B.2 Greenhouse effect flashcards: IB Physics SL and HL

All 70 B.2 Greenhouse effect cards

1. DefinitionSL & HLData booklet: No – memorise

   Define the solar constant and state its accepted value.

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   The solar constant S is the (mean) intensity of solar radiation arriving at the top of the Earth's atmosphere on a surface held perpendicular to the Sun's rays, at the mean Earth–Sun distance. S ≈ 1360 W m⁻² (1.36 × 10³ W m⁻²); it is an intensity, so the unit is W m⁻² (power per unit area), a scalar. Exam tip: the mark is lost for saying "the energy from the Sun reaching the Earth" — you must say power per unit area, perpendicular to the beam, above the atmosphere. It is not the average intensity over the Earth's surface: that is S/4 ≈ 340 W m⁻².

2. DefinitionSL & HLData booklet: Yes

   Define albedo.

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   Albedo α is the ratio of the total scattered (reflected) power to the total incident power for a body or surface: α = total scattered power / total incident power. It is a dimensionless ratio between 0 and 1 (sometimes quoted as a percentage), a scalar. Earth's mean global albedo ≈ 0.3, so about 30% of incident solar power is reflected back to space. Exam tip: students write "reflected light divided by incident light" — say power (or intensity), and say total, since albedo includes scattering by clouds, aerosols and the surface, not just specular reflection. Albedo depends on surface type, cloud cover, season and angle of incidence.

3. DefinitionSL & HL

   Define emissivity and state its value for a black body.

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   Emissivity e is the ratio of the power radiated per unit area by a body to the power radiated per unit area by a black body at the same temperature. It is dimensionless, 0 ≤ e ≤ 1, and e = 1 for a perfect black body; Earth's surface is often modelled with e ≈ 1 in the infrared. Exam tip: the common loss is omitting "at the same temperature" or "per unit area". By Kirchhoff's law a good emitter is a good absorber, so a high-emissivity surface also absorbs strongly at the same wavelengths — do not confuse e with albedo α, which refers to reflection of incoming solar radiation.

4. DefinitionSL & HL

   Define a black body.

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   A black body is a (theoretical) body that absorbs all the electromagnetic radiation incident on it at every wavelength, reflecting and transmitting none; it is therefore also the best possible emitter at every wavelength for a given temperature, with emissivity e = 1. Its emitted spectrum depends only on its absolute temperature. Exam tip: "a body that absorbs all radiation" alone often scores, but add "at all wavelengths" and "perfect emitter" for the second mark. A black body need not look black — the Sun and the filament of a lamp are treated as black bodies. Its albedo is zero, since nothing is scattered.

5. DefinitionSL & HL

   Define a greenhouse gas and name the four required by the syllabus.

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   A greenhouse gas is an atmospheric gas that strongly absorbs infrared radiation emitted by the Earth's surface and re-emits it in all directions, including back towards the surface. The four named in the 2025 syllabus are methane CH₄, water vapour H₂O, carbon dioxide CO₂ and nitrous oxide N₂O. Exam tip: students lose marks by naming oxygen, nitrogen or ozone-depleting CFCs as "the" greenhouse gases; O₂ and N₂ are symmetric diatomic molecules with no changing dipole, so they do not absorb IR. Water vapour is the most abundant natural greenhouse gas; CO₂ is the main anthropogenic contributor by radiative forcing.

6. DefinitionSL & HL

   Define the (natural) greenhouse effect.

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   The greenhouse effect is the warming of a planet's surface caused by greenhouse gases in its atmosphere absorbing infrared radiation emitted by the surface and re-emitting it in all directions, so that a fraction returns to the surface and raises its equilibrium temperature above the value it would have with no atmosphere. Exam tip: the mark scheme wants three ideas — surface emits IR, greenhouse gases absorb it, re-emission occurs in all directions including back down. Answers saying the atmosphere "traps heat like a blanket" or "reflects the radiation back" score zero: absorption and re-emission, not reflection. Without it Earth would sit near 255 K rather than 288 K.

7. DefinitionSL & HL

   Define the enhanced greenhouse effect.

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   The enhanced greenhouse effect is the additional warming of the Earth's surface caused by the increase in the concentration of greenhouse gases in the atmosphere due to human activity — principally the burning of fossil fuels (CO₂), agriculture and livestock (CH₄, N₂O) and deforestation. More absorption and back-radiation of infrared raises the surface equilibrium temperature. Exam tip: students must distinguish it from the natural greenhouse effect, which is essential for life; the word "enhanced" means extra, not "the greenhouse effect". Do not confuse it with ozone-layer depletion or with acid rain — a very common Paper 2 error. Evidence: ice cores, Keeling curve, correlated global mean temperature rise.

8. DefinitionSL & HL

   Define radiative (thermal) equilibrium for a planet and state what it implies.

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   A planet is in radiative equilibrium when the total power it absorbs from its star equals the total power it radiates into space, so its average surface temperature stays constant. Formally, absorbed power (1 − α)SπR² = emitted power eσ(4πR²)T⁴. Exam tip: the mark is for "power in = power out", not "energy in = energy out" without a time reference; and for stating the consequence — constant average temperature. If absorbed power exceeds emitted power the planet warms until the higher T⁴ restores balance, which is why a temperature rise is self-limiting rather than runaway in this simple model. Areas differ: πR² intercepts, 4πR² emits.

9. DefinitionSL & HL

   Define resonance as it applies to infrared absorption by greenhouse gas molecules.

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   Resonance occurs when the frequency of the incident radiation matches a natural frequency of vibration of the molecule, so energy transfer to the molecule is a maximum and the photon is absorbed. Greenhouse gas molecules have natural vibrational (bending and stretching) frequencies in the infrared, matching the IR emitted by the Earth's surface at ~288 K. Exam tip: the required marking points are "natural frequency of vibration of the molecule" and "matches the frequency of the infrared radiation", giving maximum energy transfer. Saying the gas "absorbs IR because it is a greenhouse gas" is circular and scores nothing. Symmetric molecules (N₂, O₂) have no IR-active modes with a changing dipole moment.

10. DefinitionSL & HL

    Define intensity of radiation and explain why the average incident intensity on Earth is S/4.

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    Intensity is the power of radiation incident per unit area of a surface perpendicular to the radiation, I = P/A, unit W m⁻², a scalar. The Earth intercepts solar power over its circular cross-section, area πR², but that power is spread over the whole rotating spherical surface, area 4πR². Hence the average intensity over the surface is SπR²/(4πR²) = S/4 ≈ 1360/4 = 340 W m⁻². Exam tip: the mark scheme demands both areas be identified — πR² for interception and 4πR² for emission. Writing "divide by 4 because half the Earth is dark" scores zero; that would give S/2.

11. DefinitionSL & HL

    Define luminosity and apparent brightness.

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    Luminosity L is the total power radiated by a body (e.g. a star) in all directions, unit W. Apparent brightness b is the power received per unit area at the observer, measured perpendicular to the direction of the source, unit W m⁻²; b = L/(4πd²). Both are scalars. Exam tip: "brightness" without "per unit area" loses the mark, and luminosity must be "total power radiated", not "energy given out". For the Sun, b at the top of Earth's atmosphere is exactly the solar constant S ≈ 1360 W m⁻². Luminosity is intrinsic to the source; apparent brightness depends on distance.

