Greenhouse effect HL Papers 1& 2

B.2 — Greenhouse Effect
HL Practice: Paper 2

15 extended-response questions • 1 lab investigation • Expandable answers

This original IB-style practice bank covers emissivity, albedo, solar irradiance, Wien’s law, energy balance, molecular absorption, and the enhanced greenhouse effect. The first seven questions retain and develop the supplied examples. Later questions add derivations, comparisons, feedback, uncertainty, and model evaluation for HL practice. The one-layer atmosphere questions explicitly supply an extension model.

Instructions and data

Attempt each question before opening its solution. Show equations, substitutions, units, and explanations.
Use $\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ and Wien’s constant $b=2.90\times10^{-3}\,\mathrm{m\,K}$ .
Use kelvin in radiation laws. Give final numerical answers to three significant figures unless uncertainty dictates otherwise; keep intermediate values unrounded.
Distinguish total power $P$ in watts from power per unit area $I$ or $F$ in $\mathrm{W\,m^{-2}}$ .
Bracketed marks are suggested practice marks. This bank is not a full official examination paper.

Notation: $\alpha$ is solar albedo; $\varepsilon$ is emitter infrared emissivity; $\varepsilon_a$ is atmospheric infrared absorptivity/emissivity in the supplied layer model; $b$ denotes Wien’s constant here.

Useful relationships

$\displaystyle I_{\mathrm{emitted}}=\varepsilon\sigma T^4,\qquad\alpha=\frac{P_{\mathrm{reflected}}}{P_{\mathrm{incident}}}$ $\displaystyle S=\frac{L}{4\pi r^2},\qquad F=\frac{(1-\alpha)S}{4},\qquad\lambda_{\max}T=b$ For a uniform single-emitter model with no explicit greenhouse atmosphere: $\displaystyle T=\left[\frac{(1-\alpha)S}{4\varepsilon\sigma}\right]^{1/4}$ Solar albedo and thermal-infrared emissivity refer to different spectral ranges; do not assume $\varepsilon=1-\alpha$ .

Extended-response questions

Question 1 — Emissivity and radiation flux [5 marks]

(a) A planet’s surface is at $210\,\mathrm{K}$ and emits a total radiant power per unit area of $110\,\mathrm{W\,m^{-2}}$ . Define emissivity. [1 mark]

(b) Calculate the surface emissivity, assuming a grey-body model. [2 marks]

(c) Explain why the quoted power per unit area must be distinguished from total power and from net radiative transfer. [2 marks]

Show solution and marking guidance

(a) [1 mark] Emissivity is the ratio of a surface’s emitted radiant power to that of a black body with the same area and temperature; $0\leq\varepsilon\leq1$ .

(b) [2 marks] Write the emitted flux as $I=\varepsilon\sigma T^4$ : $\displaystyle \varepsilon=\frac{110}{(5.67\times10^{-8})(210)^4}=0.9975\approx0.998$ The supplied numbers describe an almost black surface; they do not give $0.78$ .

(c) [2 marks] The quantity $I$ is measured in $\mathrm{W\,m^{-2}}$ ; total emitted power is $P=IA$ in watts. Net outward radiative transfer is emitted power minus absorbed incoming radiation, so it cannot be inferred from emission alone.

Question 2 — Albedo and absorbed solar radiation [5 marks]

(a) The globally averaged incident solar irradiance at the top of Earth’s atmosphere is $340\,\mathrm{W\,m^{-2}}$ . The reflected solar flux is $102\,\mathrm{W\,m^{-2}}$ . Calculate the planetary albedo. [2 marks]

(b) Calculate the globally averaged absorbed solar flux. [1 mark]

(c) Explain how increased snow and ice cover could affect the absorbed solar flux, assuming the incident flux is unchanged. [2 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle \alpha=\frac{I_{\mathrm{reflected}}}{I_{\mathrm{incident}}}=\frac{102}{340}=0.300$ Albedo is dimensionless.

