B.2 Greenhouse effect – IB Physics exam questions with mark scheme

1Hard7 marks

The solar constant at the Earth is $1.36\times10^3\ \mathrm{W\,m^{-2}}$ and the albedo of the Earth is 0.30. Assume the Earth has no atmosphere and that its surface radiates as a black body (emissivity 1). $\sigma=5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.

(a) Show that the average intensity absorbed per unit area of the Earth's surface is about $240\ \mathrm{W\,m^{-2}}$. [2]

(b) Calculate the equilibrium surface temperature of the Earth under these assumptions. [2]

Only 70% of the incident radiation is absorbed: $1360\times(1-0.30)$ (= 952 W m$^{-2}$) (allow the two steps in either order, e.g. $1360/4=340$ then $\times0.70$)(1)
Radiation is intercepted by a disc of area $\pi R^2$ but absorbed over a surface of area $4\pi R^2$, so divide by 4: $952/4 = 238\ \mathrm{W\,m^{-2}}$ (must see 238 or more s.f.; 240 alone earns no credit)(1)
At equilibrium, power radiated = power absorbed: $\sigma T^4 = 238$ (allow 240)(1)
$T=(238/5.67\times10^{-8})^{1/4} = 255$ K (allow 254–256 K)(1)
Greenhouse gases (e.g. CO₂, CH₄, H₂O, N₂O) in the atmosphere absorb infrared radiation emitted by the surface (allow: IR frequency matches natural frequency of molecular vibration / resonance; reject: answers about the ozone layer, ozone hole or UV; reject 'reflect' / 'trap heat' without absorption of IR)(1)
They re-emit infrared in all directions, so some returns to the surface (OWTTE)(1)
The surface therefore receives more power than from the Sun alone, so it warms to a higher temperature before energy balance is reached (allow: equilibrium reached at higher T)(1)
Answer: (a) 238 W m$^{-2}$ (b) 255 K (c) greenhouse gases absorb and re-radiate IR back towards the surface
Examiner tip: In (c) the mark needs gases absorbing infrared emitted by the Earth's surface: 'trapping heat', 'reflecting' or the ozone layer/UV earn nothing.

2Core4 marks

(a) Define the albedo of a planet. [1]

(b) A surface has an emissivity of 0.6. State what this means in terms of the power it radiates. [1]

(c) Methane ($\mathrm{CH_4}$) and nitrous oxide ($\mathrm{N_2O}$) are both greenhouse gases. For each gas, state one source that results from human activity. [2]

(total) scattered/reflected power ÷ (total) incident power (on the planet); accept intensity in place of power; do not accept "light reflected" without a ratio(1)
it radiates 0.6 / 60% of the power per unit area that a black body at the same temperature would radiate; accept $P/A = 0.6\sigma T^4$; "same temperature" or $T$ must be referred to(1)
$\mathrm{CH_4}$: livestock (cattle) digestion / rice paddies / landfill (decomposing waste) / leaks from natural-gas extraction or coal mining(1)
$\mathrm{N_2O}$: nitrogen-based fertilisers (in agriculture) / animal manure / some industrial processes (e.g. nylon or nitric acid production) / burning fuels in vehicles; do not accept a natural source such as wetlands or soil bacteria alone(1)
Answer: (a) albedo = scattered (reflected) power ÷ incident power (b) it emits 0.6 of the power per unit area of a black body at the same temperature (c) CH$_4$: livestock / landfill / rice farming; N$_2$O: fertilisers
Examiner tip: In (b) the comparison must be with a black body at the same temperature; "radiates 60% of the energy" on its own does not score.

3Core5 marks

Solar radiation of intensity $850\ \mathrm{W\,m^{-2}}$ falls on a flat area of sea ice. The ice scatters $680\ \mathrm{W\,m^{-2}}$ back into the sky.

