Cooler objects able to increase the temperature of warmer objects?

[JR Wakefield]

Not sure this should go here (you guys need a thermodynamics section). But there is quite the debate going on about cooler objects being able to increase the temperature of warmer objects here:
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1

They are claiming that the IR from the colder object is absorbed by molecules of the warmer object, thereby thermalizing the molecules more, increasing the temp of the warmer object.

Would like some feedback here about this.

Thanks.

 

[Dale]

A cold object cannot warm a warm object without work being done.

 

[JR Wakefield]

A cold object cannot warm a warm object without work being done.

Yes I know that. The point in that debate is IR from the cooler object must be being absorbed by the warmer object. The question is, what happens to the IR photons from the cooler object when they come in contact with the atoms in the hotter object?

 

[DrStupid]

A cold object cannot warm a warm object without work being done.

Of course it can. The temperatur of an object in thermal equilibrium can be increased by any additional heat source, including colder objects.

 

[JR Wakefield]

Of course it can. The temperatur of an object in thermal equilibrium can be increased by any additional heat source, including colder objects.

[sigh]

No it cant. Entropy forbids a colder object from giving energy to a hotter object. Why is this such an issue? This is basic thermodynamics.

 

[anorlunda]

Heat flows from warm to cold. The warm object will send more IR to the cold one than it receives.

 

[JR Wakefield]

Heat flows from warm to cold. The warm object will send more IR to the cold one than it receives.

These are the arguments I've been dealing with.

Warm to cold is a NET flow of energy.
Cold IR photons get absorbed by the warmer atoms which get "thermalized".

So what happens to the warmer atoms when they encounter IR photons from a colder object? I say nothing happens. If anything, the atom immediately re-emits that energy.

 

[anorlunda]

Your over thinking it. Thermodynamics is the study of averages of many particles. You can not apply it one photon at a time.

Just believe what @Dale said.

 

[russ_watters]

Not sure this should go here (you guys need a thermodynamics section). But there is quite the debate going on about cooler objects being able to increase the temperature of warmer objects here:
https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/comment-page-1

They are claiming that the IR from the colder object is absorbed by molecules of the warmer object, thereby thermalizing the molecules more, increasing the temp of the warmer object.

The article doesn't say that (and "thermalizing" isn't a word). It says a warm object will be kept warmer by exchanging radiation with a cold object that is blocking an even colder object than if the cold object weren't there...which is kinda a "duh".

Here's the point of the article:

Can a cold object leave a warm object warmer than it would be without the cold object?

While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view.

 

[Dale]

Of course it can. The temperatur of an object in thermal equilibrium can be increased by any additional heat source, including colder objects.

Do you have a reference for this?

 

[JR Wakefield]

Your over thinking it. Thermodynamics is the study of averages of many particles. You can not apply it one photon at a time.

Just believe what @Dale said.

The article doesn't say that (and "thermalizing" isn't a word). It says a warm object will be kept warmer by exchanging radiation with a cold object that is blocking an even colder object than if the cold object weren't there...which is kinda a "duh".

Here's the point of the article:

Thermalizing is the term Willis used. I know it's not a word. And if you read through the comments, Willis et al state categorically that the colder objects emitting IR which when interacting with a warmer object, will increase the energy of that warmer object, and hence it's temperature.

It is clear reading the comments that people do not understand the difference between energy, heat and temperature. One person even objected to the official definition of heat!

 

[russ_watters]

Thermalizing is the term Willis used.

No, he doesn't, unless he's posting under an alias in the comments section -- it looks to me like you are reading through comments to the article (and "thermalz..." is used there several times). Please quote exactly what you are referring to.

And if you read through the comments...

No, that isn't how this works. In general, you need to quote exactly what you are referencing, but for this case, we're not going to argue by proxy with the comments section to an article. That's a waste of our time.

..Willis et al state categorically that the colder objects emitting IR which when interacting with a warmer object, will increase the energy of that warmer object, and hence it's temperature.

You're skipping part of the article's claim, per my previous post.

It is clear reading the comments that people do not understand the difference between energy, heat and temperature. One person even objected to the official definition of heat!

I really don't care -- is that what this is about? A proxy argument with the people commenting on that article? We don't do that here. We'll help you understand what the author is saying, but that's it. If you have a question of your own, ask it yourself.

 

[Dale]

Warm to cold is a NET flow of energy.

This is correct, and it doesn’t matter if the mechanism of heat transfer is through thermal radiation or through thermal conduction. In both cases on a microscopic level there is energy going both ways and on a statistical level the energy goes from hot to cold.

So what happens to the warmer atoms when they encounter IR photons from a colder object? I say nothing happens.

They can be absorbed, reflected, scattered, or transmitted, based on the spectral characteristics of the object. IR photons are not particularly different from other photons.

 

[DrStupid]

the colder objects emitting IR which when interacting with a warmer object, will increase the energy of that warmer object, and hence it's temperature.

