Audit surface, atmosphere and space fluxes

A large internal flux does not create energy

At the default settings, the surface emits about 396.9583 W/m² even though absorbed sunlight is only 238.175 W/m². Before deciding that energy has been created, inspect the boundary around the surface. It receives sunlight and downward thermal emission from the atmospheric layer. A budget that omits one input cannot explain that surface’s output.

Draw three separate boundaries on paper: surface only, atmospheric layer only, and the combined system. Arrows crossing one boundary may be internal to another. The UBC single-layer model illustrates the importance of distinguishing direct surface radiation from atmospheric radiation reaching space. Use its conceptual separation without treating this lab as a measured flux record.

The lab assumes all absorbed solar flux reaches the surface, a transparent atmosphere for sunlight, a blackbody surface, and a single grey thermal layer. These assumptions make three ledgers solvable. They do not describe every route by which energy moves through real air and water.

Audit the surface boundary

Write Q for absorbed sunlight, F for surface emission and D for downward layer emission. At the default, Q = 238.175, D = 158.7833 and F = 396.9583 W/m². The surface residual is Q + D − F, which is zero apart from rounding.

A residual is incoming minus outgoing flux for the chosen boundary. A positive residual would mean energy accumulation there; a negative residual would mean energy loss. Here equilibrium is solved directly, so a near-zero numerical residual confirms the accounting rather than demonstrating an observed adjustment process.

Change atmospheric emissivity while keeping albedo at 30%. Record both F and D. Their difference remains Q. A growing gross exchange can therefore coexist with an unchanged external supply. Explain that relationship using your ledger, not the phrase “extra energy” without a named source.

Audit the layer and space boundaries

The layer absorbs εF and emits D upward plus D downward. At ε = 0.80, absorption is 317.5667 W/m² and total layer emission is twice 158.7833 W/m². The layer residual εF − 2D is zero.

Only the transmitted surface contribution, (1 − ε)F, reaches space directly. At the default this is 79.3917 W/m². Add upward layer emission of 158.7833 W/m² to get total outgoing 238.175 W/m². The planetary residual is Q − [(1 − ε)F + D], again zero.

When you combine the surface and layer boundaries, their internal exchanges cancel. This cancellation is why you must not add downward radiation to the planetary external input. It crosses an internal boundary, not the boundary between the planet and space.

Distinguish opposing radiation from net exchange

The layer is cooler than the surface in this equilibrium model, yet it emits downward radiation that the surface absorbs. Both bodies emit radiation; the direction of one arrow does not determine the direction of net energy exchange.

The surface-to-layer exchange is εF upward absorbed by the layer, minus D downward absorbed by the surface. Because D = εF/2, the net is εF/2, nonnegative. At the default it is 158.7833 W/m² from the warmer surface toward the cooler layer. The downward arrow does not imply net heat transfer from cold to hot.

For an independent paper check, choose ε = 1 at the same albedo. There is no direct surface transmission. F = 476.35 W/m², and the layer emits 238.175 W/m² each way. Confirm all three balances and describe why the surface flux can be twice Q while total outgoing remains Q.

Transfer by repairing a ledger

A learner calculates planetary outgoing as F + D using the default values. Identify the double-counting and write the correct expression. Then calculate net surface-to-layer exchange.

Check the radiative-flows transfer

F includes radiation absorbed by the layer, so it cannot all be counted as direct escape. Planetary outgoing is (1 − ε)F + D = 238.175 W/m². Net surface-to-layer exchange is εF − D = 158.7833 W/m², upward from the warmer surface.