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Homework Help: Second Law of Thermodynamics - Radiation

  1. Apr 6, 2016 #1
    Hello!
    This is an old problem from the Estonian-Finnish open physics olympiad (2013) and I hope you will be able to lend me some help.

    1. The problem statement, all variables and given/known data

    Sun-rays are focused with a lens of diameter d = 10 cm and focal length f = 7 cm to a black thin plate. Behind the plate is a mirror. Angular diameter of the sun is α = 32' and its intensity on the surface of Earth is I = 1000 W/m2
    i) Find the temperature of the heated point of the plate.
    ii) Using thermodynamic arguments, estimate the maximal diameter of the lens for which this model can be used.

    2. Relevant equations
    The lens formula: ## \frac{1}{a} + \frac{1}{b} = \frac{1}{f} ##
    Magnification: ## \frac{a}{b} = \frac{y'}{y} ##
    Stefan-Boltzmann law: ## \frac{P}{A} = \sigma T^4 ##

    3. The attempt at a solution
    i) Is not a problem, however I got really stuck on ii), so I checked the solution and could read the following: " Due to the second law of thermodynamics, it is impossible to direct heat energy from a lower temperature body to a higher temperature body. Hence, the image temperature cannot exceed the temperature of the Sun.". This statement was followed by easy calculations, but I have not really been able to accept this statement completely and would appreciate if you could help me out.

    When two bodies are placed in contact, it is intuitive that the hotter one will not get hotter while the colder cools down. It is also understandable with a microscopic perspective, because the molecules will transport energy from the hotter body to the cooler by collisions, and not the other way around. However, I have never previously encountered the second law of thermodynamics being applied to radiation. Here, the microscopic picture consists of photons with different energies hitting a plate. Assuming that the plate absorbs all radiation, I can't see why more incoming photons couldn't heat the material even more (for instance if we increase the diameter of the lens). I mean, there is no property other than the spectral distribution of the photons showing what temperature the body they are radiated from have. However, if we look at the photons one by one nothing can tell us how hot that body is, and consequently they should be able to give of their energy to the plate and heat it above the temperature of the sun. Where am I thinking wrong?

    Thank you in advance! :)
     
  2. jcsd
  3. Apr 6, 2016 #2

    DrClaude

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    Don't think in terms of photons, it is not helpful here.

    You gave the Stefan-Boltzmann law, which will apply also to the black plate (for simplicity, we can take it as a perfect blackbody). At any given temperature, it is radiating energy, including back through the lens towards the sun. There is thus an exchange of energy between the two bodies, mediated by radiation.

    If you were able to focus all the emission of the sun towards the plate, and all the emission of the plate towards the sun, at equilibrium they would reach the same temperature, which would basically be the temperature of the sun (since the plate doesn't produce its own energy, and is much smaller than the sun). This imposes an upper limit on the temperature of the plate.
     
  4. Apr 6, 2016 #3

    haruspex

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    I do not see how that explains it. Alettix is effectively considering the case where the target area is so small that the total power focussed onto it exceeds the power from the same area of the Sun's surface. For the power balance to be achieved, it would have to be hotter than the Sun's surface.
    I would have thought it was more to do with the impossibility of focussing all the power from wavelengths longer than the width of the target area.
     
  5. Apr 8, 2016 #4
    Thank you for your answeres!
    I don't see either how Stefan-Boltzmanns law explains this, as the area of the two bodies are different, thus the total radiated power will be different at the same temperature.
    I can understand that wavelengths longer than the diameter d of the point cannot be focused into in. However, I am having trouble with formulating this as a "thermodynamic argument" as asked in the question, to me it appears as very many photons corresponding to a wavelength smaller than d could heat up the point as well...
     
  6. Apr 8, 2016 #5

    haruspex

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    Then I misunderstood your difficulty.
    The thermodynamic argument comes from one of its fundamental laws... which one?
    I thought your problem was squaring this with the intuitive view that you could focus all the power onto an arbitrarily small area.
     
  7. Apr 9, 2016 #6
    Well, according to the solution given, it is the second law of thermodynamics, which states that a colder body cannot heat a hotter body. However, I do not understand why or how this is applicable to two bodies which are not in contact. My (probably very incorrect) intuition says that even if we cannot focus the long wavelengths into the image point, the energy carried by a large amount of short wavelegth photons should be enough to heat the imagepoint over the temperature of the sun.
     
  8. Apr 9, 2016 #7

    DrClaude

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    There are three was heat can be exchanged between bodies: conduction, convection, and radiation. What you have here is an example of radiative transfer. Again, fro the thermodynamic point of view, the only thing that is important is that they can exchange heat energy.

