Several posts have recently made the above statement and attempted to use it to argue against the greenhouse effect. However, I haven't the faintest idea what this statement is intended to mean. I have made several requests for an explanation but none have been forthcoming. Any thoughts?
Lets answer the idiot..
Photon temperature is defined by its EM wavelength signature. (The temperature of a black body radiating)
-80 deg C is the rough mean of 12um-16um bandwidth. At 16um its temperature is calculated to 78.1C and at 12um its temperature is 80.9C.. ON this graph we have included the power curve of the photons emitted, which shows why the photons have little effect in our atmosphere without water.
/Thread Dead...
Hahahahaha, this is priceless!
BillyBoob finally got around to attempting to explain his statement. Was it -80F or the -80C like I suggested?
Then he doubled down on stupid with his fake precision and total lack of understanding the basics. The shorter the wavelength, the more energy the photon is composed from.
Water in its gaseous form, water vapour, is very much like CO2 because they both have three atoms and two bonds that react with IR. It is only water's ability to change phase at terrestrial temperatures that adds a different pathway for energy transport.
BillyBoob is not studying to be an atmospheric physicist. He is just too ignorant and confused. He may have a parent, sibling or child that has tried to explain certain aspects but he has totally garbled them to the point where even if he says something correct it is only by accident. If you mix a gallon of shit with a gallon of ice cream you get two gallons of shit.