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

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1. Nov 26, 2017

### Staff: Mentor

Then you are introducing an extra thermal reservoir and it doesn’t simply provide a power P but has some temperature dependent heat transfer. You are talking about something substantially different than what I was talking about.

Even in your scenario I would not say that the >0 K cold reservoir increased the temperature of the system. I would just say that it kept it from cooling as much as a 0 K cold reservoir would. I wouldn’t say that failure to decrease the temperature as much as a 0 K reservoir is the same increasing temperature. After all, I say that my clothing keeps me warm, not that my clothing warms me.

As long as we are considering alternative configurations, if you have the hot-reservoir/system/cold-reservoir configuration and you remove the cold reservoir then the system temperature goes up and if you remove the hot reservoir then the system temperature goes down. So I think that the “consider alternative configurations” approach is inconclusive.

I do apologize for failing to specify the scenario I had in mind completely.

Last edited: Nov 26, 2017
2. Nov 27, 2017

### DrStupid

That is completely irrelevant. It doesn't matter where the heat comes from. Absorbed sunlight, thermal radiation from the atmosphere or geothermal energy contribute in the same way to the heat balance. It doesn't matter if there is an external thermal reservoirs or a temperature dependent heat transfer. The temperatures of the heat sources just determines their individual power and the sum of the power received by an object (no matter how that happens in detail) and it's own emission profile determines it's equilibrium temperature.

I'm talking about the topic of this thread which was clearly given by the OP and the provided link.

The colder body is no "cold reservoir" (this term makes no sense because there is no such thing like "cold" in physics) but a heat source. It emits thermal radiation which is partially absorbed by warm body, increasing its equilibrium temperature (see my calculation above).

That depends on your definition of "cooling". If you man the release of heat than you are wrong because

1. The emission of thermal radiation of the warm object is independent from its environment. It only depends on the shape and temperature of the warm object itself according to the Stefan-Boltzmann law and therefore cannot be altered by external objects or the 2.7 K background.

2. The heat release will not be reduced but increased if the temperature rises.

If you mean net heat release (including incoming heat) than you are partially right. The warm body will absorb more heat due to the additional thermal radiation of the cold body. But in the steady state the net heat flow is (of course) always zero - with or without the cold body.

If you mean temperature decay, than you are partially right. If the warm body starts above its equilibrium temperature, the resulting temperature decay will be reduced and stopped at a higher equilibrium temperature. But the temperature can also be increased (see my calculation above). That can’t be explained with the ability to keep the temperature from decreasing.

3. Nov 27, 2017

### Staff: Mentor

Sure it does. If the system is the same temperature as the hot reservoir then P=0. So if P is not work then it cannot be independent of the system’s temperature.

The term “cold reservoir” is pretty standard in thermodynamics. Particularly for heat engines. It is the heat sink not a heat source. I can provide references for this usage, if you like.

Yes, by the word “cooling” I mean heat leaving the system. And by “warming” I mean heat entering the system.

You seem to be confusing energy flow and heat. Energy flows both ways, but heat only goes from hot to cold by definition. Suppose the system gives 7 W of radiation to the hot reservoir and recieves 10 W from it, then the heat is 3 W. Neither the 10 W nor the 7 W is heat. The Stefan Boltzmann law describes power, not heat.

A cold reservoir cannot be a heat source, by definition.

Interestingly, this is wrong if P is a thermal source instead of work. If P is work then the equilibrium heat release is constant, but if P is thermal then the equilibrium heat release decreases as the system temperature rises. In neither case is the heat flow increased.

Frankly, I think that we mostly agree on the physics, but we seem to disagree pretty strongly on the terminology. I can provide references to support my terminology if you wish, but you have not provided any references to support yours, despite my request.

Sure it can. If I have a bathtub with a faucet and a drain I can easily describe the action of closing the drain as keeping the water level from decreasing. The water level may raise without me needing to claim that the drain is filling the tub. The faucet only fills and the drain only empties.

Last edited: Nov 28, 2017
4. Nov 28, 2017

### itfitmewelltoo

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

5. Nov 28, 2017

### bob012345

This paper may relate to the question at hand. It refers to a different but I think equivalent situation. Here is the reference;

Found Phys
DOI 10.1007/s10701-014-9781-5

6. Nov 28, 2017

### bob012345

I'm sure the laws of thermodynamics apply to all physics but applying them correctly is not necessarily easy.

7. Nov 29, 2017

### 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.

8. Nov 30, 2017

### swampwiz

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).

9. Nov 30, 2017

### Staff: Mentor

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.

10. Nov 30, 2017

### bob012345

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.

11. Nov 30, 2017

### Staff: Mentor

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...

12. Nov 30, 2017

### itfitmewelltoo

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.

13. Nov 30, 2017

### Staff: Mentor

I have not heard of one that is that effective, but I have heard of this approach in general:

14. Dec 1, 2017

### bob012345

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;

15. Dec 1, 2017

### Staff: Mentor

Can you show that calculation? Or otherwise show where you got the 300W/m^2 number that you posted?

16. Dec 1, 2017

### bob012345

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.