Cooling of a sphere by radiation

[fluidistic]

Homework Statement

A copper sphere of radius d is put inside walls very close to 0K. The initial temperature of the sphere is T_0. The surface of the sphere is totally black and we assume it loses energy only due to radiation.
1)Find an expression for the time passed till the temperature of the sphere decreases by a factor \eta.
2)Calculate explicitly t when \eta =2 and T_0 =300 K.

2. The attempt at a solution
I've found out http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html#c2.
So I searched for all the extra data I haven't been provided in the exercise, namely for the molar mass of copper, its density and the value of all the constants in the formula.
For part 1) I get t_{\eta}=\frac{mN_Ak_B (\eta ^3 -1)}{8M \sigma \pi d^2 T_0^3}.
For part 2), I used part 1) and plugged \eta =2, T_0=300 and I obtained t=891445d.
In all my arithmetics, I always used the SI units. So if d=1m, I get a time of 891445 s>10 \text{days} which seems to me way too long.
Further, if I double the radius of the sphere, it only double the time of cooling which doesn't seem right to me (I'd expect a 4 times bigger time of cooling), but I'm not 100% sure.
Am I doing something wrong? If not, what answer do you get?
P.S.: I used the density 8940 kg/m^3 and the molar mass M equal to 0.06354kg/mol.

 

[Filip Larsen]

For what its worth, I tried to derive the formula from scratch using constant heat capacity, and I got 11,6 days with your values.

if I double the radius of the sphere, it only double the time of cooling which doesn't seem right to me

If you double the radius you get eight times the heat content passing through four times the area, so I'd say you should expect it to take twice as long.

 

[fluidistic]

For what its worth, I tried to derive the formula from scratch using constant heat capacity, and I got 11,6 days with your values.



If you double the radius you get eight times the heat content passing through four times the area, so I'd say you should expect it to take twice as long.

Thank you very much. Impressive.

 

[fluidistic]

I just redid this problem now, but instead of taking E=\frac{3k_BT}{2}, I took \Delta E =mc \Delta T.
I reach a cooling time of t_f = \frac{mc}{\sigma A} \left ( \frac{1}{T_0}-\frac{1}{T_f} \right ).
I checked out the units, they are fine. This gives me a time of 2498.8 d days which is by far different from my previous answer. Can someone explain me what's wrong?
I used c=\frac{380J}{kg K}.

 

[fluidistic]

Here's my work in details:
\Delta Q=mc\Delta T=\Delta E \Rightarrow dE=mcdT.
\frac{dE}{dt}=P=\varepsilon A\sigma (T_{sphere}-T_{ambient})^4=A\sigma T_{sphere} ^4.
\frac{dE}{dt}=\frac{dE}{dT} \cdot \frac{dT}{dt}=mc \frac{dT}{dt}= A \sigma T_{sphere} ^4 \Rightarrow dt= \frac{dTmc}{\sigma A T_{sphere}^4}.
Now I integrate both sides: \int _{t_0}^{t_f}=t_f=\int_{T_0}^{T_f} \frac{mcdT}{\sigma A T_{sphere}}=\frac{mc}{\sigma A} \left ( \frac{1}{T_0^3} - \frac{1}{T_f^3} \right ).
With A=4 \pi d^2. m=\frac{4\pi d^3}{3} \cdot 8940 \frac{kg}{m^3}=37447.8 kg \frac{d^3}{m^3}. c=\frac{380J}{kgK}.
\Rightarrow t_f= \left ( \frac{1}{(300K)^3}- \frac{1}{(150K)^3} \right ) \left ( \frac{37447.8d^3 \cdot 380 J}{m^3 \cdot kg \cdot K} \right ) \left ( \frac{m^2K^4}{4\pi d^2 \cdot 5.67 \times 10 ^{-8}W} \right )=\frac{2.16 \times 10^8s\cdot d}{m}.
Which is more than 200 times bigger than when I considered the change of internal energy as \frac{3 k_B dT}{2}.

 

[fluidistic]

I've redone it once again, this time I get a way too low cooling time.
In function of \eta, I get t_f=\frac{d}{3} \cdot \frac{380J}{kg\cdot K}\cdot \frac{1-\eta ^3}{(300K)^3 \sigma}. For \eta =2, I reach t_f=-d579.18 s. I don't understand why there's a negative time and why the value obtained is so low.
Basically the integral I solved was t_f =\frac{mc}{A \sigma} \left [ - \frac{1}{T^3} \right ] ^{T_0/\eta }_{T_0}=\frac{mc (1-\eta ^3)}{A \sigma T_0 ^3}.

If you can solve the problem by considering \Delta E =mc \Delta T, and you get around 10 days mutiplied by the radius of the sphere, please let me know how you solved the problem. It seems like I'm going crazy on this one.