# Homework Help: Cooling of a sphere by radiation

1. Sep 10, 2010

### fluidistic

1. The problem statement, all variables and given/known data

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.

2. Sep 11, 2010

### 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 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.

3. Sep 11, 2010

### fluidistic

Thank you very much. Impressive.

4. Nov 29, 2011

### 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 wich is by far different from my previous answer. Can someone explain me what's wrong?
I used $c=\frac{380J}{kg K}$.

5. Nov 30, 2011

### 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 $\frac{3 k_B dT}{2}$.

6. Nov 30, 2011

### 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.

Last edited: Nov 30, 2011