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Radiative cooling, discussion of paper in front of face experiment

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I believe the current desciption of the activity of interjecting a piece of paper between your face and a cloudless night sky is misleading. It states that the reason your face feels warmer is that the paper is actually heating your face due to its much higher radiative temperature than the 3 degrees kelvin of outer space. I think it is more correct to say that your face is radiating to the much cooler outer space and therefore has a certain radiative heat loss and feels cool. When you interject the paper your face no longer 'sees' a large radiative delta T, rather a very small radiative delta T and therefore does NOT loose as much heat due to radiative heat loss and therefore does not feel cool. In other words, the action is a relative loss of cooling function due to lower radiative heat transfer due to lower radiative delta T, not a heating function from the paper.

At least this is the way I remember it from my school daze. —Preceding unsigned comment added by 58.137.89.33 (talk) 04:22, 28 August 2009 (UTC)[reply]

I know this is a 3 year old comment, but I'm a physicist so I didn't want to leave it here uncorrected. An object radiates heat at the same rate no matter what its environment is; your face radiates at the same rate whether exposed to a dark sky or a sheet of paper. The rate of cooling, however, depends on that and how much energy is radiated back at it. The paper radiates more power than the sky, which is why it might make your face feel warmer. 108.76.71.43 (talk) 02:36, 15 August 2012 (UTC)[reply]

Of interest to engineers

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QS removed the asphalt bit, correctly noting it was not about RC. I've now removed all of

Radiative cooling has become interesting to construction engineers who are exploring ways of increasing radiative cooling for new construction projects, particular in urban areas, where there is increased concern about urban thermal plumes and urban heat islands. Strategies being considered are the inclusion of more green areas.

for the same reasons. Most things are black in the IR so it is pretty hard to increase radiative cooling at all. Green areas don't; they increase evaporative (latent-heat) cooling. Replacing black asphalt with white would reduce net SW at the sfc. So I think this para has a place in an article about sfc energy balance, but not this one William M. Connolley (talk) 21:37, 3 May 2010 (UTC)[reply]

I agree. However, there should be something about satellites and the Moon. The current article is a bit too Earth/climate change centric. Q Science (talk) 22:22, 3 May 2010 (UTC)[reply]
You could argue that the Sun is clearest example of radiative cooling :-) William M. Connolley (talk) 21:08, 4 May 2010 (UTC)[reply]

Nocturnal ice making breaks the second law of thermodynamics

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"Nocturnal ice making" section describes process of which only outcome is refrigerating a body with no work input. Between stratosphere and water in tray is quite a layer of warmer air. I don't see how could water in the tray emit more radiation to the stratosphere than it receives from surrounding air, as amount of radiation produced depends only on area and temperature of the body. Moreover water does not "know" temperature of stratosphere and it has absolutely no influence on how much heat is radiated from the water. That means, mentioning how cold it is up there may be misleading. 89.75.162.72 (talk) 17:55, 12 July 2012 (UTC)[reply]

amount of radiation produced depends only on area and temperature of the body - no, it depends on the emissivity too William M. Connolley (talk) 21:29, 12 July 2012 (UTC)[reply]
OK, my bad. But that's still breaking the second rule of thermodynamics. 89.75.162.72 (talk) 13:38, 13 July 2012 (UTC)[reply]
Yes, warm air (actually moisture in it) radiates back as well, but not as much ... thats why you cant radiatively cool to 3K space backround temperature, but with good isolation (ie. solar thermal panels at night) can cool some teen degreees lower that surrounding air. Dont remember exact amount, and i suppose it is dependant on amount of moisture in atmosphere and its thickness ... Wait, found it ... damn, it can be cooled even more than that! http://people.csail.mit.edu/jaffer/cool/cool.pdf ... that and accidents of night freezing of thermal panels quite above ice melting point should be enough of proof, that its feasible — Preceding unsigned comment added by 78.8.106.96 (talk) 00:41, 27 July 2012 (UTC)[reply]
See my comment in the above section. The water is radiating the same amount of power regardless of its environment, but it cools down faster if less power is radiated back at it (clouds vs dark sky). It does not violate the second law of thermodynamics, because that law doesn't say that work must be done to cool a body. It says work must be done to move heat from a cold body to a warmer one, but in this case heat is moving from the warm water to the cold photon gas of the universe, which is fine. A pool of water floating in deep space will cool down for the same reason. 108.76.71.43 (talk) 02:40, 15 August 2012 (UTC)[reply]
89.75.162.72 You are making the classic mistake of treating the bulk atmosphere gas as a black body. The IR emissivity of oxygen and nitrogen is so close to zero as makes no odds. The only back radiation from the atmosphere will be that from the greenhouse gases - water vapour, CO2 etc. --Anteaus (talk) 08:01, 6 June 2015 (UTC)[reply]

Missing the point?

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The title of the article is Radiative Cooling. I was expecting a textbook tutorial on radiative heat transfer, i.e., Q= σ A ε T^4. The article should discuss emissivity and absorptivity, view factors, cosmic background temperature, maybe the solar flux constant, etc. Discussions of ice making in ancient India seem very inappropriate. I suggest changing the content, or changing the title. — Preceding unsigned comment added by 130.76.64.120 (talk) 03:49, 31 January 2013 (UTC)[reply]

"textbook tutorial on radiative heat transfer": WP:NOTTEXTBOOK, the rest is covered by thermal radiation.
Icemaking is an application of radiative cooling. Paradoctor (talk) 15:38, 7 March 2014 (UTC)[reply]
Equation should be mentionned in brief, at least the dependency with temperature (T^4) and materials (color? ) is relevant to grasp the subject correctly. There's some momentum in that field and I still don't understand most of it. http://www.sciencemag.org/news/2017/02/cheap-plastic-film-cools-whatever-it-touches-10-c — Preceding unsigned comment added by 155.69.3.165 (talk) 06:35, 15 April 2018 (UTC)[reply]