- #1
Wimpalot
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Homework Statement
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
Homework Equations
P=eσAT4
Q = mCΔT
ρ = m/V
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?