12. DefinitionSL & HL

    Define positive feedback and describe the ice–albedo feedback.

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    Positive feedback is a process in which a change in a system produces an effect that reinforces (amplifies) the original change. Ice–albedo feedback: warming melts highly reflective ice and snow (α ≈ 0.6–0.9), exposing darker ocean and land (α ≈ 0.1–0.2); the mean albedo falls, so more solar power is absorbed, causing further warming and further melting. Exam tip: you must complete the causal loop back to the starting quantity; stopping at "more radiation is absorbed" loses the final mark. Contrast with negative feedback, e.g. a warmer surface radiates more strongly (σT⁴), which opposes further warming. Other positive feedbacks: methane release from permafrost, reduced CO₂ solubility in warmer oceans.

13. DefinitionSL & HL

    Define the equilibrium (effective) temperature of a planet and state its value for Earth.

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    The equilibrium or effective temperature is the uniform surface temperature a planet would have if it behaved as a black body in radiative equilibrium with its star, with no atmospheric greenhouse effect: T = [S(1 − α)/(4σ)]^¼. For Earth with S = 1360 W m⁻² and α = 0.30, T ≈ 255 K (−18 °C), compared with the observed mean surface temperature of about 288 K (15 °C). Exam tip: the 33 K difference is the natural greenhouse effect and is the standard follow-up question; quote both numbers. Marks are lost for forgetting the fourth root, for omitting (1 − α), or for using S instead of S/4.

14. DefinitionSL & HLData booklet: Yes

    State the Stefan–Boltzmann law and define every quantity in it.

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    The power radiated by a body is proportional to its surface area and to the fourth power of its absolute temperature: P = eσAT⁴. P is the total power radiated in W, e is emissivity (dimensionless, e = 1 for a black body), σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ is the Stefan–Boltzmann constant, A is the total emitting surface area in m², and T is the absolute (kelvin) temperature. Exam tip: T must be in kelvin — using °C is the single most common error, and doubling the Celsius temperature does not multiply P by 16. For a spherical planet or star A = 4πr², so L = 4πr²σT⁴.

15. DefinitionSL & HLData booklet: Yes

    State Wien's displacement law and define its terms.

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    Wien's displacement law states that the wavelength at which the emitted power per unit wavelength of a black body is a maximum is inversely proportional to its absolute temperature: λ_max T = 2.9 × 10⁻³ m K. λ_max is in metres and T in kelvin; the constant is Wien's constant. Exam tip: λ_max is the peak of the spectral curve, not the only or the average wavelength emitted — that phrasing loses the mark. Sun at ~5800 K gives λ_max ≈ 5.0 × 10⁻⁷ m (500 nm, visible); Earth at 288 K gives λ_max ≈ 1.0 × 10⁻⁵ m (10 μm, infrared). This difference is why greenhouse gases matter.

16. DefinitionSL & HL

    Distinguish correlation from causation in the context of climate data.

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    A correlation is a statistical association between two measured variables (e.g. atmospheric CO₂ concentration and global mean surface temperature over time); causation means a change in one variable physically produces the change in the other. Correlation alone does not establish causation: a third variable, or reverse causation, may be responsible. Exam tip: this is the standard Nature of Science mark. A good answer states that the causal link here is supported by an independent physical mechanism — molecular resonance of CO₂ with infrared — plus laboratory absorption spectra and climate models that reproduce observations only when anthropogenic emissions are included. Ice cores give the long-term CO₂–temperature record; the Keeling curve gives the modern rise.

17. EquationSL & HLData booklet: Yes

    State the data-booklet equation for albedo, define its terms and give a sanity check.

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    Booklet: albedo = total scattered power / total incident power, usually written α. Both powers are in W; α is dimensionless with 0 ≤ α ≤ 1. Valid for any surface or whole planet, over the relevant (solar, short-wave) waveband and averaged over angle. Data-booklet status: printed. Common misuse: using α as the fraction absorbed — the absorbed fraction is (1 − α); another is applying planetary albedo to the outgoing infrared, where emissivity, not albedo, applies. Sanity check: fresh snow α ≈ 0.85, ocean α ≈ 0.06, thick cloud α ≈ 0.7, Earth mean α ≈ 0.3, so 0.3 × 340 ≈ 102 W m⁻² is reflected and 238 W m⁻² absorbed.

18. EquationSL & HLData booklet: Yes

    State P = eσAT⁴, define every symbol with its SI unit, and give the forms you must be able to use.

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    Booklet: P = eσAT⁴. P = power radiated (W); e = emissivity (no unit); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; A = surface area (m²); T = absolute temperature (K). Valid for radiation from any body; e = 1 for a black body. Derived forms: intensity emitted per unit area I = eσT⁴; for a sphere L = 4πr²eσT⁴; ratio form P₂/P₁ = (T₂/T₁)⁴ when e and A are fixed. Data booklet: printed. Common misuse: using Celsius, or using cross-sectional area πr² instead of 4πr² for emission. Sanity check: Earth at 288 K emits σT⁴ = 5.67 × 10⁻⁸ × 288⁴ ≈ 390 W m⁻².

19. EquationSL & HLData booklet: Yes

    State Wien's displacement law as printed, define its terms and give the rearranged forms.

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    Booklet: λ_max T = 2.9 × 10⁻³ m K. λ_max = wavelength of maximum spectral emitted power (m); T = absolute temperature (K); the constant has unit m K. Valid for black-body (and approximately grey-body) emitters. Rearrangements you must use: T = 2.9 × 10⁻³/λ_max and λ_max = 2.9 × 10⁻³/T. Data booklet: printed. Common misuse: substituting λ in nm or μm without converting to metres, or using T in °C. Sanity check: T = 5800 K → λ_max = 2.9 × 10⁻³/5800 = 5.0 × 10⁻⁷ m; T = 288 K → 1.0 × 10⁻⁵ m, i.e. 10 μm in the infrared.

20. EquationSL & HLData booklet: Yes

    State b = L/(4πd²), define its terms and explain its role in finding the solar constant.

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    Booklet: b = L/(4πd²). b = apparent brightness, the power received per unit area (W m⁻²); L = luminosity, total power radiated by the source (W); d = distance from source to observer (m). Valid for a point (or spherical) source radiating isotropically into a transparent medium with no absorption. Data booklet: printed. Common misuse: forgetting the square, or using radius of the star instead of the distance to it. Applied to Earth, b is the solar constant: with L_Sun = 3.85 × 10²⁶ W and d = 1.50 × 10¹¹ m, b = 3.85 × 10²⁶/(4π × (1.50 × 10¹¹)²) = 1.36 × 10³ W m⁻². Inverse-square: doubling d quarters b.

21. EquationSL & HLData booklet: No – derive

    Write the expression for the solar power absorbed by a planet and define each term.