(b) [1 mark] $\displaystyle F=340-102=238\,\mathrm{W\,m^{-2}}$

(c) [2 marks] Snow and ice generally reflect more sunlight than darker land or open water. More snow or ice can increase planetary albedo and reduce the absorbed fraction, provided other effects such as cloud changes are held fixed.

Question 3 — Solar constant and spherical geometry [6 marks]

(a) The Sun has luminosity $L=3.85\times10^{26}\,\mathrm{W}$ . Earth is $r=1.50\times10^{11}\,\mathrm{m}$ from the Sun. Calculate the solar constant, assuming isotropic emission and negligible absorption between Sun and Earth. [2 marks]

(b) Explain what is meant by solar constant and why it differs from globally averaged incident solar irradiance. [2 marks]

(c) Use your result to calculate the globally averaged incident solar irradiance. [2 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle S=\frac{L}{4\pi r^2}=\frac{3.85\times10^{26}}{4\pi(1.50\times10^{11})^2}\approx1.36\times10^3\,\mathrm{W\,m^{-2}}$

(b) [2 marks] The solar constant is incident solar power per unit area on a surface perpendicular to the rays at the planet’s orbital distance, outside its atmosphere. The planet intercepts sunlight over projected area $\pi R^2$ , while the global mean is taken over surface area $4\pi R^2$ .

(c) [2 marks] $\displaystyle \overline{I}_{\mathrm{incident}}=\frac{S\pi R^2}{4\pi R^2}=\frac{S}{4}\approx340\,\mathrm{W\,m^{-2}}$

Question 4 — Wien’s law and spectral comparison [6 marks]

(a) Calculate the wavelength of maximum emission for Earth modelled as a black body at $254\,\mathrm{K}$ . Identify the spectral region. [2 marks]

(b) Calculate the peak wavelength for the Sun modelled as a black body at $5800\,\mathrm{K}$ . [2 marks]

(c) Explain how the difference between these spectra helps explain the greenhouse effect. [2 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle \lambda_{\max}=\frac{b}{T}=\frac{2.90\times10^{-3}}{254}=1.14\times10^{-5}\,\mathrm{m}=11.4\,\mu\mathrm{m}$ This is infrared radiation.

(b) [2 marks] $\displaystyle \lambda_{\max,\mathrm{Sun}}=\frac{2.90\times10^{-3}}{5800}=5.00\times10^{-7}\,\mathrm{m}=500\,\mathrm{nm}$ The peak is in the visible region.

(c) [2 marks] Much incoming solar radiation lies at shorter wavelengths, including visible light, whereas Earth emits mainly longer-wavelength infrared. Greenhouse gases absorb and emit in particular infrared bands, affecting Earth’s outgoing radiation. Neither body emits only at its peak wavelength.

Question 5 — Radiative equilibrium and effective temperature [7 marks]

(a) Earth’s globally averaged absorbed solar flux is $238\,\mathrm{W\,m^{-2}}$ . Assume a uniform black-body emitter with emissivity $1.00$ . Calculate the radiative equilibrium temperature. [3 marks]

(b) State two assumptions of this model. [2 marks]

(c) Explain why the result need not equal Earth’s average surface temperature, approximately $288\,\mathrm{K}$ . [2 marks]

Show solution and marking guidance

(a) [3 marks] At equilibrium, outgoing emitted flux equals absorbed solar flux: $\displaystyle 238=\sigma T_e^4$ $\displaystyle T_e=\left(\frac{238}{5.67\times10^{-8}}\right)^{1/4}=254.536\,\mathrm{K}\approx255\,\mathrm{K}$ This is about $-18.6\,^{\circ}\mathrm{C}$ .

(b) [2 marks] Accept two: global radiative equilibrium; emissivity $1$ ; uniform effective emitting temperature; negligible internal energy source; no explicit atmospheric layers or non-radiative transfers.

