IB Physics Topic 2.2: Greenhouse effect | Notes & Flashcards | Aimnova

IB Physics — Greenhouse effect

Topic 2.2 of IB Physics covers Greenhouse effect, which is part of Unit 2: The particulate nature of matter. Students explore key concepts including Solar radiation, intensity and the solar constant, Black-body radiation: Stefan-Boltzmann and Wien, Albedo and Earth's energy balance, Greenhouse effect and greenhouse gases. A strong understanding of greenhouse effect is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Greenhouse effect

Key Idea: This topic follows the Sun's energy from source to planet: how its intensity spreads and weakens with distance, the black-body laws that fix how much power a hot body radiates and at which colour, and how albedo and the greenhouse effect decide how warm a planet ends up. It is examined on both papers — quick Paper 1A multiple-choice (inverse-square scaling, how a black-body curve shifts, the absorbed fraction) and longer Paper 2 structured questions ('show that' the solar constant or ~240 W m⁻², a black-body radius comparison, the greenhouse mechanism in words).

📋 Key formulas

Four of these are given in the data booklet (look for the booklet badge). The inverse-square and S ÷ 4 forms are just the given equations applied to a sphere.

I = \frac{P}{A}

Intensity = power ÷ area. The defining equation — given in the data booklet.

$I$: intensity — radiation power per unit area (W m⁻²)
$P$: radiation power passing through the area (W)
$A$: area the power spreads over (m²)

I = \frac{P}{4\pi d^{2}}

Power radiated equally in all directions spreads over a sphere (A = 4πd²) — the inverse-square law I ∝ 1/d². Not separately given; it is I = P/A with A = 4πd².

$I$: intensity at distance d (W m⁻²)
$P$: total power radiated by the source (W)
$d$: distance from the source (m)

L = \sigma A T^{4}

Stefan-Boltzmann law — total power (luminosity) radiated by a black body. Given. T must be in kelvin; power rises as T⁴.

$L$: luminosity — total power radiated (W)
$\sigma$: Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
$A$: surface area of the body (m²; a sphere has A = 4πr²)
$T$: absolute surface temperature (K — kelvin)

\lambda_{max}\,T = 2.9\times10^{-3}\ \text{m K}

Wien's displacement law — locates the peak wavelength. Given. Hotter body → shorter (bluer) peak; rearrange to λ_{max} = 2.9 × 10⁻³ ÷ T.

$\lambda_{max}$: wavelength of peak (brightest) emission (m)
$T$: absolute surface temperature (K — kelvin)

\text{albedo} = \frac{\text{total scattered power}}{\text{total incident power}}

Albedo — the fraction of arriving sunlight reflected back. Given. The ABSORBED fraction is (1 − albedo).

$\text{albedo}$: fraction of incident sunlight reflected (no unit, 0 to 1)
$\text{scattered power}$: power reflected/scattered back to space (W)
$\text{incident power}$: power arriving from the Sun (W)

⚖️ The four ideas side by side

🌍 Where the Sun's average energy goes (Earth)

Sunlight is captured on the disc Earth shows the Sun (area πr²) but shared over the whole sphere (4πr²): πr² ÷ 4πr² = 1/4. Earth's surface (~290 K) is far cooler than the Sun (~5800 K), so by Wien's law it radiates at a much longer (infrared) peak wavelength — and it is that outgoing infrared the greenhouse gases trap.

✏️ Worked exam-style questions

IB-style question — power radiated by a star (Stefan-Boltzmann)

A star behaves as a black body. Its radius is 9.0 × 10⁸ m and its surface temperature is 4.5 × 10³ K. Taking σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and treating the star as a sphere (A = 4πr²), determine the total power it radiates.

Solution:

Start with the given Stefan-Boltzmann law and the sphere area:
$L = \sigma A T^{4} = \sigma (4\pi r^{2}) T^{4}$
Find the surface area:
$A = 4\pi (9.0\times10^{8})^{2} = 1.02\times10^{19}\ \text{m}^{2}$
Substitute (σ, A, T = 4.5 × 10³ K); evaluate T⁴ first:
$L = (5.67\times10^{-8})(1.02\times10^{19})(4.5\times10^{3})^{4}$
Work it out — keep the unit:
$L = 2.4\times10^{26}\ \text{W}$

Final answer:

L ≈ 2.4 × 10²⁶ W. Putting T in kelvin and squaring the radius (for A = 4πr²) are the two steps students most often slip on.

