IB · B · extended response
A planet's energy balance with an albedo feedback
The particulate nature of matter · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Blackbody emitted flux
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Emitted flux / W m⁻²
Horizontal axis (x): Temperature / K
Read the exact data as a table
| Temperature / K | Emitted flux / W m⁻² |
|---|---|
| 240 | 188.117 |
| 255 | 239.742 |
| 270 | 301.327 |
| 288 | 390.079 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
A planet receives stellar irradiance 1360 W m⁻² on a surface perpendicular to incoming light. Its albedo is initially 0.30. Assume a spherical planet and uniform effective emission σT⁴, where σ = 5.67×10⁻⁸ W m⁻² K⁻⁴. The supplied graph gives emitted flux for several effective temperatures. The actual surface is warmer than this simple emission temperature. Supplied data (Temperature / K; Emitted flux / W m⁻²): (240, 188.117), (255, 239.742), (270, 301.327), (288, 390.079).
- (a)
Calculate globally averaged absorbed stellar flux and equilibrium effective temperature, explaining the factor of four.
3 marks - (b)
Albedo drops to 0.25. Find the new absorbed flux and temperature, and distinguish the greenhouse mechanism from the albedo change.
3 marks
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(a)
Calculate globally averaged absorbed stellar flux and equilibrium effective temperature, explaining the factor of four.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The planet intercepts light over a disc πR² but emits across a spherical area 4πR², so averaged incoming irradiance is S/4.
- 2
Absorbed flux is (1−0.30)(1360)/4 = 238 W m⁻².
- 3
Equilibrium requires σT⁴ = 238, giving T = (238/σ)¹/⁴≈254.6 K. This is the effective emission temperature, not automatically the surface temperature.
(b)
Albedo drops to 0.25. Find the new absorbed flux and temperature, and distinguish the greenhouse mechanism from the albedo change.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
With albedo 0.25, absorbed flux is 0.75(1360)/4 = 255 W m⁻², an increase of 17 W m⁻².
- 2
New equilibrium temperature is (255/σ)¹/⁴≈259 K, about 4.4 K above the initial effective temperature.
- 3
Albedo changes reflected incoming light. Greenhouse gases absorb and emit outgoing infrared radiation and alter energy transfer between surface and atmosphere. Melting reflective ice can lower albedo and reinforce warming; this simple model does not quantify the complete coupled climate response.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- The planet intercepts light over a disc πR² but emits across a spherical area 4πR², so averaged incoming irradiance is S/4.
- Absorbed flux is (1−0.30)(1360)/4 = 238 W m⁻².
- Equilibrium requires σT⁴ = 238, giving T = (238/σ)¹/⁴≈254.6 K. This is the effective emission temperature, not automatically the surface temperature.
- With albedo 0.25, absorbed flux is 0.75(1360)/4 = 255 W m⁻², an increase of 17 W m⁻².
- New equilibrium temperature is (255/σ)¹/⁴≈259 K, about 4.4 K above the initial effective temperature.
- Albedo changes reflected incoming light. Greenhouse gases absorb and emit outgoing infrared radiation and alter energy transfer between surface and atmosphere. Melting reflective ice can lower albedo and reinforce warming; this simple model does not quantify the complete coupled climate response.
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