Topic B.2 sits at the intersection of thermal physics, electromagnetism, and climate science. It puts Stefan-Boltzmann and Wien's law from B.1 to work on a real physical system — the Earth — and forces students to balance incoming solar radiation against outgoing infrared. The rewards for getting it right are heavy: examiners use Greenhouse Effect questions as predictable Paper 2 long answers and as Paper 3 data-analysis stems.
This cheatsheet condenses every formula, definition, trick, and trap from Topic B.2 into one page. The layout mirrors the Photon Academy classroom workflow — formula box first, then the trick that saves time, then the trap that quietly costs marks. The worked example walks through an equilibrium-temperature calculation in IB mark-scheme rhythm, and the FAQ block addresses the conceptual questions most often asked in our small-group sessions.
§1 — Black-Body Radiation & Emissivity Topic B.2
Stefan-Boltzmann (black body)
Wien's displacement law
Sun ($T \approx 5800\,\text{K}$): peak in visible light. Earth ($T \approx 288\,\text{K}$): peak in IR around $10\,\mu\text{m}$.
Emissivity
$\varepsilon = 1$: perfect black body. Earth surface: $\varepsilon \approx 0.95\!-\!0.99$.
§2 — Albedo & Solar Constant Topic B.2
Albedo
Fraction absorbed $= 1 - \alpha$. Earth global average: $\alpha \approx 0.30$.
Solar constant & mean intensity
Solar constant $S \approx 1361\,\text{W m}^{-2}$ at Earth–Sun distance, just outside the atmosphere.
Factor of 4: intercepting cross-section $\pi R^2$ vs. total sphere area $4\pi R^2$.
Typical albedo values
| Surface | Albedo $\alpha$ |
|---|---|
| Fresh snow | 0.80 – 0.90 |
| Clouds | 0.40 – 0.90 |
| Ice | 0.50 – 0.70 |
| Desert | 0.25 – 0.40 |
| Forest | 0.08 – 0.15 |
| Ocean | 0.06 – 0.10 |
| Earth (global avg) | ≈ 0.30 |
| Moon | ≈ 0.10 |
§3 — Earth's Energy Balance Topic B.2
Equilibrium temperature
At equilibrium, power absorbed = power radiated:
$$(1-\alpha)\dfrac{S}{4} = \varepsilon \sigma T_e^4 \quad\Longrightarrow\quad T_e = \left[\dfrac{(1-\alpha)S}{4\varepsilon\sigma}\right]^{1/4}$$
With $\alpha = 0.30$, $\varepsilon = 1$, $S = 1361\,\text{W m}^{-2}$: $T_e \approx 254\,\text{K} \approx -19\,^\circ\text{C}$. Actual mean surface temperature: $+15\,^\circ\text{C}$. The $\sim$34 °C difference is the natural greenhouse effect.
Reading energy-balance diagrams
- Absorbed solar at top of atmosphere $= (1 - \alpha) \times S/4$.
- At equilibrium each layer obeys: power in = power out.
- Surface radiation = solar absorbed at surface + atmospheric back-radiation.
- Apply energy conservation to each layer separately.
§4 — Greenhouse Gases & Mechanism Topic B.2
The four greenhouse gases (IB syllabus)
| Gas | Formula | Natural source | Human source |
|---|---|---|---|
| Carbon dioxide | CO$_2$ | Respiration, volcanism | Fossil fuels, deforestation |
| Water vapour | H$_2$O | Evaporation | Cooling towers, irrigation |
| Methane | CH$_4$ | Wetlands | Livestock, landfills |
| Nitrous oxide | N$_2$O | Soil bacteria | Fertilisers |
Not greenhouse gases: O$_2$, N$_2$. Their symmetric vibrations don't change the molecular dipole moment, so they don't absorb IR.
Mechanism — molecular energy levels
- GHG molecules have vibrational modes with energy gaps in the IR range.
- Earth's surface emits IR (~10 µm). GHG molecules absorb IR by resonance ($h\nu = \Delta E$).
- The molecule transitions to a higher vibrational energy level.
- It re-emits an IR photon in a random direction. About half is back-radiated towards the surface.
- Surface temperature rises above the no-atmosphere equilibrium.