(c) [2 marks] This is the effective radiating temperature of the planet as viewed from space. The atmosphere absorbs and emits infrared; much radiation reaching space originates above the surface. The surface can therefore be warmer than the effective radiating temperature.

Question 6 — Changing effective emissivity [6 marks]

(a) A simplified single-emitter model has absorbed flux $238\,\mathrm{W\,m^{-2}}$ . Its effective emissivity decreases from $1.00$ to $0.950$ . Calculate the new equilibrium temperature. [2 marks]

(b) Use a ratio to determine the percentage temperature increase without separately rounding the two temperatures. [2 marks]

(c) Explain the direction of the change and state a limitation of using this model to represent a greenhouse atmosphere. [2 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle T_{\mathrm{new}}=\left(\frac{238}{(0.950)(5.67\times10^{-8})}\right)^{1/4}=257.821\,\mathrm{K}\approx258\,\mathrm{K}$

(b) [2 marks] $\displaystyle \frac{T_{\mathrm{new}}}{T_{\mathrm{old}}}=\left(\frac{1}{0.950}\right)^{1/4}=1.01291$ The increase is $1.29\%$ , or about $3.29\,\mathrm{K}$ .

(c) [2 marks] At unchanged temperature, a lower effective emissivity means less emission to space. Warming raises emission until it balances absorption. This effective parameter is not the infrared emissivity of an actual atmospheric layer: increased greenhouse-gas absorption and emission require a layered or wavelength-dependent model.

Question 7 — Resonance and molecular energy levels [7 marks]

(a) Explain infrared absorption by carbon dioxide using a resonance model, including the condition needed for interaction with the radiation’s electric field. [3 marks]

(b) Explain absorption using a molecular energy-level model. [2 marks]

(c) Explain why nitrogen is a very weak absorber of terrestrial infrared in this comparison, even though nitrogen molecules have vibrational energy levels. [2 marks]

Show solution and marking guidance

(a) [3 marks] Incident radiation can transfer energy efficiently when its frequency matches an infrared-active molecular vibrational mode. Such a mode changes the molecule’s electric dipole moment, enabling coupling to the oscillating electric field. Carbon dioxide’s bending and asymmetric stretching modes meet this condition.

(b) [2 marks] Photon energy must match the difference between allowed molecular energy levels: $\displaystyle hf=E_2-E_1$ Absorption excites an allowed transition; frequency matching alone is insufficient if the transition cannot couple to the radiation.

(c) [2 marks] Nitrogen is a homonuclear diatomic molecule. Its ordinary vibration does not change its electric dipole moment, so strong electric-dipole infrared transitions are absent. It is incorrect to say that nitrogen has no vibrational energy levels. Weak collision-induced absorption is outside this simplified comparison.

Question 8 — Deriving a planetary energy-balance equation [9 marks]

(a) A spherical planet has radius $R$ , solar constant $S$ , albedo $\alpha$ , and uniform infrared emissivity $\varepsilon$ . With no greenhouse atmosphere and negligible internal heating, write expressions for the total absorbed solar power and total emitted thermal power. [2 marks]

(b) Derive an expression for the equilibrium temperature and explain why radius cancels. [3 marks]

(c) Calculate the temperature for $S=1360\,\mathrm{W\,m^{-2}}$ , $\alpha=0.300$ , and $\varepsilon=1.00$ . [2 marks]

(d) For $R=6.37\times10^6\,\mathrm{m}$ , calculate the total absorbed solar power. [2 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle P_{\mathrm{abs}}=(1-\alpha)S\pi R^2$ $\displaystyle P_{\mathrm{emit}}=4\pi R^2\varepsilon\sigma T^4$ The first area is the projected disc; the second is the entire emitting surface.

(b) [3 marks] Equate the powers and cancel $\pi R^2$ : $\displaystyle T=\left[\frac{(1-\alpha)S}{4\varepsilon\sigma}\right]^{1/4}$ Both powers scale with $R^2$ , so radius does not affect equilibrium temperature under these assumptions.