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    P_absorbed = (1 − α)SπR², where α = albedo (no unit), S = solar constant at that planet's orbit (W m⁻²), R = planet radius (m), so P is in W. Per unit surface area this is (1 − α)S/4. Valid when the planet is treated as a sphere at a fixed distance with a uniform mean albedo. Data booklet: not printed — derive from albedo and geometry. Common misuse: using 4πR² for the interception area, or omitting (1 − α). Sanity check: Earth R = 6.4 × 10⁶ m, πR² = 1.29 × 10¹⁴ m², P = 0.7 × 1360 × 1.29 × 10¹⁴ ≈ 1.2 × 10¹⁷ W.

22. EquationSL & HLData booklet: No – derive

    Write the expression for the power radiated by a planet to space and state the assumption used.

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    P_emitted = eσ(4πR²)T⁴, with e = emissivity (usually taken as 1 in the infrared), σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, R = planet radius (m), T = mean surface (or effective) temperature (K); P in W. Per unit area the outgoing flux is eσT⁴. Assumption: the whole surface radiates at a single uniform temperature as a black body. Data booklet: derived from the printed P = eσAT⁴ with A = 4πR². Common misuse: using the albedo here — albedo applies to incoming short-wave radiation, emissivity to outgoing long-wave. Sanity check: at 255 K, σT⁴ = 240 W m⁻², which equals the absorbed 238 W m⁻² as required.

23. EquationSL & HLData booklet: No – derive

    Derive T = [S(1 − α)/(4σ)]^¼ and state every symbol with its unit.

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    Set absorbed power equal to emitted power: (1 − α)SπR² = σ(4πR²)T⁴ (e = 1). Cancel πR²: (1 − α)S = 4σT⁴, so T = [S(1 − α)/(4σ)]^¼. T = equilibrium (effective) temperature in K; S = solar constant (W m⁻²); α = albedo (dimensionless); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Valid for a black-body planet with no atmosphere and uniform temperature. Booklet: not printed — derive. Common misuse: forgetting the fourth root or the factor 4. Sanity check: S = 1360, α = 0.30 → 1360 × 0.70/(4 × 5.67 × 10⁻⁸) = 4.20 × 10⁹, fourth root ≈ 255 K.

24. EquationSL & HLData booklet: No – derive

    State the rearranged form used to find a planet's albedo from its measured temperature.

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    From (1 − α)S = 4σT⁴, α = 1 − 4σT⁴/S. Symbols: α = albedo (dimensionless), σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, T = effective temperature (K), S = solar constant at that orbit (W m⁻²). Valid only if the planet is in radiative equilibrium with emissivity 1 and negligible internal heat source. Booklet: not printed — rearrange. Common misuse: substituting the observed surface temperature of a planet with a strong greenhouse atmosphere (Venus, 737 K), which gives a nonsensical negative albedo — you must use the effective radiating temperature. Sanity check: T = 255 K, S = 1360 → 4 × 5.67 × 10⁻⁸ × 255⁴/1360 = 0.70, α = 0.30.

25. EquationSL & HLData booklet: No – derive

    Show how the equilibrium temperature scales with solar constant and albedo.

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    From T = [S(1 − α)/(4σ)]^¼, T ∝ [S(1 − α)]^¼, so T ∝ S^¼ at fixed albedo. Symbols as before, T in K. A fractional change gives ΔT/T ≈ ¼ × ΔS/S for small changes. Booklet: not printed — derive. Common misuse: assuming T doubles when S doubles; in fact T increases only by 2^¼ = 1.19, i.e. 255 K → 303 K. Sanity check: a 1% rise in S raises T by about 0.25%, ≈ 0.6 K. Similarly, reducing α from 0.30 to 0.28 raises (1 − α) by 2.9%, so T rises by ≈ 0.7%, about 1.8 K — the quantitative form of ice–albedo feedback.

26. EquationSL & HLData booklet: Yes

    Give the equation linking a star's luminosity, radius and surface temperature, and apply it to the Sun.

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    L = 4πr²σT⁴ (from booklet P = eσAT⁴ with e = 1 and A = 4πr²). L = luminosity (W); r = stellar radius (m); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; T = surface temperature (K). Valid for a spherical black-body emitter. Rearranged: r = √(L/(4πσT⁴)) and T = [L/(4πr²σ)]^¼. Booklet: derived from a printed equation. Common misuse: using the Earth–Sun distance for r; that belongs in b = L/(4πd²). Sanity check: r_Sun = 7.0 × 10⁸ m, T = 5800 K → L = 4π(7.0 × 10⁸)² × 5.67 × 10⁻⁸ × 5800⁴ ≈ 3.9 × 10²⁶ W.

27. EquationSL & HLData booklet: No – derive

    Write the full energy-balance equation for a planet with an emissivity less than 1 and state each term.

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    (1 − α)SπR² = eσ(4πR²)T⁴, i.e. (1 − α)S/4 = eσT⁴ per unit surface area. α = albedo (dimensionless); S = solar constant (W m⁻²); e = emissivity in the infrared (dimensionless); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; T = mean surface temperature (K); R = radius (m). Rearranged: T = [(1 − α)S/(4eσ)]^¼. Booklet: built from printed albedo and P = eσAT⁴. Common misuse: setting e = 1 − α, which confuses two independent optical properties at two different wavelength ranges. Sanity check: with α = 0.3 and e = 0.61, T = [238/(0.61 × 5.67 × 10⁻⁸)]^¼ ≈ 288 K — the observed value.

28. Graph/diagramSL & HL

    Describe the black-body spectral curves of the Sun and the Earth on the same axes.

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    Axes: x = wavelength λ/m (usually log scale, 10⁻⁷ to 10⁻⁴ m), y = intensity per unit wavelength/W m⁻² m⁻¹ (or relative intensity). Each curve rises steeply from short λ to a single peak, then falls with a long tail; curves never cross. Sun (~5800 K) peaks at λ_max ≈ 5.0 × 10⁻⁷ m in the visible; Earth (~288 K) peaks at ≈ 1.0 × 10⁻⁵ m in the infrared, and is far lower and broader. Area under a curve ∝ total power per unit area = eσT⁴. Raising T moves the peak left (Wien) and raises the whole curve (Stefan). Exam tip: label both peaks with values.

29. Graph/diagramSL & HL

    Explain how a black-body curve changes when the temperature is increased, and how to extract T from it.

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    Axes: intensity per unit wavelength (W m⁻² m⁻¹) against wavelength (m). Increasing T shifts the peak to shorter wavelength (λ_max ∝ 1/T) and increases the intensity at every wavelength, so the higher-T curve lies entirely above the lower-T one and the enclosed area grows as T⁴. To find T: read λ_max at the maximum, then T = 2.9 × 10⁻³/λ_max. To find total emitted flux, measure the area under the curve; it equals eσT⁴. Exam tip: sketch marks require a smooth asymmetric curve starting at the origin, a single maximum, and curves that do not intersect; a symmetric "bell" or crossing curves lose marks.

30. Graph/diagramSL & HL

    Describe the Earth energy-balance (Sankey/flow) diagram and the numbers you should be able to quote.