§5 — Enhanced Greenhouse Effect Topic B.2
Primary cause: burning fossil fuels (releases CO$_2$). More GHG ⇒ more IR absorbed ⇒ more back-radiation ⇒ higher surface temperature.
Two key positive-feedback loops
- Water-vapour feedback: warming ⇒ more evaporation ⇒ more H$_2$O (a GHG) ⇒ more warming.
- Ice-albedo feedback: warming ⇒ ice melts ⇒ surface albedo drops ⇒ more solar absorbed ⇒ more warming.
What increases warming vs. what reduces it
| Increases warming | Reduces warming |
|---|---|
| Burning fossil fuels | Carbon capture & storage |
| Deforestation | Nuclear / renewable energy |
| Increased volcanic activity | Afforestation |
| Livestock & agriculture (CH$_4$) | Increased CO$_2$ ocean solubility |
§6 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| "Power radiated by the Sun / a star…" | Stefan-Boltzmann $L = \sigma A T^4$ (then multiply by $\varepsilon$ if non-ideal) |
| "Peak wavelength of emission…" | Wien: $T = (2.9 \times 10^{-3})/\lambda_\text{max}$ |
| "Mean solar intensity at Earth…" | $I_\text{mean} = S/4$, then multiply by $(1-\alpha)$ for absorbed |
| "Equilibrium / effective temperature…" | $T_e = ((1-\alpha)S/(4\varepsilon\sigma))^{1/4}$ |
| "Why is Earth warmer than $T_e$?" | Greenhouse effect: GHG absorbs and re-emits IR back to surface |
| "Why aren't O$_2$ / N$_2$ greenhouse gases?" | Symmetric → vibrations don't change dipole moment → no IR absorption |
| "Increase / decrease global warming…" | Trace through: GHG concentration → IR absorption → back-radiation → $T$ |
| Energy-balance diagram (Paper 2/3) | Apply power-in = power-out at each labelled layer |
| Albedo of Earth changes… | Recompute $T_e$; expect $T \propto (1-\alpha)^{1/4}$ scaling |
| "State the mechanism…" | Use the words: absorb, re-emit, vibrational, back-radiation |
Worked Example — IB-Style Equilibrium Temperature
Question (HL Paper 2 style — 7 marks)
A planet orbits a star whose luminosity is $L_\star = 3.0 \times 10^{26}\,\text{W}$ at an orbital radius $d = 1.8 \times 10^{11}\,\text{m}$. The planet has a Bond albedo $\alpha = 0.25$ and behaves as a black body in the infrared ($\varepsilon = 1$). (a) Find the solar constant at the planet's orbit. (b) Find the planet's equilibrium surface temperature.
Solution
- Apply the inverse-square law: $S = \dfrac{L_\star}{4\pi d^2} = \dfrac{3.0 \times 10^{26}}{4\pi (1.8 \times 10^{11})^2}$. (M1)
- Evaluate the denominator: $4\pi (1.8 \times 10^{11})^2 \approx 4.07 \times 10^{23}\,\text{m}^2$ ⇒ $S \approx 737\,\text{W m}^{-2}$. (A1)
- Mean absorbed intensity: $(1-\alpha) S/4 = 0.75 \times 737/4 \approx 138\,\text{W m}^{-2}$. (M1)
- Apply energy balance with $\varepsilon = 1$: $\sigma T_e^4 = 138$ ⇒ $T_e^4 = 138/(5.67 \times 10^{-8}) \approx 2.43 \times 10^{9}\,\text{K}^4$. (R1)
- Solve: $T_e = (2.43 \times 10^{9})^{1/4} \approx 222\,\text{K}$. (A1)
- Quote with appropriate s.f.: $T_e \approx 2.2 \times 10^{2}\,\text{K}$ (or about $-51\,^\circ\text{C}$). (A1)
Examiner's note: The most common slip is using $S$ instead of $S/4$ in the energy balance (gives $T \approx 314$ K — way too high). Always remember the factor of 4 comes from spreading the cross-sectional intercept $\pi R^2$ over the full spherical surface $4\pi R^2$.
Common Student Questions
Why divide the solar constant by 4 when finding mean intensity?
Do greenhouse gases reflect or absorb infrared?
Why are O$_2$ and N$_2$ not greenhouse gases?
How do I use the equilibrium temperature formula?
Does increased CO$_2$ solubility in oceans warm or cool the planet?
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