(a) Calculate the albedo of the ice. [1]

(b) Determine the power absorbed by an area of $2.5\times10^{3}\ \mathrm{m^2}$ of the ice. [2]

(c) A dark asphalt road at a temperature of 290 K radiates $320\ \mathrm{W\,m^{-2}}$. Calculate the emissivity of the road surface. [2]

albedo $= \dfrac{680}{850} = 0.80$; no unit; do not accept 80 without %(1)
absorbed intensity $= 850 - 680 = 170\ \mathrm{W\,m^{-2}}$ OR $(1-0.80)\times850$(1)
$P = 170\times2.5\times10^3 = 4.3\times10^{5}\ \mathrm{W}$; accept $4.25\times10^{5}$ or $4.2\times10^{5}$ W; allow ECF from (a); award [1 max] for $2.1\times10^6$ W (incident power used) or $1.7\times10^6$ W (scattered power used)(1)
$e = \dfrac{P/A}{\sigma T^4} = \dfrac{320}{5.67\times10^{-8}\times290^4}$(1)
$e = 0.80$; accept 0.798; no unit(1)
Answer: (a) 0.80 (b) $4.3\times10^5$ W (c) 0.80
Examiner tip: In (b) the absorbed power uses $(1-\text{albedo})$: multiplying the incident or reflected intensity by the area is the most common slip.

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4Standard5 marks

The diagram shows parallel solar radiation of intensity $S$ arriving at Mars, a planet of radius $R$. The solar constant at the orbit of Mars is $590\ \mathrm{W\,m^{-2}}$ and the albedo of Mars is 0.25. Mars has a very thin atmosphere; treat its surface as a black body.

(a) Show that the mean intensity of solar radiation incident on the whole surface of Mars, averaged over its surface and over a day, is $\dfrac{S}{4}$. [2]

(b) Calculate the mean equilibrium surface temperature of Mars. [3]

power intercepted $= S\times\pi R^2$ (area of cross-section / disc facing the Sun)(1)
spread over whole surface area $4\pi R^2$ so mean intensity $=\dfrac{S\pi R^2}{4\pi R^2} = \dfrac{S}{4}$; must see both areas(1)
absorbed intensity $= (1-0.25)\times\dfrac{590}{4} = 111\ \mathrm{W\,m^{-2}}$; accept 110.6(1)
$\sigma T^4 = 110.6$ OR $T = \left(\dfrac{110.6}{5.67\times10^{-8}}\right)^{1/4}$(1)
$T = 210\ \mathrm{K}$; award [2 max] for 226 K (albedo ignored) or 160 K (albedo used as fraction absorbed); award [1 max] for 297 K (S not divided by 4, albedo used)(1)
Answer: (a) power $S\pi R^2$ spread over $4\pi R^2$ (b) 210 K
Examiner tip: In (a) you must name both areas: the intercepting disc $\pi R^2$ and the radiating sphere $4\pi R^2$; just writing "divide by 4" scores nothing.

5Standard6 marks

The mean temperature of the Earth's surface is 288 K and the temperature of the Sun's surface is 5800 K. Both may be treated as black bodies.

(a) Calculate the wavelength at which the intensity of radiation emitted by the Earth's surface is a maximum. [1]

(b) Calculate the corresponding wavelength for the Sun. [1]

(c) Carbon dioxide molecules strongly absorb radiation of wavelength $15\ \mathrm{\mu m}$. Calculate the frequency of this radiation. [1]

(d) Using your answers, explain, with reference to resonance, why carbon dioxide in the atmosphere absorbs much of the radiation emitted by the Earth's surface but little of the incoming solar radiation. [3]