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    The diagram shows power per unit area (W m⁻²) as arrow widths: incoming solar 340 (= S/4), of which ≈ 100 (30%) is reflected by clouds, aerosols and surface, and ≈ 240 is absorbed. Outgoing: surface radiates ≈ 390 W m⁻² (σ × 288⁴) in the infrared; greenhouse gases absorb most of it and re-radiate ≈ 330 back down, with ≈ 240 finally leaving the top of the atmosphere. Non-radiative surface losses (convection and latent heat of evaporation) account for roughly 100. Balance: at the top of the atmosphere in ≈ out. Exam tip: check arrows balance at each level; a stated imbalance of ~1 W m⁻² represents current global warming.

31. Graph/diagramSL & HL

    Describe the graph of atmospheric CO₂ concentration and global mean temperature against time, and what may and may not be concluded.

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    Axes: x = time/year (Keeling curve 1958–present, or ice-core record over 800 000 years, in ka before present); y₁ = CO₂ concentration/ppm, y₂ = global mean surface temperature anomaly/°C. The Keeling curve rises from ≈ 315 ppm to over 420 ppm with an annual sawtooth (Northern-hemisphere plant growth) superimposed on a rising, slightly concave-up trend; the temperature anomaly rises roughly in step, with more scatter. Gradient of the CO₂ curve gives the rate of increase in ppm yr⁻¹. Exam tip: the two curves being correlated does not by itself prove causation — cite the resonance absorption mechanism and laboratory spectra as the causal evidence.

32. Graph/diagramSL & HL

    Describe the linearisation used to determine Wien's constant experimentally.

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    Plot λ_max/m (y) against 1/T /K⁻¹ (x) for a filament lamp or a set of black-body sources at known temperatures. Since λ_max = (2.9 × 10⁻³)(1/T), the graph is a straight line through the origin with gradient equal to Wien's constant in m K. Determine the constant from gradient = Δλ_max/Δ(1/T), using the largest available triangle; the y-intercept should be zero within uncertainty, and a non-zero intercept indicates a systematic error (e.g. a wavelength calibration offset in the spectrometer). Exam tip: plot λ_max against 1/T, not against T — a λ_max–T plot is a hyperbola and cannot be used to find a gradient reliably.

33. Graph/diagramSL & HL

    Describe two linearisations used to verify the Stefan–Boltzmann law.

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    Method 1: plot P/W (y) against T⁴/K⁴ (x). P = eσAT⁴ gives a straight line through the origin of gradient eσA, from which σ (or e) is found if A is measured. Method 2: plot ln(P/W) (y) against ln(T/K) (x); ln P = 4 ln T + ln(eσA), a straight line of gradient 4 confirming the fourth-power dependence, with y-intercept ln(eσA). Method 2 is preferred when the exponent itself is being tested. Exam tip: quote the gradient with its uncertainty from max/min gradient lines and compare with 4; state that T must be absolute, and that background radiation σT₀⁴ from the surroundings must be subtracted for accurate work.

34. Graph/diagramSL & HL

    Describe the atmospheric absorption (transmission) spectrum and identify the atmospheric window.

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    Axes: x = wavelength/μm (about 0.1 to 70 μm, log scale), y = fractional absorption or transmission (0 to 1, dimensionless). The atmosphere transmits strongly in the visible near 0.5 μm, where the Sun's output peaks, so sunlight reaches the surface. In the infrared there are strong absorption bands due to H₂O, CO₂ (notably ≈ 15 μm), CH₄ and N₂O, separated by a region of high transmission from about 8 to 13 μm — the atmospheric window — through which surface radiation escapes to space. Adding greenhouse gases narrows this window, reducing outgoing power and forcing a higher surface temperature. Exam tip: link Earth's λ_max ≈ 10 μm to the window's position.

35. Concept/explainSL & HL

    Explain why the average solar intensity incident on the Earth is taken as S/4 rather than S.

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    • The Sun's rays reaching Earth are effectively parallel, so the Earth intercepts solar power over its cross-sectional disc of area πR²
    • Total intercepted power = SπR²
    • This energy is redistributed over the whole spherical surface of area 4πR² by rotation and by atmospheric and oceanic circulation
    • Average incident intensity = SπR²/4πR² = S/4 = 1360/4 ≈ 340 W m⁻²
    • Multiplying by (1 − α) gives the average absorbed intensity ≈ 238 W m⁻²
    • The factor 4 is purely geometric and has nothing to do with albedo or emissivity. Exam tip: incomplete answers say "only half the Earth is lit" and use S/2, or quote 340 W m⁻² with no geometric justification; the mark is for the πR² : 4πR² ratio.
36. Concept/explainSL & HL

    Explain the mechanism of the natural greenhouse effect.

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    • Incoming solar radiation is mostly short-wavelength (visible and near-IR) and passes through the atmosphere largely unabsorbed
    • It is absorbed by the surface, warming it
    • The warm surface at ≈288 K radiates long-wavelength infrared, peaking near 10 μm
    • Greenhouse gases (H₂O, CO₂, CH₄, N₂O) absorb this IR because their natural frequencies of molecular vibration lie in the infrared, giving resonance
    • The excited molecules re-emit IR in all directions, so a fraction is returned to the surface as back-radiation
    • The surface therefore receives extra power and settles at a higher equilibrium temperature, 288 K rather than 255 K. Exam tip: "the gases trap heat like a blanket" scores nothing; you must state short-wave in, long-wave out, absorption and re-emission downwards.
37. Concept/explainSL & HL

    Explain why CO₂, H₂O, CH₄ and N₂O absorb infrared strongly but N₂ and O₂ do not.

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    • Absorbing an IR photon changes a molecule's vibrational (bending or stretching) energy, which requires the vibration to change the molecule's dipole moment
    • CO₂, H₂O, CH₄ and N₂O have natural frequencies of vibration that lie in the infrared region
    • Resonance occurs when the frequency of the incident IR matches a natural frequency, giving strong absorption
    • N₂ and O₂ are symmetric diatomic molecules whose single vibration produces no change of dipole moment, so the infrared wave cannot couple to it and no resonant absorption occurs — they are effectively transparent in the infrared
    • Each gas absorbs in characteristic wavelength bands, visible as gaps in the atmospheric transmission spectrum. Exam tip: students claim greenhouse gases are "heavier" or "denser"; the mark is for natural frequency of vibration matching the IR frequency (resonance).
38. Concept/explainSL & HL

    Explain the enhanced greenhouse effect and distinguish it from the natural greenhouse effect.

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    • The natural greenhouse effect raises Earth's mean surface temperature from 255 K to 288 K and is essential for liquid water and life
    • Human activity raises greenhouse gas concentrations: burning fossil fuels and deforestation (CO₂), livestock, rice and landfill (CH₄), fertilisers (N₂O)
    • More of the IR emitted by the surface is absorbed and re-emitted downwards
    • Outgoing radiation at the top of the atmosphere temporarily falls below the absorbed solar power, giving a positive energy imbalance
    • Earth warms until balance is restored at a higher surface temperature
    • Consequences include sea-level rise, ice loss and changed weather patterns. Exam tip: always distinguish the beneficial natural effect from the human-caused enhancement; do not treat them as the same phenomenon.
39. Concept/explainSL & HLData booklet: Yes

    Explain why the Sun's radiation is mainly visible while the Earth's is mainly infrared, and why this matters.