(c) [2 marks] $\displaystyle T=\left[\frac{(0.700)(1360)}{4(5.67\times10^{-8})}\right]^{1/4}\approx255\,\mathrm{K}$

(d) [2 marks] $\displaystyle P_{\mathrm{abs}}=(0.700)(1360)\pi(6.37\times10^6)^2\approx1.21\times10^{17}\,\mathrm{W}$

Question 9 — Comparing two planets [8 marks]

(a) Planet A has solar constant $1360\,\mathrm{W\,m^{-2}}$ , albedo $0.300$ , and infrared emissivity $1.00$ . Planet B orbits the same star at $1.50$ times A’s distance and has albedo $0.600$ and infrared emissivity $0.900$ . Neither has a greenhouse atmosphere. Calculate B’s solar constant. [2 marks]

(b) Determine the equilibrium temperature ratio B divided by A. [3 marks]

(c) Calculate B’s equilibrium temperature and explain which changes tend to warm or cool it relative to A. [3 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle S_B=\frac{1360}{1.50^2}\approx604\,\mathrm{W\,m^{-2}}$

(b) [3 marks] $\displaystyle \frac{T_B}{T_A}=\left[\frac{(1-0.600)}{(1-0.300)}\frac{1}{1.50^2}\frac{1.00}{0.900}\right]^{1/4}\approx0.729$

(c) [3 marks] $\displaystyle T_B=\left[\frac{(0.400)(1360/2.25)}{4(0.900)(5.67\times10^{-8})}\right]^{1/4}\approx186\,\mathrm{K}$ Greater distance reduces irradiance and higher albedo reduces absorption; both cool B. Lower infrared emissivity tends to warm B, but does not outweigh the other changes.

Question 10 — Ice–albedo feedback [8 marks]

(a) In a no-atmosphere model, the solar constant is $1360\,\mathrm{W\,m^{-2}}$ and emissivity is $1.00$ . Albedo decreases from $0.300$ to $0.250$ . Calculate the increase in globally averaged absorbed solar flux. [2 marks]

(b) Calculate the new equilibrium temperature and its increase from the original value. [3 marks]

(c) Explain how melting ice can produce positive feedback, and distinguish feedback from an initial forcing. [3 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle \Delta F=\frac{1360}{4}[(1-0.250)-(1-0.300)]=17.0\,\mathrm{W\,m^{-2}}$

(b) [3 marks] $\displaystyle T_{\mathrm{new}}=\left[\frac{(0.750)(1360)}{4(5.67\times10^{-8})}\right]^{1/4}\approx259\,\mathrm{K}$ Using unrounded temperatures: $\displaystyle \Delta T=258.964-254.536\approx4.43\,\mathrm{K}$

(c) [3 marks] An initial warming can melt reflective ice, expose darker surfaces, lower albedo, and increase solar absorption. The additional warming reinforces the initial change, giving positive feedback. The initial forcing is the imposed disturbance; the ice–albedo response is a feedback. These fixed-albedo calculations alone do not describe a complete feedback cycle.

Question 11 — An idealised one-layer greenhouse model [10 marks]

(a) A planet absorbs globally averaged solar flux $F=238\,\mathrm{W\,m^{-2}}$ entirely at its black surface. A single atmospheric layer is transparent to solar radiation, absorbs all surface infrared, and emits black-body infrared equally upward and downward. Neglect convection and evaporation. Write the atmospheric energy-balance equation in terms of surface temperature $T_s$ and atmospheric temperature $T_a$ . [2 marks]

(b) Write the surface energy-balance equation and combine the equations to find the surface emission flux. [3 marks]

(c) Calculate the equilibrium atmospheric and surface temperatures. [3 marks]

(d) Explain why downward infrared radiation does not violate energy conservation. [2 marks]

Show solution and marking guidance

(a) [2 marks] The atmosphere absorbs $\sigma T_s^4$ and emits from both sides: $\displaystyle \sigma T_s^4=2\sigma T_a^4$

(b) [3 marks] The surface absorbs solar flux plus downward atmospheric radiation: $\displaystyle F+\sigma T_a^4=\sigma T_s^4$ Thus $F=\sigma T_a^4$ and $\sigma T_s^4=2F=476\,\mathrm{W\,m^{-2}}$ .

