1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    In the immortal words of Basil Fawlty, "A satisfied customer! We should have him stuffed!"
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Problem involving thermal radiation and specific heat
  1. Radiation problem (Replies: 2)