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    • Both bodies radiate approximately as black bodies
    • Wien's law λ_max T = 2.9 × 10⁻³ m K gives peak wavelength ∝ 1/T
    • Sun's photosphere ≈5800 K → λ_max ≈ 5.0 × 10⁻⁷ m (500 nm, visible)
    • Earth's surface ≈288 K → λ_max ≈ 1.0 × 10⁻⁵ m (10 μm, infrared)
    • The atmosphere is largely transparent at 500 nm but strongly absorbing at 10 μm because of CO₂ and H₂O bands
    • This wavelength mismatch is the entire basis of the greenhouse effect
    • The area under the Sun's curve is far greater since P/A = σT⁴. Exam tip: sketched curves must show the Earth curve peaking far to the right AND with a much lower peak; drawing the two curves the same height loses the mark.
40. Concept/explainSL & HLData booklet: No – derive

    Explain the significance of the difference between 255 K and 288 K for the Earth.

    Show answer

    • A black-body Earth with no atmosphere and α = 0.30 balances at T = [S(1 − α)/(4σ)]^¼ ≈ 255 K (−18 °C)
    • The measured mean surface temperature is 288 K (15 °C), a difference of 33 K
    • The extra 33 K is due to the natural greenhouse effect: back-radiation from greenhouse gases supplements the solar power absorbed at the surface
    • 255 K is the effective radiating temperature of the planet as seen from space, roughly the temperature of the upper troposphere that emits to space
    • Overall energy balance still holds: 238 W m⁻² in = 238 W m⁻² out at the top of the atmosphere. Exam tip: say that 255 K applies at the top of the atmosphere; do not simply call the model "wrong".
41. Concept/explainSL & HLData booklet: Yes

    Explain what albedo is and what determines the Earth's mean value.

    Show answer

    • Albedo α = total scattered (reflected) power / total incident power; a dimensionless ratio between 0 and 1
    • Earth's mean planetary albedo ≈ 0.30, so 30% of incident solar power is returned to space without being absorbed
    • It depends strongly on surface type: fresh snow and ice ≈0.8–0.9, thick cloud ≈0.5–0.7, desert sand ≈0.4, forest ≈0.1, open ocean ≈0.06
    • It also depends on cloud cover, season, latitude and angle of incidence (oceans reflect much more at low Sun angles)
    • A higher albedo means less absorbed power and a lower equilibrium temperature. Exam tip: albedo is a ratio of powers, has no unit, and is not the same as emissivity, which describes infrared emission.
42. Concept/explainSL & HL

    Explain the ice–albedo feedback and identify what kind of feedback it is.

    Show answer

    • Warming melts sea ice and snow cover, exposing darker ocean and land beneath
    • The surface albedo falls, so a greater fraction of the incident solar power is absorbed rather than scattered
    • The extra absorbed power raises the temperature further, which melts more ice
    • The loop reinforces the original change, so it is a positive feedback
    • It is strongest at high latitudes, producing polar amplification of warming
    • Contrast with a negative feedback, e.g. warming → more evaporation → more low cloud → higher albedo → cooling, which opposes the change. Exam tip: students describe the melting but never state "positive feedback" or say which quantity increases; always close the loop back to the starting cause.
43. Concept/explainSL & HL

    Discuss why the correlation between CO₂ concentration and temperature in ice-core data is not by itself proof of causation.

    Show answer

    • Ice cores from Antarctica and Greenland trap air bubbles giving past CO₂ concentration, while oxygen isotope ratios give past temperature
    • Records spanning ~800 000 years show CO₂ and temperature varying together — a strong correlation
    • Correlation alone does not establish causation: a third factor (orbital variation) or reverse causation (warming oceans releasing dissolved CO₂) could produce it
    • Causation is supported by an independent physical mechanism — IR absorption by CO₂ measured in the laboratory — and by models that reproduce the record only when greenhouse gases are included
    • Carbon isotope ratios identify the modern rise as fossil-fuel in origin. Exam tip: quoting the correlation alone is not evidence of cause; you must supply the mechanism.
44. Concept/explainSL & HLData booklet: No – derive

    Explain, in terms of energy balance, how adding greenhouse gases raises the Earth's mean surface temperature.

    Show answer

    • In equilibrium the power absorbed equals the power radiated to space, so the mean temperature is constant
    • Absorbed: S(1 − α)/4 ≈ 238 W m⁻²; emitted: σT_eff⁴ with T_eff ≈ 255 K
    • Adding greenhouse gases reduces the outgoing IR escaping at the top of the atmosphere, so absorbed > emitted and there is a net positive imbalance, currently ≈1 W m⁻²
    • The excess energy raises the internal energy of oceans, land and ice; over 90% goes into the oceans
    • The surface warms, increasing σT⁴, until outgoing radiation again equals the input at a higher temperature. Exam tip: energy is always conserved — the greenhouse effect changes the temperature at which balance occurs, it does not create energy.
45. Concept/explainSL & HLData booklet: No – derive

    Explain how volcanic and industrial aerosols affect the Earth's mean temperature.

    Show answer

    • Eruptions and industrial emissions inject sulfate aerosol particles into the stratosphere
    • These scatter incoming solar radiation back to space, raising the planetary albedo α
    • Since T = [S(1 − α)/(4σ)]^¼, a larger α reduces the absorbed intensity and lowers the equilibrium temperature
    • Mount Pinatubo (1991) cooled the Earth by ≈0.5 K for about two years
    • Aerosols are washed out within a few years, whereas CO₂ persists for centuries, so the cooling is only temporary
    • This is the basis of proposed geoengineering, which would not remove CO₂ or reverse ocean acidification. Exam tip: link the change explicitly through T ∝ (1 − α)^¼ rather than simply asserting that the Earth gets colder.
46. Concept/explainSL & HLData booklet: Yes

    Explain the meaning of emissivity and its role in P = eσAT⁴ for planetary calculations.

    Show answer

    • Emissivity e = power radiated by a body / power radiated by a black body of the same area at the same temperature; dimensionless with 0 ≤ e ≤ 1
    • A perfect black body has e = 1
    • Earth's surface (ocean, soil, vegetation) is an excellent infrared emitter, e ≈ 0.95, and is usually taken as 1 in IB energy-balance questions
    • By Kirchhoff's law emissivity equals absorptivity at the same wavelength, so a good absorber is a good emitter
    • A surface can have a high albedo in the visible yet a high emissivity in the infrared — fresh snow is the classic example. Exam tip: emissivity is not "1 − albedo"; the two quantities refer to different wavelength ranges and different processes.
47. Concept/explainSL & HL

    Outline the mechanisms of sea-level rise associated with the enhanced greenhouse effect.

    Show answer

    • Thermal expansion: as ocean water absorbs the excess energy its temperature rises and its volume increases, raising sea level
    • Melting of land ice — mountain glaciers and the Greenland and Antarctic ice sheets — adds water that was previously stored on land
    • Melting floating sea ice does not appreciably change sea level, because by Archimedes' principle it already displaces its own weight of water
    • Thermal expansion dominated 20th-century rise; land-ice melt is now a rapidly growing contribution
    • Effects: coastal flooding, salt contamination of groundwater, loss of low-lying land, storm surges reaching further inland. Exam tip: the floating-sea-ice point is a frequent discriminator — state Archimedes' principle explicitly rather than just asserting no effect.
48. Concept/explainSL & HL

    Evaluate the simple energy-balance model that predicts an Earth temperature of 255 K.