    You can't concentrate the light such that you get a point whose temperature is greater than that of the source. Have a look at http://what-if.xkcd.com/145/ and a recent discussion on a similar subject: https://www.physicsforums.com/threa...lliptical-cavities-and-the-second-law.857722/
     
  9. Apr 9, 2016 #8
    Thank you Sir! This really helped!
     
  10. Apr 9, 2016 #9

    haruspex

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    It might help to put some numbers to it. Suppose you captured the light falling on a square metre of Earth. What area would you need to focus it down to to reproduce the light intensity at the Sun's surface? What wavelengths would that limit you to? How much of the Sun's power is at longer wavelengths?
     
  11. Apr 10, 2016 #10
    The light intensity at the sun's surface is approximately 64 MW/m^2 and at Earth roughly 1 kW/m^2, which means that the light should be focused to an area of ca 1.56 * 10-5 m^2. Assuming that it has a square shape, the longest wavelength which can be absorbed is ca 0.004 m.

    So the total energy which can be absorbed should be given by the integral of Planck's radiation law from 0 to 0.004. If I have done it right this is 7.3*107 W, which is 99% of the total emittance of the sun (?). Something must be wrong with that integral, mustn't it? But even if we had the amount of energy, wouldn't we need the heat capacity of the point we are focusing the light to to determine its temperatue?
     
  12. Apr 10, 2016 #11

    haruspex

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    Why? If we are treating it as a black body, it is just a question of the temperature at which the power would balance.
    Anyway, you have so far found that trying to focus it back to get the same temperature doesn't quite do it. But what you really want to know is why you cannot exceed the Sun's surface temperature by focussing it even smaller. So try focussing it on a quarter of that area. What power can you collec nowt, and what temperature does that correspond to?
    Now, I don't know where this will lead. Maybe it will say you can beat the Sun's temperature, in which case my solution to the paradox is not adequate.
     
  13. Apr 12, 2016 #12
    Sorry, I forgot about this. My calculations show that this temperature will be ca 5800 K (200 degrees lower than the temperature of the sun). (in this calculation I used that 0.99*1000 W will reach the surface, and that this must be balanced by it by thermal radiation according to the Stefan-Boltzmann law).

    The quarter of the area is ## 3.9 \cdot 10^{-6} m^2##, and the maximal wavelenght (assuming a square) ca 0.0020 m. The emittance from the which falls inside this range is still ca ## 7.3 \cdot 10^7 ## W/m (Am I calculating this right?), which is 99.6% of the total emittance, and this results in a temperature of 8200 K...

    That the integral doesn't change considerably must be because the wavelength for with the emittance is at its maximum is ## 4.8 \cdot 10^{-7} ## m, thus quiet far away from 0.002 m ⇒ the intensity for both 0.002m and 0.004 m is very low.
     
  14. Apr 12, 2016 #13

    haruspex

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    It's not an area I know enough about to be checking your integral, so I have to trust that. From what you have found, it seems my explanation is not adequate. @Chestermiller might be able to .... ahem ... shed some light on this.
     
  15. Apr 12, 2016 #14
    I'm not an expert on radiation transfer, so probably what I am going to say is pretty naive. Wouldn't the heated surface conduct heat away from the target spot? Also, when it did this, wouldn't the effective area to radiate heat from the surface be greater than that of the target spot?
     
  16. Apr 12, 2016 #15

    haruspex

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    Not sure... would a theoretical perfect insulator or perfect reflector violate thermodynamic laws? Anyway, we could allow for double the emitting surface, but it still doesn't sound as though it would get us over the line.
     
  17. Apr 12, 2016 #16

    haruspex

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    Another thought...
    It would not necessarily violate the laws to produce some small patch at a higher temperature than the original. We have to look at the whole picture. What has happened to the light we did not catch?
     
  18. Apr 13, 2016 #17

    256bits

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    It's not really a paradox if one considers that the sun's rays are just about parallel, but not parallel. Focusing to a point, or to an arbitrarily small area is impossible.
    One would have to use optics of lenses to show that since the sun is not a point source but has an arc width when viewed from the earth ( or at any distance ), the smallest area that the sun's rays can be focused is a function of the diameter of the sun, and the distance to the sun. In other words what one can hope to accomplish is only to form an image of the sun, and the temperature of the image can only approach a maximum temperature, which would be that of the surface of the sun.
     
  19. Apr 13, 2016 #18

    haruspex

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    Of course! Thankyou.
     