IB-style question — peak wavelength and scaling intensity

A planet's star has a surface temperature of 7.25 × 10³ K. (a) Find the wavelength at which the star radiates most strongly. (b) At the planet the star's intensity is 2.0 × 10³ W m⁻². A moon orbits three times further from the star than the planet. Find the intensity at the moon.

Solution:

(a) Rearrange the given Wien law for the peak wavelength:
$\lambda_{max} = \frac{2.9\times10^{-3}}{T} = \frac{2.9\times10^{-3}}{7.25\times10^{3}}$
(a) Work it out:
$\lambda_{max} = 4.0\times10^{-7}\ \text{m}\ (400\ \text{nm, violet})$
(b) Intensity follows the inverse-square law, so scale by the distance ratio squared:
$I_{moon} = I_{planet}\left(\frac{1}{3}\right)^{2} = \frac{I_{planet}}{9}$
(b) Substitute the planet's intensity:
$I_{moon} = \frac{2.0\times10^{3}}{9} = 2.2\times10^{2}\ \text{W m}^{-2}$

Final answer:

(a) λₘₐₓ = 4.0 × 10⁻⁷ m. (b) Iₘₒₒₙ ≈ 2.2 × 10² W m⁻² — three times further → one ninth the intensity, not one third.

IB-style question — solar panel output from the solar constant

A spacecraft near Earth's orbit, where the intensity is the solar constant S = 1.36 × 10³ W m⁻², carries a solar array of area 12 m² that is 25% efficient. Determine the useful electrical power the array produces.

Solution:

Incident power = intensity × area (from I = P/A rearranged):
$P_{in} = I A = (1.36\times10^{3})(12)$
Evaluate the incident power:
$P_{in} = 1.63\times10^{4}\ \text{W}$
Useful output = incident power × efficiency (as a decimal):
$P_{out} = 0.25 \times (1.63\times10^{4})$
Work it out — keep the unit:
$P_{out} = 4.1\times10^{3}\ \text{W}$

Final answer:

Pₒᵤₜ ≈ 4.1 × 10³ W (about 4.1 kW). Useful output = intensity × area × efficiency — write the percentage as the decimal 0.25, never 25.

IB-style question — absorbed intensity and the greenhouse mechanism

(a) A planet receives an average solar intensity of S ÷ 4 = 340 W m⁻² and has an albedo of 0.32. Show that it absorbs about 230 W m⁻². (b) Outline how greenhouse gases keep its surface warmer than this balance alone would suggest.

Solution:

(a) Absorbed fraction = 1 − albedo:
$1 - 0.32 = 0.68$
(a) Absorbed intensity = absorbed fraction × incoming:
$I_{abs} = 0.68 \times 340 = 231 \approx 230\ \text{W m}^{-2}$
(b) Mark 1 — the warm surface emits infrared, which greenhouse-gas molecules absorb (their bonds resonate at infrared frequencies).
(b) Mark 2 — the molecules re-emit infrared in all directions, so some is sent back down, keeping the surface warmer than with no atmosphere.

Final answer:

(a) Iₐbₛ ≈ 230 W m⁻² (and at equilibrium the planet must radiate this same amount away). (b) Gases absorb the surface's outgoing infrared and re-emit some back down — sunlight still passes straight in.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Intensity is W m⁻²: divide power by the AREA it spreads over. A source radiating in all directions spreads over a sphere, so A = 4πd² (the 4π is easy to drop).
Inverse-square scaling beats finding the Sun's power: I₂ = I₁ (d₁/d₂)². Double d → quarter I; triple d → ninth I.
Always put T in KELVIN (K = °C + 273) before Stefan-Boltzmann or Wien. Power goes as T⁴ (×16 for a doubled T), not T.
Compare two black bodies by writing L = σAT⁴ for each and dividing — σ (and 4π for spheres) cancels, leaving a clean ratio of radii² and T⁴.
Albedo is the REFLECTED fraction; the absorbed (or transmitted) fraction is 1 − albedo. The whole-planet average uses S ÷ 4, but the Sun directly overhead uses the full S.
Greenhouse effect: sunlight passes IN; the gases trap the OUTGOING infrared. The two-mark mechanism is 'absorb the surface's infrared' + 're-emit some back down'.
Solar-panel useful output = incident intensity × panel area × efficiency, with the efficiency written as a decimal.

