How could heat be concentrated ?

[swansont]

Hmm...

 

I still would like to see a better explanation... but hopefully we're getting close I'm feeling a bit stubborn because I keep asking, but I honestly still don't understand why a 2nd object cannot become warmer than another object that radiates photons (IR, light, whatever radiation), when you use a lens.

 

And you understand this violates the 2nd law, but don;t see how it violates it.

 

[quote name=CaptainPanic;501046

The radiation originating from any object includes' date=' as was already mentioned, pretty much all wavelengths. Let's assume a perfect black body (see: graph), which does radiate all wavelengths.

 

I still don't see why with a perfect lens you cannot concentrate the light so much that it will heat up an object to a temperature higher than the source of the photons. Just focus it on a really really tiny spot. The flux of photons should become so great that a higher temperature can be achieved?

 

I understand perfectly that the energy balance will not be breaking the 2nd law of thermodynamics. The total energy of all photons originating from the source of photons will always be greater than on the "really really tiny spot".

 

But why wouldn't it reach a higher temperature? That does not necessarily go against the 2nd law (look at heat pumps and your fridge: it's possible).

 

I think that for determining the temperature that is achieved somewhere we must look at the flux (photons / receiving surface area) and the wavelength. This flux can be concentrated immensely using lenses.

 

As insane_alien has noted, it's because you're adding energy. The objection of the 2nd law is for spontaneous processes only, and passive optical systems.

 

And here's the issue: you insist that you focus the light down. You can't. You can see this with ray tracing: in order to get a small focus, you need parallel light, but you don't have that. You can try to get it with two lenses - one to capture light and the other to concentrate it. Put an object at the focal point of the other lens, and it will give you parallel light. But that means the source has to be infinitely small, and you've already said that it was bigger.

 

Put another way: only one point on the source can be at the focal point of the lens. The rest must be away from the focal point, and these all map to different points near the focal point of lens #2. The spot you get will be larger than you have assumed.

 

(This all ignores the fact that you can only collect half the light with a lens, but adding mirrors doesn't solve the problem.)

Edited June 30, 2009 by swansont
add clarification

[proton]

Assumptions

For the sake of the discussion, let's assume the perfect IR lens to exist, ...

I did a search and found this

http://www.thorlabs.com/navigation.cfm?Guide_ID=132&gclid=CJal_OmEs5sCFQxM5Qod0At2Qg

IR Optics

 

These plano-convex lenses introduce minimal spherical aberration and are commonly used for IR applications in the 1 to 16 µm spectral range. Choose from silicon, germanium, or zinc selenide and focal lengths from 20 to 200 mm. These lenses have been used in CO2 lasers, as collimators, and for biomedical, security, or military imaging.

So it seems that lens for IR do exist, at least in the 1 to 16 µm spectral range. Cool! I never knew that! That's what I love about posting in forums. You never know what you might learn simply because I never asked a particular kind of question.

 

Thanks for asking this question Externet!

[Klaynos]

I work in both the near, and far IR. So I can only really comment on them...

 

For very near IR most good quality optics will work fine (it does depend on the glass).

 

For far IR you can buy lenses but they are really rather expensive, it's actually more sensible to use curved mirrors, which if you pick the right metal have good reflectivity over a very wide ranged into the very far IR.

 

You can find something to do lensing for most frequency bands. Mirrors are generally easier, but for one of our second year UG labs some of the students doding an extended lab designed a microwave lens.

[Externet]

If a heat source cannot be concentrated to a hotter temperature spot, S

 

M

M

M

M

M

M

 

Radiation from the sun S above heats and illuminates this target ---> + to -say 50C- with the intensity of one sun.

The radiation from the sun S hits mirrors "M" at left aligned to all reflect onto target + to a total of 7 suns intensity.

 

If the mirrors were covered with an infrared pass filter that blocks visible light, the target + will not be hotter than 50C ?

