Greenhouse effect
IB Physics HL· Theme B.2: Greenhouse effect· 12 min read
1. Core Physical Mechanism of the Greenhouse Effect★★☆☆☆⏱ 3 min
Short-wavelength solar radiation (peak ~500 nm, visible spectrum) passes unabsorbed through the lower atmosphere to heat the Earth's surface. The warm surface then emits longer-wavelength infrared (IR) radiation, which interacts with greenhouse gas molecules in the atmosphere.
IR-active greenhouse gas
A gas molecule with an asymmetric dipole moment that can be excited to a higher vibrational energy level by absorbing an IR photon of matching energy.
Example:
CO₂, H₂O, CH₄, N₂O are all IR-active; symmetric diatomics N₂ and O₂ are not.
Calculate the approximate peak wavelength of radiation emitted by the Earth's 290 K surface, using Wien's displacement law (\lambda_{max} T = 2.9 \times 10^{-3} \text{ m K})
- 1
Rearrange Wien's law to solve for peak wavelength
- 2\(\lambda_{max} = \frac{2.9 \times 10^{-3}}{290}\)
- 3
Calculate the result
- 4\(\lambda_{max} = 1.0 \times 10^{-5} \text{ m} = 10 \, \mu \text{m}\)
- 5
This falls in the long-wavelength IR range that is strongly absorbed by CO₂ molecules.
Confirm your understanding of molecular absorption:
Which of these gases is a primary greenhouse gas?
N₂
O₂
CO₂
Ar
Reveal answer
CO₂ —CO₂ has an asymmetric vibrational mode that absorbs IR radiation, unlike the symmetric diatomics N₂ and O₂.
2. Earth Energy Balance and Equilibrium Temperature★★★★☆⏱ 4 min
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Derive the Earth's no-atmosphere equilibrium temperature
Incoming total solar power absorbed by Earth equals total power radiated out to space at steady state
- 1
Average incident solar power per unit area across the full spherical Earth surface is the solar constant divided by 4
- 2\(P_{in} = \frac{S}{4}(1-\alpha)\)
- 3
Total power radiated per unit area from the Earth's surface follows the Stefan-Boltzmann law
- 4\(P_{out} = \sigma T^4\)
- 5
Set incoming and outgoing power equal for equilibrium
- 6\(\frac{S}{4}(1-\alpha) = \sigma T^4\)
- 7
Rearrange to solve for temperature T
- 8\(T = \sqrt[4]{\frac{S(1-\alpha)}{4\sigma}}\)
For S=1360 W/m², α=0.3, this gives T=255 K (-18°C) for a no-atmosphere Earth.
Calculate the new equilibrium temperature if the global average albedo drops from 0.3 to 0.25 due to Arctic sea ice melt.
- 1
Substitute all known values into the equilibrium temperature formula
- 2\(T = \sqrt[4]{\frac{1360 \times (1-0.25)}{4 \times 5.67 \times 10^{-8}}}\)
- 3
Simplify the numerator and denominator terms
- 4\(T = \sqrt[4]{\frac{1020}{2.268 \times 10^{-7}}} = \sqrt[4]{4.497 \times 10^9}\)
- 5
Calculate the final temperature
- 6\(T = 259 \, \text{K} = -14 ^\circ \text{C}\)
- 7
This 4 K temperature rise demonstrates the positive ice-albedo feedback effect.
3. Albedo Variations and Feedback Loops★★★☆☆⏱ 2 min
Surface type | Typical albedo value |
|---|---|
Fresh snow | 0.8 - 0.9 |
Open ocean water | 0.06 |
Tropical forest | 0.15 |
Desert sand | 0.4 |
Cumulus cloud | 0.7 |
Positive feedback loops occur when an initial temperature change drives a secondary effect that amplifies the original change. The most commonly examined example is Arctic ice melt: rising temperatures reduce sea ice coverage, exposing low-albedo ocean water that absorbs more solar radiation, causing further warming and more ice melt.
Calculate the percentage change in absorbed solar power per unit area if 1 m² of albedo 0.8 sea ice is replaced by albedo 0.1 open ocean.
