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Black Body Radiation

  1. Oct 31, 2008 #1
    ***** SHEETS two ideally black of the area 1:5 mETER SQUARE are located opposite of each other in vacuum.
    The first SHEET P1 has the FIXED temperature T1 = 0o C, the second SHEET P2 has the
    fIXED temperature T2 = 100o C.

    Now an additional black panel PM of the same size is brought between the panels P1 and P2.
    a) What will be the temperature TM of the panel PM in the radiation equilibrium?

    Can anyone just give me some hints ?
    Last edited: Oct 31, 2008
  2. jcsd
  3. Oct 31, 2008 #2


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    Welcome to PF - this should be posted in the homework forum and you should make an attempt at the question first.

    It will reach an equilibrium temperature where the energy it receives is equal to the energy it emits. What's the equation linking blackbody power and temperature?
  4. Oct 31, 2008 #3
    Ok, Thanks alot for the hint . I tried it please take a look at it ...

    According to Stefans LAW


    P = Power radiated in W (J/s)
    s = Stefan's Constant 5.67 x 10-8 W m-2 K-4
    A = Surface area of body (m²)
    T = Temperature of body (K)

    To Find out the temperature of Pm :-

    P =SAT2^(4)---(A)

    Where Sigma = 5.67 x 10-8 W m-2 K-4
    Area = 1.5 m^2
    Temperature in Degrees Kelvin = 373.15

    Therefore according to eqn A,
    P = 1649.068 W ...

    Now My Question is ,

    Is the calculated power is from panel P2 ?
    If it is then how i will approach to the temperature of Pm ?

    Boundary effects shall be negligible.
  5. Oct 31, 2008 #4


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    Correct, The power flows both ways, a larger amount flows from hot->cold and a smaller amount back. The amount of power is just the difference in emmission for the two temperatures.
    ie. P = sA(T14 - T24) for each panel.
    So you have,

    1, between T1 and the new panel
    P = sA(TM4 - T14)

    2, between the new panel and T2
    P = sA(T24 - TM4)

    Now for an equilibrium the two powers must be equal - otherwise the muiddle one would be changing in temperature. So simply set the two equations equal to each other and do a bit or re-aranging.
    Hint, would you expect the answer to depend on the area if all three are the same size?

    Note - depending on how idealised the question is you might want to consider what power flows off the other sides of the outer 2 plates. Since you aren;t given a temperature for the surroundings you shoudl probably assume they are at an infitie distance or zero temperature.
  6. Oct 31, 2008 #5
    Thanks alot for the help ,
    sA(TM4 - T14) = sA (T24 - TM4) By canceling sA

    (TM4 - T14) = (T24 - TM4)

    Tm4 = (T24 - T14)/2

    Tm4 = 6.9106 10^9 .

    How does the radiation balance of P2 change with the insertion of PM?

    Thanks again
    Last edited: Oct 31, 2008
  7. Oct 31, 2008 #6


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    Think you might have a small typo.
    Tm4 = (T24+T14)/2

    Remember that because you are multiplying it has to be absolute temperature.

    And before you type in any numbers, what temperature would you expect - roughly?
  8. Oct 31, 2008 #7

    Ohh Sorry ..

    Now i got it ,

    Tm4 = (T24 + T14)/2
    Tm = 334.21 Kelvin

    How does the radiation balance of P2 change with the insertion of PM?

    What i m getting is radiation balance of P2 with P1 will change after the insertion but how can we define it ?
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