I waited until my answer was posted in the WUWT thread. I have replicated it here as well.
Ok, so the point of this thought experiment was to engage the public in a consideration of the differences between the greenhouse effect and an actual greenhouse. Most here already know that the name itself is a misnomer, but by considering the physics of what is going on we can better understand the argument and better present our opinions on the subject. The majority of the answer below was emailed to Anthony yesterday before he ran the post at WUWT with the note- just to make sure I can’t back out! I wrote the reply in terms of energy Joules, it would be more accurate to say power Joules/Sec or Watts but since the system is stabilized the answer works fine.
I know everyone is wondering what my answers will be to the two greenhouse situations, we’ll see how many will be convinced to change their opinions – or insist that I change mine 😉 . It turns out that both problems are fairly straightforward when considered from an engineering standpoint. In the thermally stabilized systems of the example where temperature is not changing, energy into the system is equal to energy out. We’ll cover the greenhouse vs free air plant situation first. Both plants receive the same energy but the ability to remove heat from the system is limited in the greenhouse plant through convection and evaporation. So in the case of the free air plant, although it is receiving the same energy it has 3 methods of cooling: convection, evaporation and radiation. In the case of the greenhouse, we can consider evaporation and convection negligible so the only option to release the energy is through radiation. Since our camera is only measuring radiation, and since both plants must emit the same energy they receive, the free air plant radiation will sum like this:
Measured EM radiation = Energy in – convection energy – evaporation energy
the greenhouse will sum like this
Measured EM radiation = Energy in – zero convection energy – zero evaporation energy
Therefore the camera will show the greenhouse plant as warmer (brighter) than the free air plant. This holds true even if we include non-zero convection and evaporation for the greenhouse because they are still reduced values requiring a higher EM emission to balance the energy equations.
So now we have the situation where we have two planets, one with more CO2 than the other. We know that CO2 absorbs certain outgoing wavelengths of light. We also know energy in is equal to energy out for both planets. Although this is called the greenhouse effect, it is actually quite different. For both the high and low CO2 planets, the only available cooling mechanism is EM radiation. All the energy coming in has to escape by EM radiation to space, so the equations balance like this.
Measured EM radiation = Energy in
That’s it really. Both planets will measure exactly the same to our camera yet one has a higher surface temperature. The reason this works is that the average energy emission altitude has gone up, allowing a warmer surface yet the net flow is the same.
Caveats: Now I warned that some will get tied in knots over the nuance of this example. There are all kinds of subtleties of the situation which cause minute differences in the planet example. For instance, increasing CO2 will increase the albedo to incoming light, reducing reflected energy and we get a microscopically higher energy in and therefore were our camera of perfect accuracy we could measure a very slightly higher measured radiation from the warmer planet. If this is your explanation, we are in agreement. There are other details as well, but in bulk the answers are A – greenhouse plant is brighter, and B – both are the same.
I read several comments which got the right answer, Carrick was the first to write the correct answer in the comments at tAV, although he kept the answers subtle enough that people had to read it carefully. If you were one who got them both, congratulations. If you are unconvinced by my explanations, ask away and I’ll do my best.