- 1
Calculate absorbed power for sea ice
- 2\(P_{ice} = \frac{S}{4}(1-0.8) = 340 \times 0.2 = 68 \text{ W m}^{-2}\)
- 3
Calculate absorbed power for open ocean
- 4\(P_{ocean} = 340 \times (1-0.1) = 306 \text{ W m}^{-2}\)
- 5
Find the percentage increase
- 6\(\Delta P \% = \frac{306 - 68}{68} \times 100 = 350 \%\)
4. Enhanced Greenhouse Effect and Impacts★★☆☆☆⏱ 2 min
Identify two anthropogenic activities that directly increase atmospheric CO₂ concentrations and drive the enhanced greenhouse effect.
- 1
- Combustion of fossil fuels for electricity generation, transport and industry releases geologically sequestered carbon as CO₂ into the atmosphere.
- 2
- Deforestation removes forest carbon sinks, reducing the rate at which atmospheric CO₂ is absorbed via photosynthesis.
5. Common Pitfalls
Wrong move:
Stating greenhouse gases absorb incoming short-wavelength visible solar radiation
Why:
Visible solar radiation passes almost unimpeded through the atmosphere; only outgoing long-wavelength terrestrial IR is absorbed
Correct move:
Explicitly note the absorption targets radiation emitted by the Earth's surface, not incoming solar radiation
Wrong move:
Forgetting to divide the solar constant by 4 in equilibrium temperature calculations
Why:
The solar constant is defined for a flat surface perpendicular to sunlight, but the spherical Earth distributes this power across 4x its cross-sectional area
Correct move:
Always use average incident power per unit area of S/4 = ~340 W/m² for global energy balance
Wrong move:
Listing N₂ and O₂ as primary greenhouse gases
Why:
Symmetric diatomic molecules have no net dipole moment, so their vibrational modes cannot absorb IR photons
Correct move:
Only name IR-active asymmetric molecules: CO₂, H₂O, CH₄, N₂O as primary greenhouse gases
Wrong move:
Confusing albedo and emissivity in the energy balance equation
Why:
Albedo describes reflected incoming radiation, emissivity describes the fraction of black body radiation emitted to space
Correct move:
Keep separate terms for reflected fraction (α) and emitted fraction (e) in all calculations
Wrong move:
Claiming the greenhouse effect is entirely anthropogenic
Why:
The natural greenhouse effect warms the Earth by 33°C from -18°C to 15°C, which is essential for all known life
Correct move:
Clearly distinguish the natural life-sustaining effect from the anthropogenic enhanced effect driven by excess emissions
6. Quick Reference Cheatsheet
Quantity | Definition | Standard Value | Exam Formula |
|---|---|---|---|
Solar constant S | Incident solar power at top of atmosphere | 1360 W m⁻² | (P_{in} = \frac{S}{4}(1-\alpha)) |
Albedo α | Fraction of incoming radiation reflected | 0.3 (global average) | (\alpha = P_{reflected} / P_{incident}) |
Stefan-Boltzmann constant σ | Black body radiation proportionality | 5.67 × 10⁻⁸ W m⁻² K⁻⁴ | (P_{out} = \sigma T^4) |
Equilibrium temperature | Steady state surface temperature | 288 K (15°C) | (T = \sqrt[4]{\frac{S(1-\alpha)}{4\sigma}}) |
When this came up on past exams
AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.
- 2024 · Paper 2
Greenhouse gas absorption mechanism
- 2023 · Paper 2
Equilibrium temperature calculation
- 2022 · Paper 3
Albedo and ice melt feedback loop
Going deeper
What's Next
Now that you have mastered the greenhouse effect and energy balance model, you can apply these core thermal physics principles to related high-weight IB HL exam topics. You will next explore conduction, convection and radiation as the three fundamental thermal energy transfer mechanisms, which builds directly on your understanding of radiative heat exchange between the Earth and its atmosphere. You can then progress to study the physics of climate mitigation strategies, including carbon capture technologies and low-carbon renewable energy generation, which are common 6-8 mark extended response topics in Paper 2 Section B. This content also transfers directly to the Astrophysics optional unit, where you will calculate equilibrium temperatures for exoplanets to assess habitability. Mastery of this sub-topic guarantees you almost all available marks for the greenhouse effect questions that appear in nearly every IB Physics HL exam cycle.