IB Physics — Greenhouse effect

Key concepts in Greenhouse effect

Key Idea: This topic follows the Sun's energy from source to planet: how its intensity spreads and weakens with distance, the black-body laws that fix how much power a hot body radiates and at which colour, and how albedo and the greenhouse effect decide how warm a planet ends up. It is examined on both papers — quick Paper 1A multiple-choice (inverse-square scaling, how a black-body curve shifts, the absorbed fraction) and longer Paper 2 structured questions ('show that' the solar constant or ~240 W m⁻², a black-body radius comparison, the greenhouse mechanism in words).

📋 Key formulas

Four of these are given in the data booklet (look for the booklet badge). The inverse-square and S ÷ 4 forms are just the given equations applied to a sphere.

I = \frac{P}{A}

Intensity = power ÷ area. The defining equation — given in the data booklet.

$I$: intensity — radiation power per unit area (W m⁻²)
$P$: radiation power passing through the area (W)
$A$: area the power spreads over (m²)

I = \frac{P}{4\pi d^{2}}

Power radiated equally in all directions spreads over a sphere (A = 4πd²) — the inverse-square law I ∝ 1/d². Not separately given; it is I = P/A with A = 4πd².

$I$: intensity at distance d (W m⁻²)
$P$: total power radiated by the source (W)
$d$: distance from the source (m)

L = \sigma A T^{4}

Stefan-Boltzmann law — total power (luminosity) radiated by a black body. Given. T must be in kelvin; power rises as T⁴.

$L$: luminosity — total power radiated (W)
$\sigma$: Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
$A$: surface area of the body (m²; a sphere has A = 4πr²)
$T$: absolute surface temperature (K — kelvin)

\lambda_{max}\,T = 2.9\times10^{-3}\ \text{m K}

Wien's displacement law — locates the peak wavelength. Given. Hotter body → shorter (bluer) peak; rearrange to λ_{max} = 2.9 × 10⁻³ ÷ T.

$\lambda_{max}$: wavelength of peak (brightest) emission (m)
$T$: absolute surface temperature (K — kelvin)

\text{albedo} = \frac{\text{total scattered power}}{\text{total incident power}}

Albedo — the fraction of arriving sunlight reflected back. Given. The ABSORBED fraction is (1 − albedo).

$\text{albedo}$: fraction of incident sunlight reflected (no unit, 0 to 1)
$\text{scattered power}$: power reflected/scattered back to space (W)
$\text{incident power}$: power arriving from the Sun (W)

⚖️ The four ideas side by side

🌍 Where the Sun's average energy goes (Earth)

Sunlight is captured on the disc Earth shows the Sun (area πr²) but shared over the whole sphere (4πr²): πr² ÷ 4πr² = 1/4. Earth's surface (~290 K) is far cooler than the Sun (~5800 K), so by Wien's law it radiates at a much longer (infrared) peak wavelength — and it is that outgoing infrared the greenhouse gases trap.

✏️ Worked exam-style questions

IB-style question — power radiated by a star (Stefan-Boltzmann)

Solution:

Start with the given Stefan-Boltzmann law and the sphere area:
$L = \sigma A T^{4} = \sigma (4\pi r^{2}) T^{4}$
Find the surface area:
$A = 4\pi (9.0\times10^{8})^{2} = 1.02\times10^{19}\ \text{m}^{2}$
Substitute (σ, A, T = 4.5 × 10³ K); evaluate T⁴ first:
$L = (5.67\times10^{-8})(1.02\times10^{19})(4.5\times10^{3})^{4}$
Work it out — keep the unit:
$L = 2.4\times10^{26}\ \text{W}$

Final answer:

L ≈ 2.4 × 10²⁶ W. Putting T in kelvin and squaring the radius (for A = 4πr²) are the two steps students most often slip on.

