Solve for an effective radiative equilibrium

Find balance without confusing the temperature definitions

Set planetary albedo to 30%. The model absorbs 238.175 W/m². Ask what effective emitting temperature would produce exactly that outgoing thermal flux. This is an equilibrium calculation: solve for the temperature at which two energy-transfer rates match.

An effective emitting temperature is the temperature of an equivalent unit-emissivity emitter that produces the same outgoing flux. It is not automatically the mean surface temperature. It is also not the arithmetic average of temperatures across all real emitting regions. Different temperatures contribute nonlinearly to radiation, so averaging temperatures first generally changes the result.

The lab represents the planet with this equivalent emitter. It has no separate surface and atmosphere temperatures, no greenhouse layer and no ocean heat reservoir. Those omissions limit the calculation; they do not assert that Earth’s atmosphere or greenhouse effect is absent in nature. NASA’s energy-budget account describes the broader physical system.

Use an absolute temperature in the fourth power

Outgoing flux F is σT⁴, where T is temperature in kelvin and σ is the Stefan–Boltzmann constant. The lab uses σ = 5.670374419 × 10⁻⁸ W m⁻² K⁻⁴, consistent with the NIST constant listing. Unit thermal emissivity is built into this equivalent-emitter definition.

The notation T⁴ means T × T × T × T. At 255 K, this gives about 239.7576 W/m². Do not insert −18.15, the corresponding Celsius temperature, into the fourth power. Celsius and kelvin share the same size increment but have different zero points; the radiation law uses absolute temperature.

For a paper comparison, consider 200 K and 300 K. The temperature ratio is 1.5, so outgoing flux differs by 1.5⁴ = 5.0625. It does not merely increase by 50%. This ratio exercise explains the curved outgoing-flux graph without requiring a long multiplication for each temperature.

Solve the balance equation

At equilibrium, σTₑ⁴ = (1 − α)S/4. The subscript e marks the equilibrium effective emitting temperature. Divide by σ and take the fourth root: Tₑ = [(1 − α)S/(4σ)] raised to the power 1/4. Taking the fourth root is equivalent to taking a square root twice.

At α = 0.30, Tₑ is about 254.5781 K, or −18.5719°C after subtracting 273.15. The slider moves in 5 K increments, so the default 255 K is close to equilibrium but not exact. Its net is slightly negative, about −1.5826 W/m². Do not round the continuously calculated equilibrium to the nearest slider position and then call the resulting net exactly zero.

Make a paper table for selected temperatures 250, 255 and 260 K while keeping albedo at 30%. Predict each net sign using its position below or above 254.5781 K, then check the model. This tests reasoning about the crossing rather than searching for a displayed zero.

Read the crossing as a condition, not a timeline

Below equilibrium, the equivalent emitter radiates less than the absorbed input; above equilibrium, it radiates more. These signs describe energy accumulation or loss at the selected state. The lab does not move the temperature forward in time or calculate how long adjustment takes.

Likewise, the displayed equilibrium is not a promise that every place on Earth reaches that temperature. A single global flux balance supplies neither geographic temperatures nor the surface–atmosphere energy exchanges needed to infer them.

Transfer with a flux ratio

A hypothetical equivalent emitter radiates 100 W/m² at its equilibrium temperature. A second case requires 1600 W/m² with the same emissivity and constant. What is the equilibrium temperature ratio? Is the same ratio valid if temperatures are expressed in Celsius?

Check the radiative-equilibrium transfer

The flux ratio is 16, so the absolute temperature ratio is the fourth root of 16, or 2. The ratio applies to kelvin temperatures. Celsius ratios do not preserve the radiation law because their zero point is shifted.