Of course that's what happens. Just let’s do the math for a simple example: A spherical black body with the radius $r$ (e.g. Earth), heated by a heat source with the power $P$ (e.g. absorbed light from the Sun). In the dynamic equilibrium the body will emit as much energy as it absorbs from the heat source. Assuming a homogeneous temperature distribution this results in the equilibrium temperature

T_{eq}^4 = \frac{P}{{4 \cdot \pi \cdot \sigma \cdot r^2 }}

according to the Stefan–Boltzmann law. That’s the warm object.

Now let’s see what happens if we add another black body as additional heat source – let’s say a spherical shell (e.g. the atmosphere) with the temperature $T_{add}$ around the first body. This shell will also emit heat radiation. If this radiation hits the first body, it will be absorbed, resulting in the additional heat

P_{add} = 4 \cdot \pi \cdot r^2 \cdot \sigma \cdot T_{add}^4

and therefore in the new equilibrium temperature

\frac{P}{{4 \cdot \pi \cdot \sigma \cdot r^2 }} + T_{add}^4 > T_{eq}^4

no matter if $T_{add}$ is above, equal or below $T_{eq}$.

Claiming that this is not possible is a usual argumentation in climate change denial. I can’t believe that something like this is supported in this forum!

 

[mfb]

To thermalize is a word.

There are two different things discussed here, and the lack of separation between them leads to the confusion.

* If you look at an existing system and add a new object to it, that can increase the temperature of an existing object, even if the new object is colder than the existing object.
* The net power flow between these objects (any pair of objects in fact) will be from hot to cold.

Both statements are true, and they don't contradict each other.

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.

 

[Dale]

heated by a heat source with the power P

It is this power source that warms the object, not the colder object. This is work, as I mentioned above.

Assuming a homogeneous temperature distribution this results in the equilibrium temperature

T4eq=P4⋅π⋅σ⋅r2Teq4=P4⋅π⋅σ⋅r2T_{eq}^4 = \frac{P}{{4 \cdot \pi \cdot \sigma \cdot r^2 }}

according to the Stefan–Boltzmann law.

Here you areassuming a cold reservoir at 0 K.

and therefore in the new equilibrium temperature

P4⋅π⋅σ⋅r2+T4add>T4eqP4⋅π⋅σ⋅r2+Tadd4>Teq4\frac{P}{{4 \cdot \pi \cdot \sigma \cdot r^2 }} + T_{add}^4 > T_{eq}^4

no matter if TaddTaddT_{add} is above, equal or below TeqTeqT_{eq}.

Yes, if you have two different systems, one with a cold reservoir at 0 K and the other with a cold reservoir at a higher temperature then the equilibrium temperature will be higher for the system with the warmer cold reservoir. This is not an example of a cold reservoir increasing the temperature of a hot reservoir without work. Without P the temperature will drop to the temperature of the cold reservoir, not increase.

Claiming that this is not possible is a usual argumentation in climate change denial.

I stay out of climate change arguments. What I said is correct, regardless of what other groups of people may misuse the concepts. Please do not attempt any “guilt by association” here.

 

[DrStupid]

It is this power source that warms the object, not the colder object.

No, the object is warmed by the power source and by the colder object. Claiming that the colder object doesn't warm the warmer object is simply wrong. Any heat source contributes to the warming regardless of its temperature.

This is work, as I mentioned above.

The absorbed sunlight provides heat only. Tere is no work involved.

Without P the temperature will drop to the temperature of the cold reservoir, not increase.

That doesn't justify your general claim that "A cold object cannot warm a warm object without work being done."

What I said is correct

It would be correct if you would say that a cold object cannot warm a warm object without work being done if no additional heat sources and no heat sinks are involved. But you didn't do that.

 

[mfb]

Claiming that the colder object doesn't warm the warmer object is simply wrong.

It depends on your perspective. You can also say it cools, but it cools less than whatever else it blocks e.g. an even colder object).

 

[Dadface]

Objects radiate energy to the surroundings and absorb energy from the surroundings. Because of temperature differences the hot object will be a net radiator and cool down and the cold object a net absorber and heat up.

 

[DrStupid]

It depends on your perspective. You can also say it cools, but it cools less than whatever else it blocks e.g. an even colder object).

Yes, I could say that, but we are talking about warming and not about cooling and it is no matter of perspective that any heat source contributes to the warming.

 

[Dale]

The absorbed sunlight provides heat only. Tere is no work involved.

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.

It would be correct if you would say that a cold object cannot warm a warm object without work being done if no additional heat sources and no heat sinks are involved. But you didn't do that.

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

I would not use your description, even given your scenario. But that is admittedly semantics.

 

[DrStupid]

Then you are introducing an extra thermal reservoir and it doesn’t simply provide a power P but has some temperature dependent heat transfer.

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.

You are talking about something substantially different than what I was talking about.

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

Even in your scenario I would not say that the >0 K cold reservoir increased the temperature of the system.

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).

I would just say that it kept it from cooling as much as a 0 K cold reservoir would.

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.

 

[Dale]

That is completely irrelevant. It doesn't matter where the heat comes from.

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 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.

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.

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

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

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.

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 receives 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.

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

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.

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.

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.

 

[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

 

[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

 

[bob012345]

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.

 

[swampwiz]

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).

 

[Dale]

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]

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]

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.

 

[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.

 

[Dale]

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]

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]

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]

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.