IB Physics SLTopic B.2 — Climate & the Greenhouse EffectPaper 1 & 2Albedo · Emissivity · Stefan–Boltzmann~7 min read
Albedo & Emissivity
Before you can model how the Earth heats up, you need two ratios that describe how a surface handles radiation: emissivity, which compares a real surface to a perfect black body, and albedo, which measures how much incoming radiation simply bounces back off.
📘 What you need to know
Emissivity (e) compares the power a real surface radiates to the power a perfect black body of the same size and temperature would radiate.
A perfect black body has e = 1 — this isn’t given in the data booklet, so it has to be memorised.
Albedo (a) compares the power a surface scatters (reflects) to the total power arriving at it.
Both e and a are dimensionless ratios — they carry no units, and both run from 0 to 1 for real surfaces.
Earth’s average albedo is taken as roughly 0.3, but the actual value shifts daily with cloud cover, latitude, terrain and the angle the sunlight strikes at.
Emissivity — Grading a Surface Against a Black Body
Stars behave almost exactly like black bodies, but planets, rooftops and skin do not — they emit less power than an ideal radiator at the same temperature would. Emissivity puts a number on that shortfall.
Emissivitye = power radiated by the surface ÷ power radiated by a black body
The comparison black body is assumed to sit at the same temperature and have the same surface area as the real object. Combine this with the Stefan–Boltzmann law and you get the working equation for any non-ideal radiator:
Power radiated by a real surfaceP = eσAT4
where P is total power emitted (W), e is the emissivity, σ is the Stefan–Boltzmann constant, A is the total surface area (m2), and T is the absolute temperature (K).
Albedo — How Much Bounces Straight Back
Albedo doesn’t care about temperature at all — it’s purely about reflection. A surface with a high albedo throws most of the incoming radiation straight back out; a surface with a low albedo lets most of it in to be absorbed.
Albedoa = total scattered power ÷ total incident power
For a whole planet, this becomes the ratio of radiation scattered back into space to the total radiation striking it. An albedo of 0 means a surface absorbs everything; an albedo of 1 means it reflects everything.
The same incoming radiation is mostly absorbed by a dark, low-albedo surface, but mostly reflected by a bright, high-albedo surface.
Typical albedo values worth knowing
🌍 Reference values
Fresh asphalt — around 0.04, one of the lowest natural surfaces
Bare soil — around 0.17
Green grass — around 0.25
Desert sand — around 0.40
New concrete — around 0.55
Ocean ice — roughly 0.50–0.70, depending on age and coverage
Fresh snow — around 0.85, among the highest of any common surface
What Makes Earth’s Albedo Change
Earth’s albedo isn’t fixed — it drifts from day to day because of several overlapping factors:
Cloud cover and season — thicker, more widespread cloud reflects more sunlight before it ever reaches the ground.
Latitude — polar regions, covered in ice and snow, reflect far more than the darker oceans near the equator.
Terrain — forests, deserts and cities all reflect light by very different amounts.
Angle of incidence — radiation striking a surface at a shallow angle tends to be reflected more strongly than radiation striking it head-on.
Quick recap: emissivity (e) compares radiated power to a black body at the same temperature; albedo (a) compares reflected power to incoming power. Neither has units, and both sit between 0 and 1 for real surfaces.
WE 1
A metal panel at 320 K radiates a total of 850 W. A black body of the same size and temperature would radiate 1250 W. Calculate the emissivity of the panel.
Step 1 — Identify the ratio needed
Emissivity compares the real object’s output to the black body’s output.
Step 2 — Substitute the valuese = 850 ÷ 1250e = 0.68No units — it’s a pure ratio, and less than 1 as expected for a non-ideal radiator.
WE 2
A patch of ageing sea ice reflects 620 W m⁻² out of the 1000 W m⁻² striking it. Find the ratio of energy absorbed to energy reflected.
Step 1 — Find the albedoa = 620 ÷ 1000 = 0.62Step 2 — Find the fraction absorbed
If 62% is reflected, the remaining 38% must be absorbed.
absorbed fraction = 1 − 0.62 = 0.38Step 3 — Take the ratio0.38 ÷ 0.62 = 0.61ratio ≈ 0.61For every unit of energy reflected, about 0.61 units are absorbed.
💡 Top tips
Memorise e = 1 for a perfect black body — it isn’t printed in the data booklet.
Neither e nor a has units — write your answer as a plain decimal (or percentage), never with W or m2 attached.
Keep the comparison object honest: for emissivity, the black body must match both the temperature and the size of the real surface.
⚠ Common mistakes
Mixing up emissivity and albedo — emissivity is about how well a surface radiates heat, albedo is about how much light it reflects. They describe two completely different processes.
Thinking a high albedo means “absorbs a lot” — it’s the opposite. High albedo means high reflection, low absorption.
Using the object’s cross-sectional area instead of its full surface area in P = eσAT4.
Up next: The Solar Constant — where we quantify exactly how much of the Sun’s energy reaches the top of Earth’s atmosphere before albedo and emissivity even come into play.
Want this to actually stick before the exam?
Book a free session and we’ll work through albedo, emissivity and energy-balance problems until they’re second nature.