 

Again; If a heat source cannot be concentrated to a hotter temperature spot,

Does it mean the heat at the top of this tower comes only from the concentrated reflection of visible light and the heat part of the sun radiation is not a factor ?

 

----> http://www.nickelinstitute.org/multimedia/magazine/2008/December_2008/Solar/solar1-650.jpg

 

Miguel

Edited July 1, 2009 by Externet
better picture

[CaptainPanic]

And here's the issue: you insist that you focus the light down. You can't. You can see this with ray tracing: in order to get a small focus, you need parallel light, but you don't have that. You can try to get it with two lenses - one to capture light and the other to concentrate it. Put an object at the focal point of the other lens, and it will give you parallel light. But that means the source has to be infinitely small, and you've already said that it was bigger.

 

Put another way: only one point on the source can be at the focal point of the lens. The rest must be away from the focal point, and these all map to different points near the focal point of lens #2. The spot you get will be larger than you have assumed.

 

(This all ignores the fact that you can only collect half the light with a lens, but adding mirrors doesn't solve the problem.)

 

Thanks, the parts in italics were the answer for me. I feel a bit silly for needing this long to see the light (pun intended - probably the most unpopular joke with people working in optics).

 

Until now I was focusing (pun intended again) my attention on the wrong side of the lens: I sort of assumed that parallel light was available. And of course, it can be approximated by placing a source really far away... but that obviously reduces the intensity of the light.

 

Thanks for all the explanations and patience. The world makes sense again

[swansont]

If a heat source cannot be concentrated to a hotter temperature spot, S

 

M

M

M

M

M

M

 

Radiation from the sun S above heats and illuminates this target ---> + to -say 50C- with the intensity of one sun.

The radiation from the sun S hits mirrors "M" at left aligned to all reflect onto target + to a total of 7 suns intensity.

 

If the mirrors were covered with an infrared pass filter that blocks visible light, the target + will not be hotter than 50C ?

 

Again; If a heat source cannot be concentrated to a hotter temperature spot,

Does it mean the heat at the top of this tower comes only from the concentrated reflection of visible light and the heat part of the sun radiation is not a factor ?

 

----> http://www.nickelinstitute.org/multimedia/magazine/2008/December_2008/Solar/solar1-650.jpg

 

Miguel

 

All the light that is reflected will contribute. It's all heat.

 

The sun is at 6000K. That's your limit for passive heating.

[proton]

How is the blackbody radiation from the radiator not heat flow from a difference in temperature?

Heat is the energy transferred from one object to another by virtue of their difference in temperature. I suggest this definition does not include radiant energy because an object radiates electromagnetic energy regardless of whether another object is there to intercept it.

[swansont]

Heat is the energy transferred from one object to another by virtue of their difference in temperature. I suggest this definition does not include radiant energy because an object radiates electromagnetic energy regardless of whether another object is there to intercept it.

 

Yes, it includes radiant energy. The net amount it radiates depends on the temperature of the surrounding reservoir, or objects. That's why there are two temperature terms in the Stefan-Boltzmann equation, and they have opposite signs. An object at the same temperature as the surroundings radiates no net energy, because it absorbs the same amount that it radiates.

[proton]

Yes, it includes radiant energy.

In relativity heat transforms differently than energy. So one cannot merely make the assertion the blind assertion that radition is heat.

[swansont]

In relativity heat transforms differently than energy. So one cannot merely make the assertion the blind assertion that radition is heat.

 

We weren't discussing the complications that arise when you treat this relativistically.

[proton]

We weren't discussing the complications that arise when you treat this relativistically.

Obviously. But that doesn't change the fact. The reason it transforms differently is that energy in motion is defined differently than heat. Ignoring context leads to errors in application. Heat is defined as the flow of energy from one object to another caused by a difference in temperature. The cause of light from a laser is not caused by a difference in temperature so one can't call the radiant energy from a laser "heat". Heat is also the quantity Q in the relation

 

[math]\Delta U = Q + W[/math]

 

That relation defines Q and is why it transforms differently than energy. You seem to think that any energy which flows is called heat, for what reason I can't see. Since this is going nowhere so I have nothing left to say.