  20. Apr 13, 2016 #19
    Thank you very much for your help everyone! :)
     
  21. Apr 17, 2018 #20

    Charles Link

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    I can perhaps add to what has already been discussed. ## \\ ## The focused irradiance in watts/m^2 ##E_{focused}=L_{sun} \Omega_{lens} ## assuming the solid angle ## \Omega_{lens} ## is small enough that a factor of ## \cos(\theta) ## doesn't need to start to be included in the calculation. (This equation makes use of what is known as the "brightness theorem" in optics, that the image will have the same brightness ## L ## as the source, assuming 100% transmissive optics). ## \\ ## The solid angle ## \Omega_{lens}=\frac{A_{lens}}{f^2} ##. (## A_{lens}=\frac{\pi d^2}{4} ##). ## \\ ## The black surface, assuming no heat conduction, will reach a thermal equlibrium given by temperature ## T ## such that ## E_{focused}=\sigma T^4 ##. ## \\ ## The result is ## L_{sun} \Omega_{lens}=\sigma T^4 ##.## \\ ## (Note that ## \Omega_{lens} ## will never and must never exceed ## \pi ## steradians. The ## \cos(\theta) ## factor that must be included in the solid angle calculation for very wide diameter lenses will ensure that).## \\ ## ## L_{sun} ## can be computed by ## E_{sun}=L_{sun} \Omega_{sun}=(L_{sun}) (\alpha)^2 (\frac{\pi}{4}) ##, where ## E =1000 \, watts/m^2 ## as given, and ## \alpha=(32/60)(\pi/180) ## radians. ## \\ ## The answer I get, assuming my arithmetic is correct, is reasonable, but considerably less than the 5800 K posted above. ## \\ ## An alternative geometric optics approach is possible that considers image size and power collected from the lens should get the same answer. That calculation is quite straightforward: ## P_{lens}=E_{sun} A_{lens} ##, and ## A_{focused \, image}=\frac{\pi}{4} (\alpha f)^2 ##. The result is identical: ## E_{focused}=\frac{P_{lens}}{A_{focused \, image}} ##, and for equilibrium ## E_{focused}=\sigma T^4 ##. ## \\ ## The first calculation above, using solid angles, shows why the geometric optics approach doesn't work if the lens begins to fill the hemisphere, because the effective solid angle of a hemisphere, with the ## \cos(\theta) ## factor included in the calculations is ## \pi ## steradians, and not ## 2 \pi ## steradians. In addition, the formula ## \Omega=\frac{A}{f^2} ## does not give correct/accurate results for ## \frac{A}{f^2} \approx 1 ## or larger, for a couple of reasons: ## d \Omega =\frac{dA }{f^2} ## for area elements ## dA ## that are nearly on axis, but off-axis, the distance ## r>f ## and in the solid angle calculation, the area element is the projected area element: ## d \Omega= \frac{d A_{projected}}{r^2} ##. This will account for a ## \cos^3(\theta) ## factor that needs to be included in calculations of ## d \Omega ## for off-axis area elements. There is an additional ## \cos(\theta) ## factor that arises because the incident irradiance from the solid angle element ## d \Omega ## is incident at an angle ## \theta ##. The result is that off-axis, there is actually a ## \cos^4(\theta) ## factor that arises in the formula ## dE=L \, d \Omega_{effective}=\frac{L \, \cos^4(\theta) \, dA}{f^2} ##. In this problem we are nearing the limit of where the formulas ## d \Omega=\frac{dA}{f^2} ## and ## dE=\frac{L \, dA}{f^2} ## are still reasonably accurate. In this problem, where ## \Omega_{lens} \approx 1.6 ##, as computed from ## \Omega_{lens}=\frac{A_{lens}}{f^2} ##, it is already going to overestimate the result for focused irradiance ## E_{focused} ##, but not by an unreasonable amount. ## \\ ## It may also be of interest to consider the brightness of a blackbody at temperature ##T _s ## , ## L_{bb} =\frac{\sigma T_s^4}{\pi} ##, in determining why the material at the focus can never reach the temperature ## T_s ## of the source. The solid angle ## \Omega_{effective \, lens} ## has the very upper limit (if you could make the lens cover the hemisphere above the plane of focus) of ## \pi ## steradians. The upper limit of ## E_{focused}=L_{bb} \, \Omega_{effective}=L_{bb} \, \pi ##, assuming the sun is a blackbody at temperature ## T_s ##. Thereby, the computed ##T ##, where ## \sigma T^4=E_{focused}< \sigma T_s^4 ## will always have ## T<T_s ##. ## \\ ## One additional comment: ## \\## The question arises, why can't we take a very large lens and focus all of that collected power into the image, with virtually no upper limit? The answer is that for very wide angles of focusing, lens aberrations must necessarily occur, and the image will inevitably not be focused. The "brightness theorem" has an added part to it that the brightness of the image can never exceed the brightness of the source. For a very large lens, ## A_{focused \, image} ## would necessarily become considerably larger than ## \frac{\pi}{4} (\alpha f)^2 ##. Geometric optics, that is derived for paraxial rays, simply falls apart for a lens with ## d>> f ##. ## \\ ## The upper limit in this problem is to construct a lens that fills the hemisphere as viewed from the plane of focus, and in looking up from the plane of focus over the entire hemisphere, everything that is observed is at the same brightness level ## L ## as the source. Under those conditions, without any thermal conduction, ## T=T_s ##.
     
    Last edited: Apr 17, 2018
  22. Apr 17, 2018 #21

    Charles Link

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    @Alettix Please see the added last paragraph of the previous post: "One additional comment".
     
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