Why is the emissivity of metals low?

[jaumzaum]

Hello! I am having trouble to understand why the emissivity of polished metals is much lower than if they are not polished.
Consider, for example, non-polished aluminium at 300K, which is said to have an emissivity of 0.77. We put it floating in vacuum. There is an energy source near it, and in this configuration it absorbs 77W of energy (and emits 77W as it is in thermal equilibrium). Now the metal is polished. It's emissivity drops to 0.05. For it to be in thermal equilibrium the emissivity must be equal to the absorvivity, so that the new emited/absorbed energy is 5W.

If we apply Stefan-Boltzmann law:
$$\sigma T'^4 = \frac{0.05} {0.77} \sigma 300^4$$
$$T'=152K$$

That way the metal needs to drop its temeprature to -121 celsius degrees! Is this true?

I was wondering why don't we see low temperature polished metals in real world situations, is this because the metals are in contact with other surfaces such that conduction plays more? Even with conduction, I would suppose the temperature of polished metals wpuld still be some degrees lower than that of the ambient. Is it true? How much?

EDIT: Even considering conduction, if we consider a body with 1m length and surface area of 1m^2 in 300K, it would radiate 460W of energy. Consider this body is on a table that made of wood at 300K, with thermal conductivity of 0.2 W/mK, This would give a very huge temperature drop.

 

[hutchphd]

A black body both absorbs and emits radiation. In fact for any wavelength light the emissivity and absorbtivity are exactly equal (this is dictated by the second law of thermodynamics).

Given enough time any isolated object will radiatively reach temperature equilibrium with its surroundings. The emissivity /absorbtivity will determine the rate at which this happens.

In non-isolated situations (a pot on a stove) the shiny object will emit less heat (which is why pots are polished

 

[tech99]

Hello! I am having trouble to understand why the emissivity of polished metals is much lower than if they are not polished.
Consider, for example, non-polished aluminium at 300K, which is said to have an emissivity of 0.77. We put it floating in vacuum. There is an energy source near it, and in this configuration it absorbs 77W of energy (and emits 77W as it is in thermal equilibrium). Now the metal is polished. It's emissivity drops to 0.05. For it to be in thermal equilibrium the emissivity must be equal to the absorvivity, so that the new emited/absorbed energy is 5W.

If we apply Stefan-Boltzmann law:
$$\sigma T'^4 = \frac{0.05} {0.77} \sigma 300^4$$
$$T'=152K$$

That way the metal needs to drop its temeprature to -121 celsius degrees! Is this true?

I was wondering why don't we see low temperature polished metals in real world situations, is this because the metals are in contact with other surfaces such that conduction plays more? Even with conduction, I would suppose the temperature of polished metals wpuld still be some degrees lower than that of the ambient. Is it true? How much?

EDIT: Even considering conduction, if we consider a body with 1m length and surface area of 1m^2 in 300K, it would radiate 460W of energy. Consider this body is on a table that made of wood at 300K, with thermal conductivity of 0.2 W/mK, This would give a very huge temperature drop.

If the metal is polished, we can see that it creates an image of the source, which is a coherent secondary source and radiates almost as much power as it receives.
If the surface is irregular, we see many secondary sources of radiation which are incoherent - they are of random phase. So the rays do not add coherently and the radiated power is less.
You might be interested in the information on the Rayleigh Roughness Criterion under a Google search.

 

[Dale]

That way the metal needs to drop its temeprature to -121 celsius degrees! Is this true?

What? Changing the emissivity does not change the equilibrium temperature. It only changes how quickly it reaches equilibrium.

 

[hutchphd]

Changing the emissivity does not change the equilibrium temperature. It only changes how quickly it reaches equilibrium.

True enough but if you read the passage carefully (I did not originally do so!) there is an energy source near the pot so this is a steady-state but nonequilibrium situation.

 

[jaumzaum]

@hutchphd you Agee with the calculations?

 

[Dale]

True enough but if you read the passage carefully (I did not originally do so!) there is an energy source near the pot so this is a steady-state but nonequilibrium situation.

As long as the energy source is providing energy via radiation the equilibrium temperature is the same. I guess if it were providing energy by conduction it would make a difference.

 

[Dale]

@hutchphd you Agee with the calculations?

I don’t, provided the energy source is radiation. It would help if you clarify instead of rushing to seek agreement for an ambiguous scenario.

 

[hutchphd]

@hutchphd you Agee with the calculations?

To do the calculation correctly you need to include the details of the "energy source" and perhaps the details of the incoming radiation from the ambient room.
So as @Dale says you need to be much more specific...it looks like you have started OK but that is all one can say.

 

[jaumzaum]

Ok. So let's consider a energy source that radiates 2000W of energy. There are no other energy sources in the system. There is an object, floating in vacuum (so there is no convection or conduction), that receives 100W of irradiated energy from the energy source. It's initial emissivity is 0.77. In the equilibrium, 23W of the irradiated energy is reflected/transmitted and 77W is absorbed by the body. For it to be in equilibrium it needs to emit 77W. Consider the body is a cube of side 1m. We can do some calculations:
$$ 77 = 0.77 \times \sigma \times T^4$$
$$T=205K$$

Now consider this body is polished and the emissivity becomes 0.05. In equilibrium the body emits 5W:
$$ 5 = 0.05 \times \sigma \times T^4$$
$$T=205K$$

I was wrong before, it really does not seem to affect the temperature of the body if we change the emissivity. Is it right now?

