# Question with energy density

1. Jan 16, 2009

### Feldoh

1. The problem statement, all variables and given/known data
The density of the energy of radiation in a cavity at temperature T is $$u(T) = aT^4$$. Suppose the cavity is a sphere whose radius increases at a rate of:

$$\frac{dr}{dt} = v_0$$

Assuming that no energy enters or leaves the enclosure, will the temperature increase or decrease, and if so, at what rate?

2. Relevant equations
All given in the problem, I believe.

3. The attempt at a solution
I was thinking that intuitively as the sphere gets bigger the temperature would have to go down because the distribution of radiation would be more spread throughout, but then again that's just a guess at best.

I was thinking that u(T) = E/V since it's just energy density, but I can't really much get much farther than that. I also thought that perhaps obtaining $$\frac{dV}{dt} = 4\pi*r^2*v_0$$ might be useful later on but I don't know how.

Last edited: Jan 16, 2009
2. Jan 17, 2009

### CompuChip

I don't know if this is too easy, but you said: u(T) = E / V because u is an energy density. What happens to E and V according to the question? Then what happens to u(T)? So what happens to T?

3. Jan 18, 2009

### Feldoh

Yeah well the answer is simple but I'm not sure that Stefan–Boltzmann law, (Which is what I think this is) is energy per volume. But assuming that it is:

$$u(T) = \frac{3E}{4 \pi r^3}$$

$$\frac{du}{dr} = \frac{-9E v_0}{4 \pi r^4}$$

So the change in energy density is negative for all r such that r is a positive real, which means that the temperature is decreasing as the radius is increasing.

Does this sound correct?

Last edited: Jan 19, 2009