IB-style question — peak wavelength and scaling intensity

Solution:

(a) Rearrange the given Wien law for the peak wavelength:
$\lambda_{max} = \frac{2.9\times10^{-3}}{T} = \frac{2.9\times10^{-3}}{7.25\times10^{3}}$
(a) Work it out:
$\lambda_{max} = 4.0\times10^{-7}\ \text{m}\ (400\ \text{nm, violet})$
(b) Intensity follows the inverse-square law, so scale by the distance ratio squared:
$I_{moon} = I_{planet}\left(\frac{1}{3}\right)^{2} = \frac{I_{planet}}{9}$
(b) Substitute the planet's intensity:
$I_{moon} = \frac{2.0\times10^{3}}{9} = 2.2\times10^{2}\ \text{W m}^{-2}$

Final answer:

(a) λₘₐₓ = 4.0 × 10⁻⁷ m. (b) Iₘₒₒₙ ≈ 2.2 × 10² W m⁻² — three times further → one ninth the intensity, not one third.

IB-style question — solar panel output from the solar constant

Solution:

Incident power = intensity × area (from I = P/A rearranged):
$P_{in} = I A = (1.36\times10^{3})(12)$
Evaluate the incident power:
$P_{in} = 1.63\times10^{4}\ \text{W}$
Useful output = incident power × efficiency (as a decimal):
$P_{out} = 0.25 \times (1.63\times10^{4})$
Work it out — keep the unit:
$P_{out} = 4.1\times10^{3}\ \text{W}$

Final answer:

Pₒᵤₜ ≈ 4.1 × 10³ W (about 4.1 kW). Useful output = intensity × area × efficiency — write the percentage as the decimal 0.25, never 25.

IB-style question — absorbed intensity and the greenhouse mechanism

Solution:

(a) Absorbed fraction = 1 − albedo:
$1 - 0.32 = 0.68$
(a) Absorbed intensity = absorbed fraction × incoming:
$I_{abs} = 0.68 \times 340 = 231 \approx 230\ \text{W m}^{-2}$
(b) Mark 1 — the warm surface emits infrared, which greenhouse-gas molecules absorb (their bonds resonate at infrared frequencies).
(b) Mark 2 — the molecules re-emit infrared in all directions, so some is sent back down, keeping the surface warmer than with no atmosphere.

Final answer:

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Intensity is W m⁻²: divide power by the AREA it spreads over. A source radiating in all directions spreads over a sphere, so A = 4πd² (the 4π is easy to drop).
Inverse-square scaling beats finding the Sun's power: I₂ = I₁ (d₁/d₂)². Double d → quarter I; triple d → ninth I.
Always put T in KELVIN (K = °C + 273) before Stefan-Boltzmann or Wien. Power goes as T⁴ (×16 for a doubled T), not T.
Compare two black bodies by writing L = σAT⁴ for each and dividing — σ (and 4π for spheres) cancels, leaving a clean ratio of radii² and T⁴.
Albedo is the REFLECTED fraction; the absorbed (or transmitted) fraction is 1 − albedo. The whole-planet average uses S ÷ 4, but the Sun directly overhead uses the full S.
Greenhouse effect: sunlight passes IN; the gases trap the OUTGOING infrared. The two-mark mechanism is 'absorb the surface's infrared' + 're-emit some back down'.
Solar-panel useful output = incident intensity × panel area × efficiency, with the efficiency written as a decimal.

Key concepts in Greenhouse effect

📋 Key formulas

⚖️ The four ideas side by side

🌍 Where the Sun's average energy goes (Earth)

✏️ Worked exam-style questions

IB-style question — power radiated by a star (Stefan-Boltzmann)

IB-style question — peak wavelength and scaling intensity

IB-style question — solar panel output from the solar constant

IB-style question — absorbed intensity and the greenhouse mechanism

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 2.2

Study resources — 2.2 Greenhouse effect

Solar radiation, intensity and the solar constant

Black-body radiation: Stefan-Boltzmann and Wien

Albedo and Earth's energy balance

Greenhouse effect and greenhouse gases

Ready to study Greenhouse effect?

Ready to practice?

Key concepts in Greenhouse effect

📋 Key formulas

⚖️ The four ideas side by side

🌍 Where the Sun's average energy goes (Earth)

✏️ Worked exam-style questions

IB-style question — power radiated by a star (Stefan-Boltzmann)

IB-style question — peak wavelength and scaling intensity

IB-style question — solar panel output from the solar constant

IB-style question — absorbed intensity and the greenhouse mechanism

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 2.2

Study resources — 2.2 Greenhouse effect

Solar radiation, intensity and the solar constant

Black-body radiation: Stefan-Boltzmann and Wien

Albedo and Earth's energy balance

Greenhouse effect and greenhouse gases

Ready to study Greenhouse effect?

Ready to practice?