare the effect on the Earth's energy balance of this thermal energy with the effect of the increase in carbon dioxide.

(3)

(f)

(i)

Carbon dioxide has natural and human sources. State one natural source of atmospheric carbon dioxide.

(1)

(ii)

Explain why the burning of fossil fuels is a primary cause of the enhanced greenhouse effect.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Tangent drawn to the curve at 2015	✓ 1	A tangent, not a chord over a long interval, is required.
Gradient ≈ 2.2 ppm per year	✓ 1	Accept 1.9–2.5 ppm per year.
Part (b)(i)
2.2 × 7.8 × 10¹²/4.0 × 10¹³ = 0.43	✓ 1	Allow ECF from (a). Accept 0.37–0.49.
Part (c)(i)
n = 2.2 × 7.8 × 10¹²/0.044 = 3.9 × 10¹⁴ mol	✓ 1	Allow ECF from (a).
N = nN_A = 3.9 × 10¹⁴ × 6.02 × 10²³ = 2.3 × 10³⁸	✓ 1	Accept 2.0–2.6 × 10³⁸. Award [1 max] for 5.5 × 10³⁸ (all the emitted CO₂).
Part (d)(i)
C in 2015 read from the graph ≈ 400 ppm	✓ 1	Accept 397–403 ppm.
ΔF = 5.35 ln(400/280) = 1.9 W m⁻²	✓ 1	Accept 1.8–2.0 W m⁻².
Part (e)(i)
Mean power = 5.6 × 10²⁰/3.16 × 10⁷ = 1.8 × 10¹³ W	✓ 1
Per unit area of the surface: 1.8 × 10¹³/4π(6.37 × 10⁶)² = 0.035 W m⁻²	✓ 1	Using πR² (0.14 W m⁻²) scores [0] for this mark.
This is about 1/55 of ΔF: the direct thermal energy is negligible compared with the effect of the extra carbon dioxide	✓ 1	Allow ECF from (d). The comparison must be numerical.
Part (f)(i)
Any one of: respiration of animals and plants; decay of dead organic matter; volcanic eruptions; release from the oceans; natural forest fires	✓ 1	Burning of fossil fuels or deforestation scores [0].
Part (f)(ii)
Fossil fuels release carbon that has been stored underground for millions of years, adding carbon dioxide faster than natural processes (oceans, plants) remove it, so the concentration rises and the atmosphere absorbs and re-emits more of the outgoing infrared	✓ 1	Both the rise in concentration (link to (b): over half the emission is removed, but the rest accumulates) and the effect on outgoing infrared are needed.

Answers: (a)(i) 2.2 ppm per year · (b)(i) 0.43 · (c)(i) 2.3 × 10³⁸ · (d)(i) 1.9 W m⁻² · (e)(i) 0.035 W m⁻², about 1/55 of the CO₂ effect (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — that methane CH₄, water vapour H₂O, carbon dioxide CO₂, and nitrous oxide N₂O, are the main greenhouse gases and each of these has origins that are both natural and created by human activity; that the augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect (Guidance: the burning of fossil fuels is a primary cause of the enhanced greenhouse effect); the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; B.3 — amount of substance n = N/N_A; A.3 — power as the rate of energy transfer; Tools — tangent to a curve Command term: Determine

38B-2-48

The surface temperature of Venus·B.2 Greenhouse effect

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

Venus orbits the Sun at a distance of 1.08 × 10¹¹ m. Its albedo is 0.76 and the mean temperature of its surface is 737 K.

Luminosity of the Sun = 3.83 × 10²⁶ W

Single-layer model of an atmosphere: energy flows per unit area (not to scale)

(a)

(i)

Show that the intensity of sunlight at the distance of Venus is about 2.6 × 10³ W m⁻².

(1)

(b)

(i)

Determine the equilibrium temperature T_e that Venus would have if it radiated as a black body and its atmosphere had no greenhouse effect.

(3)

(c)

(i)

The corresponding temperature for the Earth (albedo 0.30, solar constant 1.36 × 10³ W m⁻²) is 255 K. Explain why T_e for Venus is lower, even though Venus is closer to the Sun.

(2)

(d)

In a simple model, the atmosphere is a single layer that is transparent to sunlight and absorbs all the infrared radiation emitted by the surface. The layer, at temperature T_a, radiates as a black body both upwards and downwards. The surface, at temperature T_s, also radiates as a black body. The diagram shows the energy flows.