(c) [3 marks] $\displaystyle T_a=\left(\frac{238}{\sigma}\right)^{1/4}\approx255\,\mathrm{K}$ $\displaystyle T_s=\left(\frac{476}{\sigma}\right)^{1/4}\approx303\,\mathrm{K}$ These are model results, not a prediction of Earth’s actual surface temperature.

(d) [2 marks] Radiation transfers energy between surface and atmosphere in both directions. Net surface-to-atmosphere radiation remains outward because the surface is hotter. The whole planet loses only the upward atmospheric flux to space, equal to the absorbed solar flux at equilibrium; the layer creates no energy.

Question 12 — Partial atmospheric absorption: an extension [12 marks]

(a) Use the same assumptions as Question 11, but let atmospheric infrared absorptivity and emissivity both be $\varepsilon_a$ , with no infrared reflection. The surface remains black. Write the atmospheric balance and show that $T_a^4=T_s^4/2$ for $\varepsilon_a>0$ . [2 marks]

(b) Write the surface balance and derive $F=\sigma T_s^4(1-\varepsilon_a/2)$ . [2 marks]

(c) Calculate $T_s$ for $\varepsilon_a=0.600$ and then for $\varepsilon_a=0.800$ , with $F=238\,\mathrm{W\,m^{-2}}$ . [3 marks]

(d) For $\varepsilon_a=0.600$ , calculate the total outgoing flux to space, including direct surface radiation and atmospheric radiation. [3 marks]

(e) State two limitations of this model. [2 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle \varepsilon_a\sigma T_s^4=2\varepsilon_a\sigma T_a^4$ For nonzero $\varepsilon_a$ , division gives $T_a^4=T_s^4/2$ .

(b) [2 marks] $\displaystyle F+\varepsilon_a\sigma T_a^4=\sigma T_s^4$ Substitute the atmospheric result: $\displaystyle F=\sigma T_s^4\left(1-\frac{\varepsilon_a}{2}\right)$

(c) [3 marks] $\displaystyle T_s=\left[\frac{F}{\sigma(1-\varepsilon_a/2)}\right]^{1/4}$ $\displaystyle T_s(0.600)\approx278\,\mathrm{K},\qquad T_s(0.800)\approx289\,\mathrm{K}$ Greater atmospheric infrared absorption produces a higher surface equilibrium temperature in this model.

(d) [3 marks] For $\varepsilon_a=0.600$ , surface emission is $238/0.700=340\,\mathrm{W\,m^{-2}}$ . Direct escape is $(1-0.600)(340)=136\,\mathrm{W\,m^{-2}}$ . Atmospheric upward emission is $(0.600)(340/2)=102\,\mathrm{W\,m^{-2}}$ . Their sum is $238\,\mathrm{W\,m^{-2}}$ , equal to absorbed solar flux.

(e) [2 marks] Accept two: one uniform-temperature layer; wavelength-independent infrared properties; no clouds or atmospheric solar absorption; no convection or latent-energy transfer; no geographical or time variation. This is an explicitly supplied extension model, not an additional formula to memorise.

Question 13 — Infrared photons and absorption bands [7 marks]

(a) A molecular transition has energy gap $1.33\times10^{-20}\,\mathrm{J}$ . Calculate the photon frequency that can excite this allowed transition. Use $h=6.63\times10^{-34}\,\mathrm{J\,s}$ . [2 marks]

(b) Calculate the wavelength using $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$ and identify its spectral region. [2 marks]

(c) Earth’s surface is modelled at $288\,\mathrm{K}$ . Calculate its peak emission wavelength and explain why a gas absorbing near $15\,\mu\mathrm{m}$ can still affect its energy loss. [3 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle f=\frac{\Delta E}{h}=\frac{1.33\times10^{-20}}{6.63\times10^{-34}}\approx2.01\times10^{13}\,\mathrm{Hz}$

(b) [2 marks] $\displaystyle \lambda=\frac{c}{f}=\frac{hc}{\Delta E}\approx1.50\times10^{-5}\,\mathrm{m}=15.0\,\mu\mathrm{m}$ This is infrared.

