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Problem involving thermal radiation and specific heat

  1. May 24, 2017 #1
    1. The problem statement, all variables and given/known data
    A satellite to reflect radar is a 3.5-m-diameter, 2.0-mm-thick spherical copper shell. While orbiting the earth, the satellite absorbs sunlight and is warmed to 50 °C. When it passes into the earth's shadow, the satellite radiates energy to deep space. You can assume a deep-space temperature of 0 K. If the satellite's emissivity is 0.75, to what temperature, in °C, will it drop during the 50 minutes it takes to move through the earth's shadow

    2. Relevant equations
    P=eσAT4
    Q = mCΔT
    ρ = m/V

    3. The attempt at a solution
    I thought I had this question worked out but my solution didn't work. In any case this is what I did:

    First some useful things. I found the surface area of the sphere to be 38.48m2 and then I calculated the volume as being:
    22.45 - 22.37 = 0.08m3
    by taking the volume of the outer sphere with radius 1.75m and subtracting the volume of the inner sphere of radius 1.748m

    Then I found the power that is radiated using P=eσAT4 as we can assume Tc=0
    P = 0.75 * 5.67*10^-8 * 38.48 * 3234
    = 17812.2W

    Then I calculated the total energy over the 50 minutes:
    17812.2 * 60 * 50 = 5.34*107J

    Then I used the density of copper and the aforementioned volume to calculate the mass of copper:
    8920 * 0.08 = 713.6kg

    And then I used the specific heat formula to calculate the final temperature after losing the energy from above. Negative because the heat is lost.

    -5.34*107 = 713.6 * 385 * (TF - 323)
    TF = 128.5K = -144.502°C

    But that is apparently wrong. I am not sure where I went wrong. Does the radiation formula not work because the temperature is changing perhaps?
     
  2. jcsd
  3. May 24, 2017 #2
    If the temperature changes that much, P will not be constant. Equate the instantaneous value of P to -mCpdT/dt and solve the differential equation.
     
  4. May 24, 2017 #3
    I am not really sure how to do that. Do I just say:
    eσAT(t)4 = -mCdT/dt?

    How do I go about solving this? And is it okay to do this because now both are with respect to time (as in the power is J/s and we have a dt on the other side)?
     
  5. May 24, 2017 #4
    Write (eσA/mC) dt = -1/T4 dT and integrate both sides.
     
  6. May 24, 2017 #5
    I got -7.79°C which is still wrong apparently
     
  7. May 24, 2017 #6
    Oh. I got -44°C. How did you do the integration?
     
  8. May 24, 2017 #7
    I got:
    5.95*10-12 * t = 1/3T3
    And then I substituted t = 3000 and solved for T
     
  9. May 24, 2017 #8

    haruspex

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    Think about the integration bounds. The result of the integration step relates the change in each side. On one one side you have a change in time, but the other side is not a change in temperature, nor is it the cube of the final temperature. What is it the change in?

    (I get -46C.)
     
  10. May 25, 2017 #9
    I am not entirely sure. Is it the total temperature difference? That doesn't sound right...?
     
  11. May 25, 2017 #10
    It's a definite integral. The solution (assuming your arithmetic is right) is
    [5.95*10-12 * t]03000 = [1/(3T3)]323Tf
     
  12. May 25, 2017 #11
    Thank you that worked perfectly and I actually understand why too.
     
  13. May 25, 2017 #12
    In the immortal words of Basil Fawlty, "A satisfied customer! We should have him stuffed!"
     
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