(i)

Show that, in equilibrium, T_s = 2^1/4T_e.

(2)

(ii)

Calculate T_s for Venus predicted by this model and compare it with the measured value.

(1)

(e)

(i)

A better model treats the atmosphere as N such layers stacked on top of each other, giving T_s = (N + 1)^1/4T_e. Determine N for Venus.

(2)

(f)

(i)

The pressure at the surface of Venus is 9.2 × 10⁶ Pa and the atmosphere there is almost pure carbon dioxide, of molar mass 0.044 kg mol⁻¹. Determine the density of the atmosphere at the surface.

(2)

(g)

(i)

Use your answer to (f) to suggest why the atmosphere of Venus behaves like a large number of absorbing layers.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
S = 3.83 × 10²⁶/(4π × (1.08 × 10¹¹)²) = 2.61 × 10³ W m⁻²	✓ 1	An answer to at least 3 s.f. or the full substitution is required.
Part (b)(i)
Mean incoming intensity = S/4 = 653 W m⁻² (projected disc πR², radiating surface 4πR²)	✓ 1	Allow ECF from (a).
Absorbed = (1 − 0.76) × 653 = 157 W m⁻²	✓ 1
T_e = (157/5.67 × 10⁻⁸)^1/4 = 229 K	✓ 1	Accept 228–231 K. Award [2 max] for 324 K (factor ¼ omitted) or 306 K (albedo used as the fraction absorbed).
Part (c)(i)
The sunlight at Venus is about 1.9 times more intense, but Venus absorbs only 24 % of it, against 70 % for the Earth	✓ 1	Allow ECF from (a) and (b).
so the absorbed mean intensity is smaller for Venus (157 W m⁻² against 238 W m⁻²), and T_e ∝ (absorbed intensity)^1/4 is lower	✓ 1	The comparison of absorbed intensities (not incident) is required.
Part (d)(i)
Only the layer radiates to space, so at the top: σT_a⁴ = absorbed solar intensity = σT_e⁴, giving T_a = T_e	✓ 1
Layer: absorbs σT_s⁴ and emits 2σT_a⁴, so σT_s⁴ = 2σT_a⁴ = 2σT_e⁴, hence T_s = 2^1/4T_e	✓ 1	ALT: surface balance σT_s⁴ = σT_e⁴ + σT_a⁴.
Part (d)(ii)
T_s = 2^1/4 × 229 = 272 K, far below the measured 737 K, so a single layer cannot explain the surface temperature of Venus	✓ 1	Allow ECF from (b). Accept 271–274 K. Both the value and the comparison are needed.
Part (e)(i)
(N + 1) = (T_s/T_e)⁴ = (737/229)⁴ = 107	✓ 1	Allow ECF from (b).
N ≈ 106 (about 100 layers)	✓ 1	Accept 100–110.
Part (f)(i)
n/V = p/RT = 9.2 × 10⁶/(8.31 × 737) = 1.50 × 10³ mol m⁻³	✓ 1	From pV = nRT.
ρ = 1.50 × 10³ × 0.044 = 66 kg m⁻³	✓ 1	Accept 65–67 kg m⁻³.
Part (g)(i)
The atmosphere contains a very large number of CO₂ molecules per unit volume (density about 50 times that of air on Earth), so infrared emitted from any level is absorbed within a short distance and re-emitted many times before it can escape to space	✓ 1	Allow ECF from (f).

Answers: (a)(i) 2.61 × 10³ W m⁻² · (b)(i) 229 K · (d)(ii) 272 K · (e)(i) 106 · (f)(i) 66 kg m⁻³ (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — the solar constant S; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4; albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; Guidance: the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants; Guidance: energy balance problems will include energy exchanged between the surface and the atmosphere of a body; the absorption of infrared radiation by the main greenhouse gases in terms of the molecular energy levels and the subsequent emission of radiation in all directions; B.1 — apparent brightness b = L/4πd²; B.3 — the ideal gas equation pV = nRT Command term: Determine

39B-2-49

Radiative forcing and an albedo-based response·B.2 Greenhouse effect

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine

In a simple climate model the Earth, together with its atmosphere, is treated as a single body at the mean surface temperature of 288 K. The body absorbs solar radiation (solar constant 1.36 × 10³ W m⁻², albedo 0.30) and radiates as a grey body with an effective emissivity e.

(a)

(i)

Show that e is about 0.61.

(2)

(b)

(i)

Explain why the value of e in (a) is less than the emissivity of the Earth's surface, which is close to 1.

