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Heat lost from face due to black body radiation

  1. Sep 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Much of the of heat lost from a cross-country skier's body is radiated from the head, since it is often uncovered. Treat a head as a 20-cm-diameter, 20-cm-tall cylinder with a flat top. If the surface temperature of a body is 35 C, what is the net rate of heat loss on a frosty 5 C day? All skin, regardless of color in the visible wavelengths of light, is effectively black in the infrared where the radiation occurs. Therefore, assume an emissivity of 0.97.

    2. Relevant equations

    Q/Δt = eσA(T4-T04)

    where:
    e=.97
    σ=5.76*10-8 W/m2k4
    A=Surface area of cylinder, SA=2∏r2+2∏rh, = .188 m2
    T=308 K
    T0=278K

    3. The attempt at a solution

    Seems like straight plug n' chug.

    Q/Δt = (.97)(5.67*10-8)(.188)(3084-2784)
    =31.29 W

    However, this isn't right. Can anyone tell me what I did wrong?
     
  2. jcsd
  3. Sep 9, 2012 #2
    Why [itex]2\pi r^2[/itex]?
     
  4. Sep 9, 2012 #3
    So would the SA just be SA=2∏r+2∏rh since the cylinder doesn't have a bottom? Which would make the SA = .754 m2, Which would give an answer of 125.5 W?
     
  5. Sep 9, 2012 #4
    So I think sometimes I may be retarded. So it should have been ∏r2+2∏rh. I got the correct answer now. Thank you
     
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