$\lambda_{max} = \dfrac{2.9\times10^{-3}}{288} = 1.0\times10^{-5}\ \mathrm{m}$ (10 μm)(1)
$\lambda_{max} = \dfrac{2.9\times10^{-3}}{5800} = 5.0\times10^{-7}\ \mathrm{m}$ (500 nm)(1)
$f = \dfrac{c}{\lambda} = \dfrac{3.00\times10^{8}}{15\times10^{-6}} = 2.0\times10^{13}\ \mathrm{Hz}$(1)
the (bonds in) CO$_2$ molecules have natural frequencies of vibration in the infrared (about $2\times10^{13}$ Hz)(1)
the Earth emits mainly infrared, close to these frequencies, so the molecules are driven at their natural frequency and resonate / absorb energy (strongly)(1)
solar radiation is mainly visible (shorter wavelength / higher frequency, about $6\times10^{14}$ Hz), far from the natural frequency, so it is not absorbed and passes through; must compare wavelengths or frequencies of the two spectra(1)
Answer: (a) $1.0\times10^{-5}$ m (b) $5.0\times10^{-7}$ m (c) $2.0\times10^{13}$ Hz (d) Earth's IR matches the natural vibration frequency of CO$_2$ so it resonates and absorbs; sunlight is mostly visible, far from resonance, so it is transmitted
Examiner tip: "Greenhouse gases trap heat" earns nothing: the marks are for linking the infrared frequency to the molecules' natural frequency of vibration.

6Hard8 marks

The diagram shows a simplified model of the global mean energy flows for the Earth. All values are in $\mathrm{W\,m^{-2}}$, averaged over the Earth's surface. X represents the energy transferred from the surface to the atmosphere by convection and evaporation.

(a) Calculate the albedo of the Earth according to this model. [1]

(b) Show that the Earth as a whole is in energy equilibrium with space. [1]

(d) The surface is in equilibrium. Calculate X. [1]

(e) Seen from space, the Earth radiates as a single body with a temperature equal to your answer to (c). Estimate the effective emissivity of the Earth. [2]

(f) The concentration of carbon dioxide in the atmosphere increases. State which labelled energy flow increases first as a direct result. [1]

albedo $= \dfrac{100}{340} = 0.29$; accept 0.294(1)
incoming $340$ = outgoing $100 + 240 = 340\ \mathrm{W\,m^{-2}}$(1)
$\sigma T^4 = 398$ OR $T = \left(\dfrac{398}{5.67\times10^{-8}}\right)^{1/4}$(1)
$T = 289\ \mathrm{K}$; accept 290 K(1)
$X = 161 + 342 - 398 = 105\ \mathrm{W\,m^{-2}}$(1)
$e = \dfrac{240}{\sigma T^4} = \dfrac{240}{5.67\times10^{-8}\times289^4}$; allow ECF from (c)(1)
$e = 0.61$; accept 0.60–0.62; no unit(1)
back radiation (342) increases; accept "the infrared absorbed by the atmosphere / re-emitted towards the surface"; do not accept "outgoing infrared 240"(1)
Answer: (a) 0.29 (b) 340 in = 100 + 240 out (c) 289 K (d) 105 W m$^{-2}$ (e) 0.61 (f) back radiation
Examiner tip: In (e) the effective emissivity uses the 240 W m$^{-2}$ that leaves to space, not the 398 W m$^{-2}$ emitted by the surface (that would give e = 1).

7Stretch7 marks

A simple model of the Earth treats the atmosphere as a single layer at temperature $T_a$ above a surface at temperature $T_s$.

The surface behaves as a black body.
The atmosphere is transparent to solar radiation, has emissivity $e$, and absorbs a fraction $e$ of the infrared radiation emitted by the surface.
The atmosphere radiates equally upwards and downwards, emitting $e\sigma T_a^4$ per unit area from each side.

The solar constant is $1361\ \mathrm{W\,m^{-2}}$, the albedo is 0.30 and $e = 0.78$.

(a) By considering the energy balance of the atmosphere, show that $T_a^4 = \dfrac{T_s^4}{2}$. [2]

(b) Show that the surface temperature predicted by the model is about 288 K. [3]