(1)

(c)

(i)

A doubling of the concentration of carbon dioxide reduces the power per unit area escaping to space by 3.7 W m⁻² while the temperature is still 288 K. The albedo does not change. Determine the increase in the equilibrium temperature predicted by the model.

(3)

(d)

(i)

Immediately after the doubling, the extra 3.7 W m⁻² warms the upper layer of the oceans. Model this as a layer of water 70 m deep covering the whole surface of the Earth. Density of sea water = 1.03 × 10³ kg m⁻³; specific heat capacity of sea water = 4.0 × 10³ J kg⁻¹ K⁻¹. Determine the initial rate of rise of temperature of the layer, in K per year. (1 year = 3.16 × 10⁷ s)

(2)

(ii)

Hence explain why the temperature takes longer than about 3 years to reach the value found in (c).

(1)

(e)

(i)

It has been proposed that small particles could be spread in the upper atmosphere to increase the Earth's albedo. Determine the increase in albedo needed to keep the equilibrium temperature at 288 K after the doubling.

(2)

(ii)

Suggest one reason why this would not exactly cancel the effect of the extra carbon dioxide everywhere on the Earth.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Absorbed mean intensity = (1 − 0.30) × 1360/4 = 238 W m⁻²	✓ 1	The factor ¼ must be seen.
e = 238/(5.67 × 10⁻⁸ × 288⁴) = 238/390 = 0.610	✓ 1	An answer to at least 3 s.f. is required.
Part (b)(i)
Greenhouse gases absorb much of the infrared emitted by the surface and re-emit part of it back down, so the power per unit area escaping to space (238 W m⁻²) is less than the surface emits (≈ 390 W m⁻²)	✓ 1	Allow ECF from (a).
Part (c)(i)
New effective emissivity e′ = (238 − 3.7)/390 = 0.601	✓ 1	Allow ECF from (a). Award this mark for the idea that the emitted power at 288 K falls by 3.7 W m⁻².
New equilibrium: e′σT′⁴ = 238 W m⁻², so T′ = (238/(0.601 × 5.67 × 10⁻⁸))^1/4 = 289.1 K	✓ 1	The absorbed intensity is unchanged, so the planet must again radiate 238 W m⁻².
Increase ≈ 1.1 K	✓ 1	Accept 1.0–1.2 K. ALT: T′ = 288 × (238/234.3)^1/4.
Part (d)(i)
Heat capacity per m² = 70 × 1030 × 4.0 × 10³ = 2.9 × 10⁸ J K⁻¹	✓ 1
Rate = 3.7/2.9 × 10⁸ = 1.3 × 10⁻⁸ K s⁻¹ = 0.41 K per year	✓ 1	Accept 0.40–0.42 K per year.
Part (d)(ii)
1.1 K at 0.41 K per year would take about 2.7 years, but as the temperature rises the body radiates more, so the imbalance (and the rate of warming) decreases and the approach to equilibrium is gradual	✓ 1	Allow ECF from (c) and (d)(i). The decreasing imbalance is required.
Part (e)(i)
The absorbed mean intensity must fall by 3.7 W m⁻²: Δα × 1360/4 = 3.7	✓ 1	The factor ¼ must be used.
Δα = 0.011 (albedo 0.311)	✓ 1	Accept 0.010–0.011. Award [1 max] for 0.0027 (factor ¼ omitted).
Part (e)(ii)
Any one: the particles reduce absorbed sunlight only on the day side and mostly at low latitudes, while the extra CO₂ reduces the outgoing infrared day and night at all latitudes; the particles fall out of the atmosphere and must be replaced continually; the CO₂ concentration keeps rising	✓ 1	Accept any valid physical reason.

Answers: (a)(i) 0.610 · (c)(i) 1.1 K · (d)(i) 0.41 K per year · (e)(i) 0.011 (the remaining parts are explanations — see the table above)

Syllabus understandingB.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature as given by emissivity = power radiated per unit area/σT⁴; albedo as a measure of the average energy reflected off a macroscopic system as given by albedo = total scattered power/total incident power; the solar constant S; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4; Guidance: the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo, emissivity, and solar or other constants; that the augmentation of the greenhouse effect due to human activities is known as the enhanced greenhouse effect (Guidance: the burning of fossil fuels is a primary cause of the enhanced greenhouse effect); B.1 — specific heat capacity Q = mcΔT Command term: Determine

B.2 Greenhouse effect: IB Physics HL exam-style questions

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