    Show answer

    • Strength: it uses only conservation of energy and Stefan–Boltzmann, needs no fitting, and correctly shows that an atmosphere is required to explain the observed 288 K
    • Weakness: it treats Earth as an isothermal sphere, ignoring the large equator–pole and day–night temperature variation, and T⁴ averaging is not the same as averaging T
    • It assumes α and e are constant, whereas both vary with cloud, ice cover, season and angle of incidence
    • It assumes instantaneous equilibrium, ignoring the decades-long thermal inertia of the oceans
    • It ignores latent heat, convection and horizontal transport by winds and currents. Exam tip: a question saying "evaluate the model" needs at least one strength as well as the limitations.
49. Worked problemSL & HLData booklet: Yes

    The solar constant is S = 1360 W m⁻² and Earth's mean albedo is 0.30. Determine the average intensity absorbed by the Earth–atmosphere system.

    Show answer

    Average incident intensity = S/4, because the disc πR² intercepts while the sphere 4πR² emits

    • S/4 = 1360/4 = 340 W m⁻²
    • Absorbed fraction = (1 − α) = 0.70
    • Absorbed intensity = 0.70 × 340 = 238 W m⁻² ≈ 2.4 × 10² W m⁻² (2 s.f.)
    • Scattered back to space = 0.30 × 340 = 102 W m⁻²
    • Check: 238 + 102 = 340 W m⁻², as required by conservation of energy. Check/Trap: do not multiply S by 0.70 and then forget the ÷4 (this gives 952 W m⁻²); the geometric factor and the albedo factor are independent — apply geometry first, then albedo.
50. Worked problemSL & HLData booklet: No – derive

    Calculate the equilibrium surface temperature of an Earth with no atmosphere, taking S = 1360 W m⁻², α = 0.30 and e = 1.

    Show answer

    Balance absorbed and emitted power: S(1 − α)πR² = eσ(4πR²)T⁴

    • πR² cancels, so T = [S(1 − α)/(4σ)]^¼
    • S(1 − α) = 1360 × 0.70 = 952 W m⁻²; divide by 4 to get 238 W m⁻²
    • 238/(5.67 × 10⁻⁸) = 4.20 × 10⁹ K⁴
    • T = (4.20 × 10⁹)^¼ = 255 K (3 s.f.), i.e. −18 °C
    • The observed mean surface temperature is 288 K, so the natural greenhouse effect accounts for 33 K. Check/Trap: R² cancels, so the Earth's radius is never needed; take the fourth root (press √ twice or use ^0.25), and give the answer in kelvin, not °C.
51. Worked problemSL & HLData booklet: No – derive

    Earth's radius is 6.37 × 10⁶ m and S = 1360 W m⁻². Calculate the total solar power intercepted by the Earth and the power absorbed if α = 0.30.

    Show answer

    Intercepting area is the cross-sectional disc: πR² = π(6.37 × 10⁶)² = 1.27 × 10¹⁴ m²

    • Intercepted power = SπR² = 1360 × 1.27 × 10¹⁴ = 1.73 × 10¹⁷ W
    • Absorbed power = 0.70 × 1.73 × 10¹⁷ = 1.21 × 10¹⁷ W (3 s.f.)
    • Cross-check by the other route: absorbed intensity 238 W m⁻² over the full surface 4πR² = 5.10 × 10¹⁴ m² gives 238 × 5.10 × 10¹⁴ = 1.21 × 10¹⁷ W, which agrees. Check/Trap: use πR² for interception and 4πR² for emission; using 4πR² for both is the commonest error and overestimates the intercepted power by a factor of 4.
52. Worked problemSL & HLData booklet: Yes

    The Sun's luminosity is 3.85 × 10²⁶ W and the Earth–Sun distance is 1.50 × 10¹¹ m. Calculate the solar constant, and the value at Mars (1.52 AU).

    Show answer

    Use b = L/(4πd²) from the data booklet

    • 4πd² = 4π(1.50 × 10¹¹)² = 2.83 × 10²³ m²
    • b = 3.85 × 10²⁶/2.83 × 10²³ = 1.36 × 10³ W m⁻²
    • So S ≈ 1360 W m⁻², the solar constant
    • Since b ∝ 1/d², at Mars S_M = 1360/1.52² = 1360/2.31 = 589 W m⁻² (3 s.f.). Check/Trap: d is the Sun–planet distance, not the Sun's radius, and 4πd² is the area of the imaginary sphere the radiation has spread over, not the planet's surface area. Convert AU to a ratio before squaring — squaring 1.52 AU in metres is unnecessary work.
53. Worked problemSL & HLData booklet: Yes

    The Earth's surface has a mean temperature of 288 K. Determine the wavelength of peak emission and the power radiated per square metre, taking e = 1.

    Show answer

    Wien's law: λ_max = 2.9 × 10⁻³/T = 2.9 × 10⁻³/288 = 1.0 × 10⁻⁵ m = 10 μm, in the infrared

    • Stefan–Boltzmann: P/A = eσT⁴ = 1 × 5.67 × 10⁻⁸ × 288⁴
    • 288⁴ = 6.88 × 10⁹ K⁴
    • P/A = 5.67 × 10⁻⁸ × 6.88 × 10⁹ = 390 W m⁻² (2 s.f.)
    • This exceeds the 238 W m⁻² absorbed from the Sun; the surplus is supplied by back-radiation from greenhouse gases. Check/Trap: T must be in kelvin — using 15 °C gives λ_max ≈ 1.9 × 10⁻⁴ m and a nonsense power. Note that 10 μm lies inside the CO₂ and H₂O absorption region, which is the point of the question.
54. Worked problemSL & HLData booklet: Yes

    The Sun radiates as a black body with λ_max = 5.0 × 10⁻⁷ m. Estimate its surface temperature and the energy of a photon of this wavelength.

    Show answer

    Wien's law: T = 2.9 × 10⁻³/λ_max = 2.9 × 10⁻³/(5.0 × 10⁻⁷) = 5.8 × 10³ K

    • Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(5.0 × 10⁻⁷)
    • = 1.99 × 10⁻²⁵/(5.0 × 10⁻⁷) = 4.0 × 10⁻¹⁹ J = 4.0 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 2.5 eV
    • By comparison an Earth-emitted 10 μm photon carries only 2.0 × 10⁻²⁰ J (0.12 eV), enough to excite molecular vibration but not electronic transitions. Check/Trap: 500 nm is green light in the middle of the visible band — a useful sanity check on the arithmetic. Keep λ in metres throughout and do not mix nm with m.
55. Worked problemSL & HLData booklet: No – derive

    Mars orbits at 1.52 AU and has an albedo of 0.25. Calculate its equilibrium temperature, taking e = 1.

