Temperature Change in a Radiating Sphere with Increasing Radius

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SUMMARY

The discussion centers on the temperature change in a radiating sphere as its radius increases at a constant rate, defined by \(\frac{dr}{dt} = v_0\). The energy density of radiation is given by \(u(T) = aT^4\). As the radius increases, the energy density decreases, leading to a decrease in temperature. The derived relationship shows that \(\frac{du}{dr} = \frac{-9E v_0}{4 \pi r^4}\) confirms that temperature decreases as the radius increases.

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  • Familiarity with energy density concepts in thermodynamics
  • Basic calculus for differentiation and rates of change
  • Knowledge of spherical geometry and volume calculations
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Feldoh
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Homework Statement


The density of the energy of radiation in a cavity at temperature T is [tex]u(T) = aT^4[/tex]. Suppose the cavity is a sphere whose radius increases at a rate of:

[tex]\frac{dr}{dt} = v_0[/tex]

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

Homework Equations


All given in the problem, I believe.

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 [tex]\frac{dV}{dt} = 4\pi*r^2*v_0[/tex] might be useful later on but I don't know how.
 
Last edited:
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Feldoh said:

Homework Statement


The density of the energy of radiation in a cavity at temperature T is [tex]u(T) = aT^4[/tex]. Suppose the cavity is a sphere whose radius increases at a rate of:

[tex]\frac{dr}{dt} = v_0[/tex]

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


Homework Equations


All given in the problem, I believe.



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 [tex]\frac{dV}{dt} = 4\pi*r^2*v_0[/tex] might be useful later on but I don't know how.

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?
 
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:

[tex]u(T) = \frac{3E}{4 \pi r^3}[/tex]

[tex]\frac{du}{dr} = \frac{-9E v_0}{4 \pi r^4}[/tex]

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:

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