(d) An increase in greenhouse gas concentration raises $e$ to 0.80. Calculate the new surface temperature. [1]

power absorbed by atmosphere per unit area $= e\sigma T_s^4$; power emitted $= 2e\sigma T_a^4$ (upwards and downwards)(1)
equating: $e\sigma T_s^4 = 2e\sigma T_a^4 \Rightarrow T_a^4 = \dfrac{T_s^4}{2}$(1)
solar intensity absorbed by surface $= (1-0.30)\times\dfrac{1361}{4} = 238\ \mathrm{W\,m^{-2}}$(1)
surface balance: $238 + e\sigma T_a^4 = \sigma T_s^4$ OR $238 = \sigma T_s^4\left(1-\dfrac{e}{2}\right)$(1)
$T_s = \left(\dfrac{238}{5.67\times10^{-8}\times0.61}\right)^{1/4} = 288\ \mathrm{K}$; answer to at least 3 s.f. needed for a "show that"(1)
$T_a = \dfrac{288}{2^{1/4}} = 242\ \mathrm{K}$(1)
$T_s = \left(\dfrac{238}{5.67\times10^{-8}\times0.60}\right)^{1/4} = 289\ \mathrm{K}$; accept 289.3 K / a rise of about 1 K(1)
Answer: (a) $e\sigma T_s^4 = 2e\sigma T_a^4$ (b) 288 K (c) 242 K (d) 289 K
Examiner tip: The atmosphere radiates from both faces, so its emitted power is $2e\sigma T_a^4$; forgetting the factor of 2 wrecks every later part.

8Standard6 marks

The graph shows the mean concentration of carbon dioxide in the atmosphere, measured in parts per million (ppm), from 1960 to 2020. Before industrialisation the concentration was about 280 ppm.

(a) Using the graph, determine the mean rate of increase of the concentration between 2000 and 2020. [2]

(b) Calculate the percentage increase in concentration between pre-industrial times and 2020. [1]

(d) Outline how the data support the claim that the enhanced greenhouse effect is accelerating. [1]

reads 370 ppm (2000) and 414 ppm (2020) from the graph; allow ±3 ppm(1)
rate $= \dfrac{414-370}{20} = 2.2\ \mathrm{ppm\,yr^{-1}}$; accept 2.0–2.4; unit needed(1)
$\dfrac{414-280}{280}\times100 = 48\%$; accept 47–49%; allow ECF from graph reading(1)
burning fossil fuels (coal, oil, gas) in power stations / vehicles / industry(1)
deforestation (fewer trees to absorb CO$_2$; burning or decay of cleared wood) OR cement production; do not accept a second fossil-fuel example for this mark(1)
the gradient increases with time, e.g. about $1.1\ \mathrm{ppm\,yr^{-1}}$ in 1960–1980 compared with $2.2\ \mathrm{ppm\,yr^{-1}}$ in 2000–2020, so CO$_2$ (and its contribution to back radiation) is being added faster; a comparison of rates is needed, not just "the concentration is increasing"(1)
Answer: (a) 2.2 ppm yr$^{-1}$ (b) 48% (c) burning fossil fuels; deforestation (d) the gradient is steeper later (1.1 → 2.2 ppm yr$^{-1}$)
Examiner tip: In (d) "concentration is rising" shows an increase, not an acceleration: you must compare gradients over two time intervals.

9Hard6 marks

Without an atmosphere, the mean surface temperature of the Earth would be about 255 K. The actual mean surface temperature is about 288 K.

Explain, in terms of the absorption and emission of radiation, why the Earth's surface is warmer than 255 K, and discuss how human activity and feedback processes may change the surface temperature in the future. [6]