    Show answer

    Solar intensity at Mars: S_M = 1360/1.52² = 589 W m⁻²

    • T = [S_M(1 − α)/(4σ)]^¼
    • S_M(1 − α)/4 = 589 × 0.75/4 = 110 W m⁻²
    • 110/(5.67 × 10⁻⁸) = 1.95 × 10⁹ K⁴
    • T = (1.95 × 10⁹)^¼ = 210 K (2 s.f.), i.e. −63 °C
    • The measured mean surface temperature of Mars is also ≈210 K, showing that Mars has almost no greenhouse warming because its CO₂ atmosphere is very thin (surface pressure ≈0.6 kPa). Check/Trap: apply the inverse-square law to get S at the planet BEFORE using the balance equation; using 1360 W m⁻² for every planet is a guaranteed zero.
56. Worked problemSL & HLData booklet: No – derive

    Venus receives S = 2600 W m⁻² and has an albedo of 0.77. Calculate its equilibrium temperature and comment, given a measured surface temperature of 737 K.

    Show answer

    Absorbed intensity = S(1 − α)/4 = 2600 × 0.23/4 = 150 W m⁻²

    • 150/(5.67 × 10⁻⁸) = 2.64 × 10⁹ K⁴
    • T = (2.64 × 10⁹)^¼ = 227 K (3 s.f.)
    • This is lower than Earth's 255 K, even though Venus is closer to the Sun, because its thick sulfuric-acid cloud scatters 77% of the incident radiation
    • The measured 737 K is 510 K higher than the calculated value, caused by a runaway greenhouse effect in an atmosphere that is ≈96% CO₂ at ≈9 MPa. Check/Trap: a high albedo LOWERS the equilibrium temperature; Venus's extreme surface temperature comes entirely from the greenhouse effect, not from its proximity to the Sun.
57. Worked problemSL & HLData booklet: Yes

    A sensor measures 1.24 kW of solar power incident on a patch of snow-covered ground of area 1.0 m² and 0.87 kW scattered from it. Determine the albedo and the absorbed intensity.

    Show answer

    Albedo α = total scattered power/total incident power (data booklet)

    • α = (0.87 × 10³)/(1.24 × 10³) = 0.70 (2 s.f.)
    • Absorbed power = 1.24 − 0.87 = 0.37 kW = 3.7 × 10² W
    • Over 1.0 m², absorbed intensity = 3.7 × 10² W m⁻²
    • For comparison, fresh snow has α ≈ 0.85 and open ocean α ≈ 0.06, which is why replacing snow with ocean absorbs far more power and drives the ice–albedo feedback. Check/Trap: albedo is a ratio of powers, so it is dimensionless and can never exceed 1 — an answer greater than 1 means you divided the wrong way round. Do not convert to percentages unless asked.
58. Worked problemSL & HLData booklet: Yes

    A land surface of area 2.5 × 10⁴ m² is at 295 K and has emissivity 0.85. Calculate the intensity and the total power it radiates.

    Show answer

    Use P = eσAT⁴ from the data booklet

    • T⁴ = 295⁴ = 7.57 × 10⁹ K⁴
    • σT⁴ = 5.67 × 10⁻⁸ × 7.57 × 10⁹ = 429 W m⁻² (black-body value)
    • Intensity = eσT⁴ = 0.85 × 429 = 365 W m⁻² ≈ 3.6 × 10² W m⁻² (2 s.f.)
    • Total power P = 365 × 2.5 × 10⁴ = 9.1 × 10⁶ W (2 s.f.). Check/Trap: raise T to the fourth power first and only then multiply by σ and e — squaring instead of fourth-powering is the standard slip. Remember that a 1% rise in T gives about a 4% rise in P, since ΔP/P ≈ 4ΔT/T for P ∝ T⁴.
59. Worked problemSL & HLData booklet: No – derive

    Model the atmosphere as a single layer transparent to solar radiation but a black body to infrared. Show that the surface temperature is 2^¼ times the effective temperature and evaluate it.

    Show answer

    Let the layer be at T_a and the surface at T_s, with 238 W m⁻² of solar power absorbed at the surface

    • Top-of-atmosphere balance: the only emission to space is from the layer, so σT_a⁴ = 238 W m⁻², giving T_a = 255 K
    • The layer radiates σT_a⁴ upwards AND σT_a⁴ downwards
    • Surface balance: σT_s⁴ = 238 + σT_a⁴ = 2 × 238 = 476 W m⁻²
    • Hence T_s⁴ = 2T_a⁴ and T_s = 2^¼ × 255 = 1.19 × 255 = 303 K (30 °C)
    • The observed value is 288 K, so one fully opaque layer slightly over-predicts. Check/Trap: the essential physics is that the layer emits from both faces; the downward half is the back-radiation that warms the surface.
60. Worked problemSL & HLData booklet: No – derive

    Ice loss reduces the Earth's albedo from 0.30 to 0.28. Estimate the resulting change in equilibrium temperature, with S = 1360 W m⁻².

    Show answer

    T = [S(1 − α)/(4σ)]^¼

    • With α = 0.30: S(1 − α)/4 = 238 W m⁻², so T₁ = (238/5.67 × 10⁻⁸)^¼ = 254.5 K
    • With α = 0.28: 1360 × 0.72/4 = 244.8 W m⁻²; 244.8/(5.67 × 10⁻⁸) = 4.32 × 10⁹ K⁴, so T₂ = 256.3 K
    • ΔT = 256.3 − 254.5 = +1.8 K (2 s.f.)
    • Shortcut: since T ∝ (1 − α)^¼, ΔT/T = ¼ × Δ(1 − α)/(1 − α) = ¼ × (0.02/0.70) = 0.71%, giving 0.0071 × 254.5 = 1.8 K, which agrees. Check/Trap: keep extra significant figures until AFTER the subtraction — rounding both temperatures to 255 K destroys the difference entirely.
61. Worked problemSL & HLData booklet: No – derive

    Doubling atmospheric CO₂ gives a radiative forcing of 3.7 W m⁻². Estimate the direct warming, ignoring feedbacks, for an Earth radiating at an effective temperature of 255 K.

    Show answer

    Outgoing intensity I = σT⁴, so the rate of change with temperature is dI/dT = 4σT³

    • 255³ = 1.66 × 10⁷ K³
    • 4σT³ = 4 × 5.67 × 10⁻⁸ × 1.66 × 10⁷ = 3.76 W m⁻² K⁻¹
    • For balance to be restored the outgoing intensity must rise by 3.7 W m⁻², so ΔT = ΔI/(4σT³) = 3.7/3.76 = 0.98 ≈ 1.0 K (2 s.f.)
    • The observed climate sensitivity is ≈3 K, the extra coming from positive feedbacks (water vapour, ice–albedo, cloud). Check/Trap: use the effective radiating temperature 255 K, not the surface 288 K, which would give 0.68 K. Quote the answer as a temperature change, where K and °C are numerically identical.
62. Worked problemSL & HLData booklet: Yes

    Earth's energy imbalance is 0.90 W m⁻² averaged over its whole surface of 5.10 × 10¹⁴ m². Calculate the energy gained in one year and the temperature rise of the top 100 m of ocean (area 3.6 × 10¹⁴ m², ρ = 1030 kg m⁻³, c = 3900 J kg⁻¹ K⁻¹).