(c) [3 marks] $\displaystyle \lambda_{\max}=\frac{2.90\times10^{-3}}{288}\approx1.01\times10^{-5}\,\mathrm{m}=10.1\,\mu\mathrm{m}$ Thermal emission has a broad spectrum. Radiation is emitted at $15\,\mu\mathrm{m}$ as well as near the peak, so absorption in that band can affect outgoing energy transfer.

Question 14 — Interpreting an energy imbalance [10 marks]

(a) A hypothetical satellite dataset gives global mean incoming solar flux $340\,\mathrm{W\,m^{-2}}$ , reflected solar flux $102\,\mathrm{W\,m^{-2}}$ , and outgoing longwave flux $236\,\mathrm{W\,m^{-2}}$ . Determine the net energy imbalance and state its sign. [2 marks]

(b) Use Earth’s surface area $5.10\times10^{14}\,\mathrm{m^2}$ to calculate the total rate of energy accumulation. [2 marks]

(c) If that imbalance persisted for $365$ days, calculate the accumulated energy. In a simplified calculation, assume it all warms ocean water of mass $1.00\times10^{21}\,\mathrm{kg}$ with $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . Find the temperature rise. [3 marks]

(d) Explain how an increase in greenhouse gases can initially produce a positive imbalance and how warming can restore radiative balance. [3 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle N=340-102-236=+2.00\,\mathrm{W\,m^{-2}}$ Positive means the planet accumulates energy.

(b) [2 marks] $\displaystyle P_{\mathrm{gain}}=NA=(2.00)(5.10\times10^{14})=1.02\times10^{15}\,\mathrm{W}$

(c) [3 marks] $\displaystyle E=(1.02\times10^{15})(365\times86400)\approx3.22\times10^{22}\,\mathrm{J}$ $\displaystyle \Delta T=\frac{E}{mc_w}\approx7.70\times10^{-3}\,\mathrm{K}$ This is only the stated uniform-ocean model, not a prediction of atmospheric warming.

(d) [3 marks] At initially unchanged temperatures, increased greenhouse-gas concentrations can reduce outgoing longwave radiation to space, creating a positive imbalance. Energy then accumulates in the climate system. As temperatures rise, thermal emission increases and may restore balance at a warmer state, with the response also affected by feedbacks.

Question 15 — Uncertainty and model interpretation [9 marks]

(a) A single-emitter planet model uses $S=(1360\pm20)\,\mathrm{W\,m^{-2}}$ , $\alpha=0.300\pm0.020$ , and $\varepsilon=0.900\pm0.030$ . Calculate $T=[S(1-\alpha)/(4\varepsilon\sigma)]^{1/4}$ . [2 marks]

(b) Estimate the maximum percentage uncertainty in temperature by adding fractional uncertainty contributions. Treat constants as exact. [3 marks]

(c) State the temperature with an absolute uncertainty. [1 mark]

(d) If the same absorbed flux $238\,\mathrm{W\,m^{-2}}$ is represented by $\varepsilon_{\mathrm{eff}}\sigma(288\,\mathrm{K})^4$ , calculate $\varepsilon_{\mathrm{eff}}$ . Explain why this is not a measurement of Earth’s surface emissivity. [3 marks]

Show solution and marking guidance

(a) [2 marks] $\displaystyle T=\left[\frac{(1360)(0.700)}{4(0.900)(5.67\times10^{-8})}\right]^{1/4}\approx261\,\mathrm{K}$

(b) [3 marks] The absolute uncertainty in $1-\alpha$ is $0.020$ . Hence: $\displaystyle \frac{\delta T}{T}\approx\frac14\left[\frac{20}{1360}+\frac{0.020}{0.700}+\frac{0.030}{0.900}\right]=0.01915$ Maximum percentage uncertainty is $1.92\%$ .

