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Power Delivered From Black Body

  1. Oct 22, 2014 #1
    1. The problem statement, all variables and given/known data
    The power delivered in narrow spectrum range near 10 µm from black body source with temperature 1000 K is 10 µW.

    Not Solved
    a) What power would be delivered in narrow spectrum range near 1 µm?

    Solved
    b) At what wavelength the power delivered takes its maximum?
    c) Find this maximum value.


    2. Relevant equations
    For b and c, you are using:

    λpeak = 2.898 x 10-3/T

    That aspect of this problem is easy and was solvable.

    For a, I would assume that one would use:

    P = σAT4

    Problem is that the question doesn't give surface area at all, only λ. Plugging what was given doesn't produce the relationship at all (even squaring the given λ = 10 µm only produces 5.6 µW). I don't know if it is a bad question or missing something obvious.

    I also tried E = σT4 and got 5.7 W/cm2. I would assume that I would have to calculate per 1 µm?

    The other alternative solution I tried was using Radiant flux density (Planck’s law):

    W = 2hc25 x 1/(e(hc/λkT)-1) and the numbers still seem off.

    3. The attempt at a solution
    From above.

    Also, based on the information above and solving part b and c that moving closer to the black body (2.898 µm) will produce the highest power and would dip off after as we approach 1 µm, but it feels like there is something missing in part a.
     
    Last edited: Oct 22, 2014
  2. jcsd
  3. Oct 22, 2014 #2

    DrClaude

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    Staff: Mentor

    You don't know the size of the black body, and therefore can't say anything about the total power emitted. I don't understand what "squaring the given λ = 10 µm" means. Do you know what λ is?

    Again, if you don't know how big the black body is, you can't get an absolute number. But this equation is a good starting place. You need to figure out how to use it correctly for your problem.


    Moving closer??? I guess you're indeed confused about what λ is.
     
  4. Oct 22, 2014 #3
    For this question, yes, I am confused about the purpose of λ.

    For all intents and purposes (and to prevent you from continuing the assumption I am a total idiot) I know that λ is wavelength, which is the distance between successive crests (high points) of any type of wave (ala period). Problem is that part a is extremely arbitrary.

    When I meant by 'moving closer' I was referring to the Power Density/Wavelength chart (http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html) where if you are starting at 10 µm and approach the λpeak of 2.898 µm, you hit the maximum Power Density. But if you move away from the peak number and continue decreasing the value to 1 µm, there is supposed to be a drop off in power density (via the Power Desnity/Wavelength chart). Problem is like you said, I don't know the size of the black body.

    Now, with that in mind, I did compute W = 2hc2/λ5 x 1/(e(hc/λkT)-1) at first using the given variables just to see if this was the right approach.

    2hc2 = 2 (6.67 x 10-34) (2.99 x 108)2 = 1.19 x 10-16

    λ5 = (10 μm)5 = 100000 μm5

    Dividing those two will get = 1.19 x 10-21

    1/(e(hc/λkT)-1) = e1.99 x 10-25/1.38 x 10-19 = e9.173 x 10-23 ≈ 1 - 1 ≈ 9.173 x 10-23 = 1.09 x 1022

    Multiplying the two will get 12.971, but given we were dealing with rounding errors and the sort, I can assume that @ 10 μm & T = 1000 K, P ≈ 10 μW
     
  5. Oct 22, 2014 #4
    That formula does not provide a power, in W.
    It is spectral density. In order to get power you need to multiply by area and wavelength interval.
    So there is no reason to expect to get 10 μW.

    But you don't need area, it is the same in both cases. And I suppose you can assume the same wavelength interval (a narrow one).
    It's just a matter of ratios.
     
  6. Oct 23, 2014 #5

    DrClaude

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    Staff: Mentor

    To make things clear, I never thought you were an idiot! For one thing, there is a different between idiocy and ignorance. And it's not like I have never been confused before :)

    This.
     
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