[swansont]

Obviously. But that doesn't change the fact. The reason it transforms differently is that energy in motion is defined differently than heat. Ignoring context leads to errors in application. Heat is defined as the flow of energy from one object to another caused by a difference in temperature. The cause of light from a laser is not caused by a difference in temperature so one can't call the radiant energy from a laser "heat". Heat is also the quantity Q in the relation

 

[math]\Delta U = Q + W[/math]

 

That relation defines Q and is why it transforms differently than energy. You seem to think that any energy which flows is called heat, for what reason I can't see. Since this is going nowhere so I have nothing left to say.

 

A laser is not a blackbody, doesn't have a well-defined temperature and the Stefan-Boltzmann equation doesn't apply. It's not a situation where energy is being transferred due to a temperature difference, as you note. So it's a non-sequitur.

 

I'm not sure how you came to the conclusion you wrote. The forms of spontaneous heat transfer include radiation. That's not the same as saying all radiation is spontaneous heat transfer (i.e. caused by a temperature difference)

Edited July 2, 2009 by swansont

[McCrunchy]

Well, sorry, I still don't get why we can't get an arbitrarily large amount of parallel rays (ie, of watts) to start off with. Let's consider for example a 1000*1000 km black grid heated to a temperature T. Let's put in front of the grid two lenses A and B , of same focal distance, but one 1 m in diameter, the other 2 m. The power at B's focal point will be 4 times that of lens A, won't it ? Once we have an arbitrarily large heat flux Q_input, the temperature of a black body heated by it should go as

T= (Q_input/A sigma)^(1/4) by the Stefan Boltzmann law (A - area of black body , sigma - SB constant) - so T can also rise arbitrarily.

 

PS : there must be a flaw in my line of thought

[swansont]

The object you're heating up radiates as well. The object can only heat up as long as the incoming power is greater than the radiated power. In the ideal case, it can only equalize.

[alex folen]

As light can be focused to a small area with the use of a lens, how could heat -say from a household radiator- be concentrated into a smaller area ?

 

Cool question, maybe simple to answer. Light can be funneled via lens made by humans. Heat radiant (non-light energy heat) however needs a device other then a lens to be funneled. Let's make one? ....but would be most probably inefficient. End of discussion. The remarkable thing, if remarkable, is that the shadow (the lack of energy) of the man made lens is diverted into to the funneling effect of the optics or lens. No energy gained here, just focused and concentrated. Hello.

Edited July 13, 2009 by alex folen

[McCrunchy]

Swansont,

 

I agree the black body we're heating also radiates heat out. At equilibrium, these fluxes equalize. This gives us : Qin=Qout, with Qout = A sigma T^4 . My argument is that as long as Qin can be made big enough by using a big enough lens to focus the rays, there's no reason why T couldn't reach an arbitrarily high value - even higher than that of the initial source of radiation.

 

We don't need to be in the IR regime - let's consider a 3000 K emission source for which we know the lenses exist and work.

[swansont]

You need an infinite source for this to work. Physically impossible. For any finite system, there is a limit to how much power you can capture.

[timetes]

again just a mother.....

but if you put something like a metal screen in front of it....wouldnt it heat the screen which is larger....and i would assume a large area is warmed. The idea is a small heat area to make a larger heat area.

Kinda like an electric stove.

 

I thought about that with solar pannels. If they put a lens dome on a solar pannel it would generate more solor energy.

 

still the problem is....storing this energy. If someone can figure that out...

 

 

As light can be focused to a small area with the use of a lens, how could heat -say from a household radiator- be concentrated into a smaller area ?

 

As to focusing invisible heat from a large source to raise temperature of a smaller body-

Miguel

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Merged post follows:

Consecutive posts merged

opps to a smaller area....sorry......ok .....a funnel...lol