incoming solar radiation is mainly short wavelength (visible / near-IR) and passes through the atmosphere largely unabsorbed(1)
the surface absorbs it, warms, and emits infrared (long wavelength, peak about 10 μm) because it is at a much lower temperature than the Sun(1)
greenhouse gas molecules (CO$_2$, H$_2$O, CH$_4$, N$_2$O) have natural frequencies of vibration in the infrared, so they absorb this radiation by resonance(1)
the molecules re-emit infrared in all directions, so part of it returns to the surface (back radiation)(1)
the surface therefore receives more power than from the Sun alone and must reach a higher temperature for $\sigma T^4$ to balance the energy it receives(1)
burning fossil fuels / deforestation / agriculture increase concentrations of CO$_2$, CH$_4$, N$_2$O: more IR absorbed and re-emitted, so the enhanced greenhouse effect raises the equilibrium temperature(1)
positive feedback: warming melts ice and snow, lowering albedo, so more radiation is absorbed(1)
positive feedback: warmer oceans hold less dissolved CO$_2$ / more water evaporates (water vapour is a greenhouse gas) / thawing permafrost releases CH$_4$(1)
effective emissivity of the Earth seen from space falls, so a higher surface temperature is needed for outgoing power to equal absorbed power(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, linked explanation of why the surface is warmer, covering the wavelength difference, absorption by resonance and re-emission towards the surface, and the resulting energy balance; human causes and at least one feedback are discussed and linked to the temperature change. Physics terms are used accurately throughout.
Level 2	3–4	A mostly correct explanation including absorption and re-emission of infrared by greenhouse gases, with some discussion of human activity or a feedback. Some links are missing or loosely stated (for example no reference to wavelength or resonance).
Level 1	1–2	Isolated relevant points, such as naming greenhouse gases or saying they "trap heat", with little or no explanation of the mechanism. Terminology may be imprecise.
No marks	0	No relevant physics.

Answer: Short-wavelength sunlight passes through; the surface emits IR, which greenhouse gases absorb by resonance and re-emit, some back to the surface, raising the equilibrium temperature. Human emissions enhance this, amplified by ice-albedo and water-vapour/CO$_2$ feedbacks.
Examiner tip: Top-band answers link each step: wavelength of the radiation → resonance absorption → re-emission downwards → new energy balance at a higher temperature; a list of facts without that chain stays in the middle band.

10Standard6 marks

A student models the greenhouse effect. Two identical sealed glass flasks, A and B, are placed 30 cm from a filament lamp. Flask A contains air; flask B contains air with added carbon dioxide. A thermometer that reads to the nearest 1 °C, with an uncertainty of $\pm0.5\ ^\circ\mathrm{C}$, is fixed inside each flask.

	Flask A (air)	Flask B (air + CO$_2$)
Initial temperature / °C	21	21
Temperature after 20 minutes / °C	27	28

(a) State two variables, other than the distance to the lamp, that the student should keep the same for both flasks. [2]

(b) The student concludes that carbon dioxide caused flask B to warm more. By considering the uncertainties in the temperature rises, evaluate this conclusion. [2]

(d) Suggest one improvement that would make the conclusion more reliable. [1]

any one: volume / mass of gas in flask; initial temperature; time of exposure; position of thermometer (in flask / shielded from direct light); lamp power / same lamp; colour / type of flask(1)
a second different variable from the list(1)
rises are $6\pm1\ ^\circ\mathrm{C}$ (A) and $7\pm1\ ^\circ\mathrm{C}$ (B); uncertainty of a difference = sum of the two absolute uncertainties(1)
ranges 5–7 °C and 6–8 °C overlap, so the difference is not significant / the conclusion is not (fully) supported by these data(1)
any one: glass absorbs much of the infrared, so the flask itself warms (like a glasshouse) / sealed flask stops convection, unlike the atmosphere / CO$_2$ concentration is far higher than in the atmosphere / thermometer absorbs radiation directly from the lamp / lamp spectrum differs from the Sun's (more infrared) / CO$_2$ has a different specific heat capacity or density from air(1)
any one: repeat the experiment (several times, swapping the flasks' positions) and average / use temperature probes or data loggers with a resolution of 0.1 °C / record temperature over time and compare graphs / shield the thermometer bulb from direct radiation; do not accept "use a more accurate thermometer" without stating resolution or type(1)
Answer: (b) rises 6 ± 1 °C and 7 ± 1 °C; the ranges overlap so the conclusion is not supported (c) e.g. sealed glass flask blocks IR and convection (d) e.g. repeat and use probes with 0.1 °C resolution
Examiner tip: A temperature rise is a difference of two readings, so its uncertainty is ±1 °C, not ±0.5 °C; with ±0.5 °C the ranges (5.5–6.5 and 6.5–7.5) only just meet and the evaluation is weakened.

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