Problem involving thermal radiation and specific heat

[Wimpalot]

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 T_c=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 * (T_F - 323)
T_F = 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?

 

[mjc123]

If the temperature changes that much, P will not be constant. Equate the instantaneous value of P to -mC_pdT/dt and solve the differential equation.

 

[Wimpalot]

If the temperature changes that much, P will not be constant. Equate the instantaneous value of P to -mC_pdT/dt and solve the differential equation.

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)?

 

[mjc123]

Write (eσA/mC) dt = -1/T⁴ dT and integrate both sides.

 

[Wimpalot]

Write (eσA/mC) dt = -1/T⁴ dT and integrate both sides.

I got -7.79°C which is still wrong apparently

 

[mjc123]

Oh. I got -44°C. How did you do the integration?

 

[Wimpalot]

Oh. I got -44°C. How did you do the integration?

I got:
5.95*10^-12 * t = 1/3T³
And then I substituted t = 3000 and solved for T

 

[haruspex]

I got:
5.95*10^-12 * t = 1/3T³
And then I substituted t = 3000 and solved for T

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.)

 

[Wimpalot]

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.)

I am not entirely sure. Is it the total temperature difference? That doesn't sound right...?

 

[mjc123]

It's a definite integral. The solution (assuming your arithmetic is right) is
[5.95*10^-12 * t]₀³⁰⁰⁰ = [1/(3T³)]₃₂₃^Tf

 

[Wimpalot]

It's a definite integral. The solution (assuming your arithmetic is right) is
[5.95*10^-12 * t]₀³⁰⁰⁰ = [1/(3T³)]₃₂₃^Tf

Thank you that worked perfectly and I actually understand why too.

 

[mjc123]

In the immortal words of Basil Fawlty, "A satisfied customer! We should have him stuffed!"