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Homework Help: Thermal Radiation

  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A wooden box is painted completely with black paint. The interior dimensions are 18”x24”x3.5”. The bottom and sides are made with 9 pieces of 2x4 douglas-fir-larch dimension wood lumber. The top of the box is open and is covered with 0.118” thick clear acrylic. The acrylic is sealed to the box and only allows light to enter the box. The air surrounding the box has negligible velocity.

    The box and the ambient air temperature are 47 degrees F. At time, t = 0, t a 500 W light bulb that is suspended 6 feet above the box is turned on. Thermal equilibrium of the air and box occurs 10 minutes later. What is the temperature of the air inside the box at this time?

    Black Paint ε = 0.80
    500 W light ε = 0.95
    T light = 3000 K

    CWood 1700 J/kg*K
    CAir 1009 J/kg*K

    ρ air = 1.225 kg/m3
    ρ wood = 530 kg/m3

    σ = 5.670373E -8 J/S*m2*K4

    2. Relevant equations

    E = A ε σ T4
    Q = m c ΔT

    3. The attempt at a solution
    To heat the Air:
    Vair = 18x24*3.5 in3 = 0.02477 m3
    ma = ρ airVair = 0.0304 kg
    Qair = ma cair ΔT

    To heat the Wood:
    Vwood = 3.5x1.5x 24 in3 = 0.0021 m3
    mw = ρ woodVwood = 1.094 kg
    Qwood = mw cwood ΔT

    To heat the wood/air:
    Qtotal = ma cair ΔT + mw cwood ΔT

    Energy from 500 W light:
    Distribute the 500 W over a sphere with a 6 ft (1.8288 m) radius, A = 4π r2 = 42.028 m2 500 W/ 42.028 m2 = 11.9 w/m2 (at the box from the light)

    The box is 18x24 in 2 (0.2787 m2) so it receives 3.317 W on the surface at that distance. 3.317 W applied for 10 minutes (600 seconds) = 1990 J applied to the air and box.

    1990 J = ma cair ΔT + mw cwood ΔT

    solve for ΔT = 1.05 K ?
  2. jcsd
  3. Mar 21, 2017 #2


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    What processes are going on once thermal equilibrium is reached?
  4. Mar 21, 2017 #3
    Just intuitively, the box is giving off as much energy as it is taking in because at that point the temp quits changing so the air and the box are in equilibrium with each other, however, there is a ΔT between the box and surrounding air so convection (minimal) and radiation (thermal) are going on. I'm more interested in the process I applied, more specifically, is it correct to distribute the 500W to the "inside surface area" of the sphere as I have done?
  5. Mar 21, 2017 #4


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    Right. So I would think this is what you need to analyse in order to arrive at the equilibrium temperature.
    Yes, and I agree with the 3.3W. (It is only an approximation, though. To get it exact you would need to compute the solid angle the rectangle subtends at the centre of the sphere, which is quite tricky.)

    However, you cannot simply multiply that by 600 seconds to find the added energy. As you note, it will be losing energy increasingly as the temperature rises.
    If you assume the only losses are radiative, and that the principal radiating surfaces are all at the internal temperature (bit of a stretch) can you figure out the effective emissivity?
  6. Mar 21, 2017 #5
    For now, the simpler approach is better. If necessary, I can refine once I get a better understanding of the procedure.
    I'm not sure I can answer your question. While at steady state, air in the box and the box are at the same temperature and this "system" is receiving 3.3W hence, it is giving off 3.3 W otherwise the temp would go up. If I knew the final temp I can get ε. I know A so I could get ε using E = A ε σ T4 . BUT, I don't know Tf.
    I could use the ΔT from my "first pass", redo the calculation, and see if Tf converges? I'm afraid I'm guessing because I don't see a direct route.
    3.3 =A ε σ T4 has 2 unknowns.
  7. Mar 21, 2017 #6


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    Sorry, I forgot - you are given the emissivities. So you just need to get the equilibrium equation right. Don't forget that without the light it was in equilibrium at Ti, and the light is an additional input of power.
  8. Mar 22, 2017 #7
    At the beginning of the problem, the box is in equilibrium with the surroundings all at a temperature of 47 degrees F (281.483 K). Which means, again, it is giving up as much as it is taking in. ΔE = Ef- Ei = A ε σ (Tf4 -Ti4) = 3.317 W → Tf = 284.21 K (51.9° F). ?
  9. Mar 22, 2017 #8


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    That's the right formula. Haven't checked your numbers... what did you get for A?
    I note you are given ε=0.95 for the light, i.e. you do not get the full 500W.

    In reality, the outsides of the box would not be at the same temperature as the inside. There would be a temperature gradient through the wood. Were you given a conductivity for the wood?
  10. Mar 22, 2017 #9
    I used the area of the box, A = 0.2787 m2 (18"x24") in the equation to arrive at Tf = 284.21 K.

    Now I have another question... shoot... If the box is radiating and absorbing at the same rate AND some it is lost to heat in conduction, say H = h A ΔT then I need to subtract H from the 3.317W being delivered?:

    3.317W - h A (Tf - TI) = A ε σ (Tf4 -Ti4) (in this case the area conducting = area radiating).
  11. Mar 22, 2017 #10


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    I wasn't sure what you were calculating as A. I see that you only used the area of the acrylic, so ignored losses through the wood.
    That's probably ok.
    To allow for conduction is a bit tricky because you have to set the outside temperature of the box to some unknown, Tb, and match the conduction through the box over that temperature gradient to the radiation from the outside of the box at that temperature. That involves solving a quartic.
  12. Mar 22, 2017 #11

    I think for now I'm good with just considering the radiation portion. I'm sure this discussion, which I enjoyed and appreciated your leadership in, can be further refined and tweaked to a more precise answer, but for now, I think I have something close that I can work with comfortably.

    Here's the rub. I built this dumb box and tested it. Don't ask why, I just did... I was getting between 4 and 5 degrees F change and was really having trouble finding how to calculate it. I appreciate the help getting to the equation in post #7 (this thread) and I did have a good laugh with the result of that calculation, 47 to 51.9, so the measured temps seem to be in reasonable line with that calculation. Who knows, if time allows, I'll make another one of different size and repeat the that calculation to confirm.
  13. Mar 23, 2017 #12


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    Congratulations on getting a result in such good agreement between theory and practice.
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