 

[hutchphd]

That seems better. Of course even in the vacuum of deep space the universe will be radiating back at you isotropically with characteristic 4K blackbody (microwave background). There will always be incoming radiation in addition to your source so you really need to subtract it to get net emission to find T.

 

[sophiecentaur]

If the metal is polished, we can see that it creates an image of the source, which is a coherent secondary source and radiates almost as much power as it receives.
If the surface is irregular, we see many secondary sources of radiation which are incoherent - they are of random phase. So the rays do not add coherently and the radiated power is less.

This is hard to understand. Why would a coherent reflection, where all the received and reflected radiation from a point ends up radiating as if from a single image (spherical or cuboidal pan) radiate any more energy, just because there is no identifiable image. Conservation of Energy could just as easily apply except when the surface is so rough as to cause multiple reflections before the radiation finally leaves. Deep surface roughness depth would / could affect the absorption according to angle.
I can't help thinking that a polished surface is cleaner, with no resistive oxide layer (10¹⁴ Ωcm) and the conductivity of the surface layer would be greater and the induced surface currents would be greater and cause more re-radiation. Would shininess (coherent reflection of visible wavelengths) be as important as flatness on a bigger scale (if at all) when dealing with much longer IR wavelengths? Microwave reflectors are not made with shiny surfaces so there must be something in my suggestion.
But :

Consider, for example, non-polished aluminium at 300K, which is said to have an emissivity of 0.77. We put it floating in vacuum. There is an energy source near it, and in this configuration it absorbs 77W of energy (and emits 77W as it is in thermal equilibrium). Now the metal is polished. It's emissivity drops to 0.05.

That seems to be a massive effect for a single molecular thickness of Aluminium Oxide[/SUP]

 

[Dale]

That seems to be a massive effect for a single molecular thickness of Aluminium Oxide

I think that the 0.77 emissivity is for anodized aluminum.

https://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html

Anodization is a few tens of microns thick, and dramatically changes the surface.

 

[sophiecentaur]

I think that the 0.77 emissivity is for anodized aluminum.

https://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html

Anodization is a few tens of microns thick, and dramatically changes the surface.

Random oxidation of the surface would probably be much more shallow though. But how much difference is due to the surface components and how much to the roughness would be interesting. How 'shiny' is 'shiny', for the purposes of that 0.05 figure, I wonder.

 

[Dale]

Random oxidation of the surface would probably be much more shallow though. But how much difference is due to the surface components and how much to the roughness would be interesting. How 'shiny' is 'shiny', for the purposes of that 0.05 figure, I wonder.

Rough aluminum is 0.07 and highly polished is 0.05. So I would estimate that the roughness/smoothness contribution is fairly small. In contrast, heavily oxidized is 0.25. So I would estimate that the oxidation contribution is the dominant one. This goes along with the fact that the 0.77 figure is for anodized aluminum which is basically just a deliberately thick layer of oxidation. So I would guess that the difference between 0.77 and 0.05 is much more to do with the simple fact that aluminum oxide has different emissivity than aluminum. Apparently aluminum oxide is much better at radiating than aluminum itself, which shouldn't be too surprising.

 

[hutchphd]

So the rays do not add coherently and the radiated power is less.

I believe (as indicated by @sophiecentaur) that this is a spurious argument.

Conservation of Energy could just as easily apply

If the rough surface itself is perfectly reflective, the presence of the roughness will certainly diminish the specular reflection, but the diffuse reflectance increases . The net reflected power is the same . As I recall there are variants of the optical theorem that show this (and I will provide a reference if I can...)
If however there are dissipative channels the roughness will often facilitate coupling to these internal degrees of freedom in the metal creating inelastic scattering of various kinds.

 

[sophiecentaur]

If however there are dissipative channels

If the small hills and valleys are steep enough then a significant amount of radiated energy can be reflected multiple times - with loss at each reflection. At the other end of the scale of things, a very good absorber / radiator can be made with deep pile black velveteen.

 

[tech99]

Rough aluminum is 0.07 and highly polished is 0.05. So I would estimate that the roughness/smoothness contribution is fairly small. In contrast, heavily oxidized is 0.25. So I would estimate that the oxidation contribution is the dominant one. This goes along with the fact that the 0.77 figure is for anodized aluminum which is basically just a deliberately thick layer of oxidation. So I would guess that the difference between 0.77 and 0.05 is much more to do with the simple fact that aluminum oxide has different emissivity than aluminum. Apparently aluminum oxide is much better at radiating than aluminum itself, which shouldn't be too surprising.

I think Aluminium Oxide is a good dielectric and can support long waves efficiently but it has a crystalline structure which scatters waves with a wavelength comparable to the size of the crystals, such as light.