# Cooler objects able to increase the temperature of warmer objects?

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bob012345
Gold Member
The arguments provided by Anthony Watts and others arises out of not understanding how the laws of thermodynamics we're developed and the systems that they describe. Thermodynamic principles define the average property of mass quantities of materials, such as steam. It is an excellent example of how being self taught often fails. They are only as good as their teachers. The laws of thermodynamics do not apply to electromagnetic energy or individual quantum particles
I'm sure the laws of thermodynamics apply to all physics but applying them correctly is not necessarily easy.

Rap
Yes the cold object will add heat to the warm object by emitting IR radiation which is absorbed by the warm object. The thing is, the warm object emits MORE IR radiation than the cold object and so it will add more heat to the cold object than it receives and that will lower the temperature of the warm object.

As an example, the presence of Mars increases the temperature of Earth a tiny bit, as some light from it reaches Earth. The heat Earth emits is independent of the existence of Mars, but the heat it gets is increased from Mars.
The proper way to say this is that in a hypothetical alternative Universe in which Mars does not exist, Earth would receive less radiation from the small solid angle that Mars would be in - with the reason being that Mars has a temperature greater than the cosmic background radiation (i.e., which is the temperature of deep space).

russ_watters and mfb
Dale
Mentor
Yes the cold object will add heat to the warm object by emitting IR radiation which is absorbed by the warm object.
The cold object will add energy to the warm object by emitting IR. Energy goes both ways, but heat only goes from hot to cold.

bob012345
Gold Member
The cold object will add energy to the warm object by emitting IR. Energy goes both ways, but heat only goes from hot to cold.
I've read about devices that direct IR radiation from objects to the night sky which is effectively a radiator to space though not as effective as actually being in space. You can make ice even when the ambient air is in the 50's F. Cleverly designed systems may be able to convert the IR radiation of the earth or of solar warmed objects such as rooftops to useful energy by passive radiation to space as the cold sink. The earth typically radiates over 300 W/m^2 at night. Of course doing all that may not be worth the cost and complexity.

berkeman
Mentor
I've read about devices that direct IR radiation from objects to the night sky which is effectively a radiator to space though not as effective as actually being in space. You can make ice even when the ambient air is in the 50's F. Cleverly designed systems may be able to convert the IR radiation of the earth or of solar warmed objects such as rooftops to useful energy by passive radiation to space as the cold sink. The earth typically radiates over 300 W/m^2 at night. Of course doing all that may not be worth the cost and complexity.
At least the last part of your post is incorrect, I believe. Do you have links to the other assertions in your post?

It looks like you are confusing the reflective albedo of the Earth during the daytime (when insolation is around 1kW/m^2) and the much lower radiation during the night...

from -- http://eesc.columbia.edu/courses/ees/climate/lectures/radiation/

The Earth's albedo.

The Earth's surface reflects (that is, returns the radiation back to space in more or less the same spectrum) part of the solar energy. This is what makes the part of the Earth lit by the sun visible from space (Figure 8) in the same way that the moon and the other members of the solar system are visible to us, despite their lack of an inner source of visible radiation. The most obvious aspect of Figure 8 is the brightness of the Earth's cloud cover. A significant part of the Earth's reflectivity can be attributed to clouds (this is but one reason why they are so important in the Earth's climate). In climate texts the reflectivity of a planet is referred to as the albedo (that is, albedo = reflectivity) and is expressed as a fraction. The albedo of Earth depends on the geographical location, surface properties, and the weather (can you tell from Figure 7 which has higher albedo, the land or the ocean?). On the average however, the Earth's albedo is about 0.3. This fraction of incoming radiation is reflected back into space. The other 0.7 part of the incoming solar radiation is absorbed by our planet.

I am wondering if the question is appropriately termed "thermodynamics". The problem is really a quantum dynamics issue. Thermodynamics is a mass matter theory and applies to steam, refrigerants, etc. It changed to statistical mechanics after Einstein's paper on Brownian motion. There is some confusion that arises from over applying the 4 laws of thermodynamics. It's a bit of a context problem. Thermo just doesn't really apply to pure IR problems. But without the context of experiment and application, without the historical context, or the context of having taken one course in thermodynamics, another in quantum mechanics, and a third in electro-dynamics, it is easy to see how they get convoluted when trying to learn it on the internet.

Dale
Mentor
I've read about devices that direct IR radiation from objects to the night sky which is effectively a radiator to space though not as effective as actually being in space. You can make ice even when the ambient air is in the 50's F.
I have not heard of one that is that effective, but I have heard of this approach in general:

https://news.stanford.edu/news/2014/november/radiative-cooling-mirror-112614.html

bob012345 and berkeman
bob012345
Gold Member
At least the last part of your post is incorrect, I believe. Do you have links to the other assertions in your post?

It looks like you are confusing the reflective albedo of the Earth during the daytime (when insolation is around 1kW/m^2) and the much lower radiation during the night...

from -- http://eesc.columbia.edu/courses/ees/climate/lectures/radiation/

The earth is a blackbody at night radiating to space or at least the surface radiates to the atmosphere which radiates to space and to the surface. I wasn't confusing the nighttime radiated power with the daytime albedo. If the surface temp is say 20C and the emissivity is say 0.9, one can compute the radiated power. As an aside, I believe panels that effectively absorb IR radiation in low earth orbit at night would be able to use that power also if they effectively coupled to the dark sky as the cold sink. Regarding the cooling, here is a link;

https://en.m.wikipedia.org/wiki/Radiative_cooling

berkeman
Mentor
e earth is a blackbody at night radiating to space or at least the surface radiates to the atmosphere which radiates to space and to the surface. I wasn't confusing the nighttime radiated power with the daytime albedo. If the surface temp is say 20C and the emissivity is say 0.9, one can compute the radiated power.
Can you show that calculation? Or otherwise show where you got the 300W/m^2 number that you posted?

bob012345
bob012345
Gold Member
Can you show that calculation? Or otherwise show where you got the 300W/m^2 number that you posted?
Assuming 20C surface temperature and an emissivity of around .9, that's 293K so using the Stephan Boltzmann law we have

8.67 10^-8 W/(m^2 K^4) * 293K^4 * 0.9 which is about 376 W/m^2 which is over the 300 W/m^2 number which was an estimate of the average.