As is well known, if the sun is shining brightly enough, you can leave a frying pan in the sun and then fry an egg on it. This works because aluminum (and many other metals) have low emissivity.
And as we also know, if an object is absorbing solar radiation, it will warm up until it is emitting the same amount of radiation energy (ignoring other forms of heat transfer, like conduction etc).
If an object (the frying pan) has low emissivity, then for a given temperature, it is emitting less radiation than another object at the same temperature with higher or 100% emissivity (a 'blackbody'). But the amount of radiation absorbed, and hence emitted, is fixed, so the frying pan reaches a higher temperature (than the ground around it) before it reaches a steady-state temperature where solar radiation absorbed = radiation emitted.
So on a hot sunny day, ground-level air temp might be 'only' 30 degrees C, but the frying pan is more than 100 degrees C, hot enough to fry an egg. Key thing to note is that ground and pan are emitting a similar amount of radiation. Which is why you have to calibrate an IR thermometer to take the emissivity of the object into account - if you don't, it will tell you that the ground and the pan are a similar temperature.
I doubt any Physics Denier looks at the hot frying pan and concludes that, because the frying pan's actual temperature is greater than the hypothetical temperature it would be if it were a 'blackbody' with 100% emissivity, this must be an example of the Greenhouse Effect.
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But this is the best (and basically only) evidence they have that there is a 33 degree Greenhouse Effect in the first place - they just compare actual temperature of oceans and land with hypothetical temperature of clouds, oceans and land if clouds, oceans and land had 100% emissivity. (Cooler clouds are missing from the first part of the comparison, this is a key part of the deceit).
There's a handy overview of cloud properties here. Note that typical or average cloud emissivity is given as 0.7. If you pick sensible estimates for typical or average cloud-top altitude, then you know how much warmer it is at sea level than at the cloud-tops (this relationship is fixed by the gravito-thermal effect), then you can work out total radiation emitted to space by clouds and cloud-free oceans/land respectively (oceans/land beneath clouds can't emit radiation directly to space, the clouds absorb, reflect and re-emit it). Take a weighted average and hey presto, the overall radiation being emitted spacewards is the same as incoming solar.
Therefore, no radiation is being blocked or absorbed by 'Greenhouse Gases'; there is no 'Greenhouse Effect' (unless you see clouds as acting like the roof of a greenhouse, which I suppose they do) and there is nothing left to explain away.
When I did the workings, I calculated that clouds emit about 47% of the total radiation reaching space, the 'official' estimate of "cloud amount weighted by the cloud IR emissivity" is 50%, so I'm not far off and I suspect the 50% is rounded.
And of course, clouds are nebulous. Nobody will ever know what the exact thickness; altitude; temperature; emissivity; or radiation being emitted by any particular cloud or cloud-top are. How many measurements would you have to make to have a fair picture of the global average over a year? Dunno, even though it wouldn't be that difficult with enough weather balloons and satellites.
But it's easy enough choosing reasonable mid-points of ranges of estimates to get sensible answers, and until these time wasters dedicate a bit of time and effort on doing actual observations of all these variables (and proving me wildly wrong), I will assume that they simply don't want to know (and they can't).
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