Greenhouse Effect, what Greenhouse Effect?

It's homework time at Science Academy. Teacher has explained to the class about solar radiation, albedo, Stefan-Boltzmann constant, all that stuff.

Teacher sets a homework: how do you reconcile incoming solar radiation (1,370 W/m2 overhead at the Equator) with the observed average temperature of the hard surface and ocean surface of the planet Earth (288 K or 15C)?

Pupil A

Pupil A makes all manner of short cuts and simplifications and knocks out this calculation on the 'bus on the way into school:

"Solar radiation averaged over surface of Earth in a 24-hour period = 342 W/m2; less light reflected (1 minus albedo 0.3 x 342 W/m2) = 240 W/m2, divide that by the S-B constant; take the fourth root of that = 255 K."

Bugger, thinks Pupil A, I'm out by 33 degrees. Not to worry, too late to re-work it, scribble something about Greenhouse Gases to make up the difference. Job done, hand it in a few seconds before the deadline.

Pupil B

Pupil B takes it a bit more seriously, and splits things up into day and night; cloud cover, land and ocean with their different altitudes and albedos; takes the mid-point of all the variables you can find on the inter-web; finds out about latent heat of evaporation; Googles some weather forecasting sites as a reality check; looks up how quickly air, rock and water cool down at night; and then calculates the weighted average:

a. On the sunlit side, average incoming solar radiation is 684 W/m2.

b. Clouds have an albedo of 0.6; they absorb 684 W/m2 x 40%; divide by S-B constant (5.67 x 10^-8); take the fourth root = 263K.

c. Clouds are at a typical altitude of 4 km above sea-level (average of low-lying fog, proper clouds and con-trails). If their actual temperature is 263 K (-10 C), then their potential temperature is about 289 K; i.e. to be in a neutral equilibrium, the air at sea-level beneath them must be about 4km x 6.5 K/km lapse rate warmer = 289 K.

d. Incoming solar radiation which hits the ocean surface = 684 W/m2; albedo 0.1; so deduct 10% reflected = 616 W/m2; about one-quarter of that doesn't go in to warming, it is 'lost' as Latent Heat of Condensation to reappear elsewhere and higher up (150g water evaporates every hour for each m2 of ocean surface, seems reasonable); [calculate as before] = 299 K.

e. Hard surface is easy, 684 W/m2 less 30% reflected [calculate as before] = 302 K.

f. Pupil B then weights this one-half cloud cover, one-third cloud-free ocean and one-sixth cloud-free land to calculate a weighted average day-time maximum for the whole surface of 294 K.

g. That's just the daytime maximum temperature. It's four in the morning by now, so Pupil B decides that the night time low is about ten or fifteen degrees lower than that based on a joyous experience of partying until dawn and a bitter experience of missing the last 'bus home (and pops head out of window just to check); and calls it 282 K for the night-time low.

h. Pupil B then takes the simple average of 294 K and 282 K = 288 K; heaves a sigh of relief; uploads the file to the school portal; turns off the computer and collapses into bed.

The results?

Pupil A gets an A* grade and a photo in the local newspaper.

Pupil B and Pupil B's parents are called into the Head of Physics' office. Pupil B is painfully aware that all the variables used were mid-points and that there was some reverse-engineering to get the required answer (but if you are asked to reconcile something you can work in either direction, can't you?); sticking your head out of the window is no substitute for proper measurements; and that a proper calculation would have run to several hundred pages; but hey, it was only a first term homework. Maybe a retake or something?

It's far worse. Head of Physics asks the parents why the Hell they are pumping their child full of this Climate Science Denier Bullshit; suggests that Pupil B "might prefer attending a local school which offers vocational qualifications" and shoves the signed-off transfer papers over the desk.