SCIENCE · INVESTIGATE AND EXPLAIN

Build a one-layer greenhouse energy balance

Separate solar transmission, thermal absorption and two-direction atmospheric emission.

Learning outcome: Solve the surface and layer equilibrium equations while handling zero emissivity without inventing a temperature.

Starting recap

The surface is a blackbody. The prescribed grey layer transmits sunlight and has equal thermal absorptivity and emissivity.

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Try

At emissivity 80% and albedo 30%, surface emission is 238.175/0.60 = 396.9583 W/m².

Open the experiment

Predict

Does a layer that absorbs and emits no thermal radiation have a radiatively determined temperature?

Your prediction and reason

Experiment

Compare emissivity 0%, 80% and 100% at fixed albedo 30%. Trace solar input, surface emission and each atmospheric contribution.

Trace a one-layer greenhouse energy balance

Worked teaching model · synthetic values · Global-mean radiative equilibrium, fixed sunlight 1361 W/m² and σ = 5.670374419 × 10⁻⁸ W m⁻² K⁻⁴. All unreflected sunlight reaches a blackbody surface. One uniform atmospheric layer is transparent to sunlight and has equal grey thermal absorptivity/emissivity ε. It emits upward and downward. No convection, spectral/vertical structure, heat capacity, time evolution, coupled feedback or CO₂-concentration mapping is included. Temperatures are solved model values, not measurements.

Atmospheric thermal emissivity: 80 % Planetary albedo: 30 %

Reset keeps your prediction, captures and practice history.

The atmospheric layer emits in both directions. Gross surface emission can exceed absorbed sunlight because downward radiation also enters the surface budget. The net radiative transfer from surface to layer is 158.783 W/m² toward the layer; downward emission alone is not the net direction of heat exchange.

Three radiative balances at the solved equilibrium (fluxes in W/m²)
Boundary	Input	Output
Surface	Sun 238.175 + air 158.783	Gross emission 396.958
Atmospheric layer	Absorbed surface IR 317.567	Up 158.783 + down 158.783
Whole system	Unreflected sunlight 238.175	Direct surface 79.392 + air 158.783

Surface, layer and effective temperatures have different meanings
Quantity	Value
Uniform modeled surface temperature	289.257 K
Uniform atmospheric-layer temperature	243.235 K
Effective emitting temperature of combined outgoing flux	254.578 K
No-layer surface reference at the same albedo	254.578 K
Surface difference from no-layer reference	34.678 K

Let Q = (1 − α)S/4 be absorbed solar power per area, F = σTₛ⁴ be gross surface emission, and D be the layer's emission in each direction. Layer balance gives 2D = εF; surface balance gives F = Q + D. Therefore F = Q/(1 − ε/2). Direct surface escape is (1 − ε)F, so radiation reaching space is (1 − ε)F + D = Q.

For ε greater than zero, atmospheric balance gives Tₐ = Tₛ/2^(1/4). At ε = 0, dividing by ε is invalid: the layer neither absorbs nor emits, and its temperature is unconstrained by these radiative equations. It is not zero kelvin. The surface then equals the no-layer reference.

The graph compares equilibrium solutions; it is not a warming timeline. Neither parameter responds to temperature. Atmospheric emissivity is a grey teaching parameter, not CO₂ concentration, and the plotted sensitivity is not real-world equilibrium climate sensitivity. Convection, atmospheric solar absorption, wavelength-dependent absorption and emission, clouds, vertical structure and recorded observations require additional models and evidence.

Modeled surface 289.257 K; effective emission 254.578 K. Modeled atmospheric layer 243.235 K.

Absorbed sunlight 238.175 W/m²; total radiation to space 238.175 W/m². Surface gross emission 396.958 W/m² includes energy returned from the layer. All three budgets balance; the comparison does not simulate adjustment through time or predict real climate sensitivity.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Add one layer and name its assumptions

Draw a blackbody surface, one atmospheric layer above it, and space above both. Sunlight that is not reflected passes through the layer and is absorbed entirely at the surface. Thermal radiation leaving the surface is partly absorbed by the layer and partly transmitted to space. The layer emits thermal radiation upward and downward.

This is an idealized separation of solar and thermal behavior, not a statement that real air absorbs no sunlight. The surface has thermal emissivity one. The layer has thermal absorptivity and emissivity ε, controlled between zero and one. A grey parameter represents the thermal band without resolving individual wavelengths. The Penn State one-layer explanation develops this type of radiative balance.

Planetary albedo α determines how much solar energy is reflected before the remaining energy enters this budget. With S = 1361 W/m², absorbed solar flux is Q = (1 − α)S/4. At α = 0.30, Q = 238.175 W/m². This remains the external energy supply when you change ε at fixed albedo.

Follow absorption and emission separately

Let F be the surface’s upward thermal emission in W/m². The layer absorbs εF, while (1 − ε)F passes directly to space. Let D be the layer’s emission in each direction. Its total emission is 2D, because there are two equal outgoing contributions.

At atmospheric radiative equilibrium, absorption equals emission: εF = 2D. Thus D = εF/2. At surface equilibrium, absorbed sunlight plus downward atmospheric radiation balances surface emission: Q + D = F. Substituting the first equation into the second gives F = Q/(1 − ε/2).

Build a paper arrow diagram with those labels before inserting numbers. Do not send all of F to space and then add upward layer emission: some of F has already been absorbed. Keeping transmitted and absorbed paths separate makes energy conservation visible.

Calculate the default layer

At ε = 0.80 and α = 0.30, the denominator is 1 − 0.80/2 = 0.60. Surface emission is 238.175/0.60 = 396.9583 W/m². The layer absorbs 317.5667 W/m² and emits 158.7833 W/m² in each direction. Direct surface emission to space is 79.3917 W/m².

The surface temperature is Tₛ = (F/σ) raised to the power 1/4, about 289.2566 K. The Stefan–Boltzmann constant is σ = 5.670374419 × 10⁻⁸ W m⁻² K⁻⁴; use kelvin in the fourth power. Its value is listed in the NIST constants table.

For ε greater than zero, layer balance also gives Tₐ = Tₛ / 2^(1/4), about 243.2348 K at the default. Here the denominator is the fourth root of two. The atmospheric layer temperature is a lumped model quantity, not a vertical sounding or a temperature assigned to a particular height.

Inspect the zero-emissivity endpoint

Set ε to zero. The layer absorbs and emits no thermal radiation, so its balance becomes 0 = 0. This equation cannot determine an atmospheric temperature. The display therefore reports it as unconstrained, not 0 K. Dividing the layer equation by ε at zero would be invalid.

The surface still has a determined temperature: about 254.5781 K at 30% albedo. All its thermal radiation reaches space directly. Compare that no-atmosphere radiative reference with the default while holding albedo fixed. The experiment changes equilibrium solutions instantly; it does not calculate the time needed to warm between them.

Transfer with a different supply

For a paper-only case with Q = 200 W/m² and ε = 0.50, calculate surface emission, absorption by the layer, each directional layer emission and direct emission to space. Check the total reaching space.

Check the greenhouse-layer transfer

F = 200/0.75 = 266.6667 W/m². Atmospheric absorption is 133.3333, each layer emission is 66.6667 and direct surface emission is 133.3333 W/m². The direct contribution plus upward layer emission is 200 W/m², equal to the absorbed solar input.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain the two balance equations and why atmospheric temperature is unconstrained at zero emissivity.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Bound climate claims with energy evidence Next: Audit surface, atmosphere and space fluxes

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