    Show answer

    Net power = 0.90 × 5.10 × 10¹⁴ = 4.6 × 10¹⁴ W

    • Energy in one year = Pt = 4.6 × 10¹⁴ × 3.15 × 10⁷ = 1.4 × 10²² J
    • Ocean volume = 3.6 × 10¹⁴ × 100 = 3.6 × 10¹⁶ m³; mass m = ρV = 1030 × 3.6 × 10¹⁶ = 3.7 × 10¹⁹ kg
    • Q = mcΔT so ΔT = 1.4 × 10²²/(3.7 × 10¹⁹ × 3900) = 0.10 K per year (2 s.f.). Check/Trap: the imbalance is quoted per square metre of the WHOLE Earth surface, so multiply by 4πR², not πR². The enormous ocean heat capacity is why the surface warms slowly — this is the thermal inertia that delays the response to emissions.
63. Exam technique/trapSL & HL

    How should answers be tailored to the command terms state, outline, describe, explain, deduce and suggest in greenhouse-effect questions?

    Show answer

    State: a bare fact with no reasoning — "the mean albedo is 0.30". Outline: a brief account, typically two linked points. Describe: say what happens without giving causes — "the temperature rises and then levels off". Explain: give the mechanism — short-wave in, long-wave out, resonant absorption by greenhouse gases, re-emission in all directions. Deduce: reach a conclusion using data already given, and quote that data. Suggest: propose a plausible physical reason in an unfamiliar context; credit is for physics consistency, not a single correct answer. Trap: writing a paragraph for "state" wastes time, while one line for a 4-mark "explain" loses three marks. Match the number of distinct points to the mark tariff.

64. Exam technique/trapSL & HL

    What conceptual confusions about the greenhouse effect are routinely penalised in mark schemes?

    Show answer

    Trap 1: confusing the greenhouse effect with ozone depletion — ozone loss concerns UV absorption by O₃ and destruction by CFCs, a separate problem. Trap 2: saying greenhouse gases "trap" or "reflect" heat; they absorb IR and re-emit it in all directions. Trap 3: claiming the effect is entirely man-made — the natural effect supplies the essential 33 K and human activity enhances it. Trap 4: assuming a glass greenhouse works the same way; a greenhouse works mainly by preventing convection. Trap 5: confusing albedo (visible scattering) with emissivity (IR emission). Correct approach: in every answer name the wavelength range, the absorbing molecules and the direction of re-emission.

65. Exam technique/trapSL & HLData booklet: No – derive

    Identify the algebraic and calculator traps in planetary energy-balance calculations and how to avoid them.

    Show answer

    Trap: using S instead of S/4 as the mean incident intensity — the factor 4 comes from πR²/4πR². Trap: omitting (1 − α), or using α itself as the absorbed fraction. Trap: taking a square root instead of a fourth root; press √ twice or use ^0.25. Trap: leaving T in °C — the Stefan–Boltzmann balance only works in kelvin. Trap: keeping R in the final expression when πR² cancels on both sides. Correct approach: write S(1 − α)πR² = eσ4πR²T⁴ in full, cancel, rearrange to T = [S(1 − α)/(4σ)]^¼, then substitute. Sanity check: any rocky-planet answer far outside 150–350 K signals an arithmetic slip.

66. Exam technique/trapSL & HLData booklet: No – derive

    The solar constant is 1360 ± 20 W m⁻² and the albedo 0.30 ± 0.02. Explain how to obtain the uncertainty in the equilibrium temperature of 255 K.

    Show answer

    Since T = [S(1 − α)/(4σ)]^¼, fractional uncertainties of the multiplied quantities add, and the power ¼ then multiplies the total. Fractional uncertainty in S = 20/1360 = 1.5%. The uncertainty in (1 − α) = 0.70 has the same absolute value 0.02, so its fractional uncertainty is 0.02/0.70 = 2.9%. Total = 4.4%. Because T ∝ (quantity)^¼, fractional uncertainty in T = ¼ × 4.4% = 1.1%. Absolute: ΔT = 0.011 × 255 = 2.8, so T = 255 ± 3 K. Trap: students take ¼ of the ABSOLUTE uncertainty, or forget that α and (1 − α) share an absolute uncertainty but not a fractional one. Quote uncertainties to 1 s.f. and match decimal places.

67. Exam technique/trapSL & HL

    Outline an experiment to compare the albedo of different surfaces, including controlled variables, limitations and improvements.

    Show answer

    Apparatus: filament lamp as source, shallow trays of sand, soil, white paper, black card, water and crushed ice, a light sensor or lux meter on a clamp stand, metre rule, darkened room. Method: fix lamp height and angle; measure incident intensity with the sensor facing the lamp, then the scattered intensity with the sensor at a fixed distance and angle above the sample; albedo ≈ scattered/incident; repeat three times and average. Controlled: lamp power, all distances, angle of incidence, sensor orientation, background light (subtract a dark reading). Limitations: the sensor samples one direction only whereas true albedo integrates over all directions; the lamp spectrum differs from sunlight. Improvements: use an integrating hemisphere and a solar-simulator lamp.

68. Exam technique/trapSL & HL

    Describe a laboratory model of the greenhouse effect and evaluate its validity.

    Show answer

    Apparatus: two identical transparent containers of soil, each with a temperature probe and datalogger, under identical lamps at equal distances; one flushed with CO₂ and sealed, the other containing air. Independent variable: gas in the container; dependent: temperature; controlled: lamp power and distance, container size, soil mass and colour, initial and ambient temperature. Record temperature every 30 s for 20 minutes and plot heating curves; the CO₂ container reaches a higher steady temperature. Limitations: sealing also suppresses convection, so the effect is not purely radiative; the IR path length is tiny compared with the atmosphere; conduction through the walls loses energy. Improvements: use IR-transparent polythene windows, an infrared sensor, and repeat at several CO₂ concentrations.

69. Exam technique/trapSL & HL

    How should a data-analysis question on ice-core CO₂ or a temperature-trend graph be tackled?

    Show answer

    Read the axes and units first (ppm, °C anomaly, years before present) and note that an anomaly is a difference from a baseline, not an absolute temperature. Describe the trend quantitatively: quote two data points and calculate a gradient with its unit, e.g. °C per century. Use the error bars: draw maximum and minimum gradient lines through the extreme ends of the first and last error bars, then uncertainty in gradient = (max − min)/2. Distinguish correlation from causation before concluding. Trap: writing "the graph goes up" with no numbers, and extrapolating far beyond the data range. Random error appears as scatter about the trend and is reduced by repeats; a systematic error (mis-calibrated sensor) shifts the whole line and is not.

70. Exam technique/trapSL & HL

    What rules on significant figures, precision and accuracy apply when quoting energy-balance results?

    Show answer

    Quote the answer to the least number of significant figures in the data: with S = 1360 W m⁻² (3 s.f.) and α = 0.30 (2 s.f.), give T as 255 K (2–3 s.f.). Keep full precision in the calculator and round only at the end — rounding 238 W m⁻² to 240 shifts T by about 0.5 K. Precision means repeated readings agree closely (small random error); accuracy means the mean is close to the true value (small systematic error). A satellite radiometer with a drifting calibration is precise but inaccurate. Trap: writing 254.9836 K, or writing 255 °C. Temperature DIFFERENCES may be quoted in K or °C interchangeably, but absolute temperatures used in σT⁴ must be in kelvin.

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