(c) [1 mark] $\displaystyle \delta T\approx(0.01915)(261.329)\approx5.0\,\mathrm{K}$ $\displaystyle T=(261\pm5)\,\mathrm{K}$

(d) [3 marks] $\displaystyle \varepsilon_{\mathrm{eff}}=\frac{238}{(5.67\times10^{-8})(288)^4}\approx0.610$ This parameter relates emission to space to a chosen surface temperature in a collapsed model. It incorporates atmospheric effects and is not the material infrared emissivity of the ground or ocean.

Lab investigation — Measuring albedo

Investigation 1 — Comparing two opaque panels [12 marks]

A laboratory compares a dark panel and a light panel under the same lamp. The quoted reflected powers are calibrated totals collected over the reflection hemisphere, with background and collection corrections already applied. Both panels are opaque and have the same illuminated area. Use the data below; uncertainties are absolute.

Incident and reflected radiant powers
Quantity	Dark panel	Light panel
Incident power / W	4.00 ± 0.08	4.00 ± 0.08
Total reflected power / W	0.800 ± 0.040	2.40 ± 0.06

(a) Describe how to measure incident power and total reflected power for the two panels. Identify controls and explain why one detector at a single angle is insufficient. [3 marks]

(b) Calculate each panel’s albedo and the power absorbed by the dark panel, assuming no transmission. [2 marks]

(c) Estimate the maximum uncertainty in the light panel’s albedo and state the result with an absolute uncertainty. [3 marks]

(d) For two otherwise identical no-atmosphere planets with these albedos, the same solar constant, and the same infrared emissivity, calculate the equilibrium temperature ratio light divided by dark. [2 marks]

(e) Explain two limitations of applying these laboratory measurements directly to Earth’s planetary energy balance. [2 marks]

Show investigation solution and evaluation

(a) [3 marks] Measure incident beam power at the sample position using a calibrated sensor, then use an integrating sphere or calibrated angular sampling with appropriate solid-angle weighting to collect the total reflected power. Subtract background signals and account for collection losses. Keep illuminated area, lamp output and spectrum, incidence angle, geometry, and surface condition fixed; repeat measurements. A detector at one angle samples only part of the reflected radiation and cannot directly determine hemispherical reflectance.

(b) [2 marks] $\displaystyle \alpha_d=\frac{0.800}{4.00}=0.200,\qquad\alpha_l=\frac{2.40}{4.00}=0.600$ $\displaystyle P_{\mathrm{absorbed},d}=4.00-0.800=3.20\,\mathrm{W}$

(c) [3 marks] $\displaystyle \frac{\delta\alpha_l}{\alpha_l}\approx\frac{0.06}{2.40}+\frac{0.08}{4.00}=0.0450$ Maximum percentage uncertainty is $4.50\%$ . $\displaystyle \delta\alpha_l=(0.600)(0.0450)=0.027$ $\displaystyle \alpha_l=0.600\pm0.027$

(d) [2 marks] $\displaystyle \frac{T_l}{T_d}=\left[\frac{1-0.600}{1-0.200}\right]^{1/4}=\left(\frac12\right)^{1/4}\approx0.841$

(e) [2 marks] The lamp spectrum and incidence angle may differ from the solar spectrum and range of angles on Earth, so the measured reflectance need not equal the solar-weighted planetary albedo. Earth also contains clouds, atmosphere, oceans, land, and changing ice cover, whereas these measurements concern two uniform panels. Visible reflectance does not by itself determine thermal-infrared emissivity.

Exam tips and common pitfalls

Use $S/4$ for a globally averaged incoming flux; $S$ applies to a surface perpendicular to the solar rays.
Albedo is dimensionless. Subtract the reflected fraction when calculating absorbed sunlight.
In $P=\varepsilon\sigma AT^4$ , include area for total power. Omit area only when calculating power per unit area.
Equilibrium means no net energy accumulation; absorption and emission still occur.
Distinguish effective radiating temperature, surface temperature, and atmospheric-layer temperature.
Carbon dioxide has infrared-active modes because those vibrations change its dipole moment. Nitrogen’s weak infrared interaction is not caused by an absence of energy levels.
Increasing atmospheric infrared emissivity is different from decreasing a collapsed planet-model effective emissivity.
State model assumptions and avoid treating an idealised model result as an exact climate prediction.

Faith connection — God’s care and our responsibility

Hebrews 1:1–3 presents Jesus as the One through whom God made all things and who sustains creation. In 1 John 1:5, God is described as light; Revelation 21:3–5 points to His presence with His people and the renewal of creation. These passages invite us to know God’s character and the love revealed in Jesus.

Studying Earth’s energy balance can encourage gratitude for a habitable world and thoughtful care for our neighbours and creation. Reflect on one practical way your school can use energy responsibly while showing Christ’s love to others.

Further reading: NASA — Earth’s radiation budget; David Archer — Greenhouse gases and molecular absorption.

B.2 — Greenhouse Effect
HL Practice: Paper 2

Instructions and data

Useful relationships

Extended-response questions

Question 1 — Emissivity and radiation flux [5 marks]

Question 2 — Albedo and absorbed solar radiation [5 marks]

Question 3 — Solar constant and spherical geometry [6 marks]

Question 4 — Wien’s law and spectral comparison [6 marks]

Question 5 — Radiative equilibrium and effective temperature [7 marks]

Question 6 — Changing effective emissivity [6 marks]

Question 7 — Resonance and molecular energy levels [7 marks]

Question 8 — Deriving a planetary energy-balance equation [9 marks]

Question 9 — Comparing two planets [8 marks]

Question 10 — Ice–albedo feedback [8 marks]

Question 11 — An idealised one-layer greenhouse model [10 marks]

Question 12 — Partial atmospheric absorption: an extension [12 marks]

Question 13 — Infrared photons and absorption bands [7 marks]

Question 14 — Interpreting an energy imbalance [10 marks]

Question 15 — Uncertainty and model interpretation [9 marks]

Lab investigation — Measuring albedo

Investigation 1 — Comparing two opaque panels [12 marks]

Exam tips and common pitfalls

Faith connection — God’s care and our responsibility

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IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Greenhouse effect HL Papers 1& 2

B.2 — Greenhouse EffectHL Practice: Paper 2

Instructions and data

Useful relationships

Extended-response questions

Question 1 — Emissivity and radiation flux [5 marks]

Question 2 — Albedo and absorbed solar radiation [5 marks]

Question 3 — Solar constant and spherical geometry [6 marks]

Question 4 — Wien’s law and spectral comparison [6 marks]

Question 5 — Radiative equilibrium and effective temperature [7 marks]

Question 6 — Changing effective emissivity [6 marks]

Question 7 — Resonance and molecular energy levels [7 marks]

Question 8 — Deriving a planetary energy-balance equation [9 marks]

Question 9 — Comparing two planets [8 marks]

Question 10 — Ice–albedo feedback [8 marks]

Question 11 — An idealised one-layer greenhouse model [10 marks]

Question 12 — Partial atmospheric absorption: an extension [12 marks]

Question 13 — Infrared photons and absorption bands [7 marks]

Question 14 — Interpreting an energy imbalance [10 marks]

Question 15 — Uncertainty and model interpretation [9 marks]

Lab investigation — Measuring albedo

Investigation 1 — Comparing two opaque panels [12 marks]

Exam tips and common pitfalls

Faith connection — God’s care and our responsibility

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B.2 — Greenhouse Effect
HL Practice: Paper 2