# Black body balloon in vacuum [closed]

The problem statement, all variables and given/known data

There is a perfectly spherical balloon with surface painted black. It is placed in a perfect vacuum. It is gently inflated with an ideal mono-atomic gas at Kelvin temperature $T_i$, slowly enough so that it reaches thermal equilibrium with the gas, and then it is sealed off. It has radius $r_i$ at this time and contains N atoms. The vacuum is large, so radiation from its walls can be ignored.

a) Show that $\frac{T}{T_i}= \left(\frac{r}{r_i}\right)^3$ if the pressure inside the balloon is independent of its radius.

b) How much energy does the balloon radiate per second when it is at radius $r$? Express your answer in terms of $r$ and constants.

c) What is the rate of change of the internal energy of the gas? Express you answer in terms of $r$, $r˙$, and constants.

Relevant equations:

$$PV = nRT$$

$$J * A = σT^4 (4\pi r^2)$$

The attempt at a solution:

I already showed a), so I don't need help with that. In part b), I wrote pretty much $J * A = σT^4 (4\pi r^2)$ but I am unsure how to get a speed of radiation. I also have no idea how to do part c), so any help on parts b) or c) is welcome. Thanks!

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## closed as too localized by Qmechanic♦Feb 15 '13 at 15:39

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I don't know what J*A is, so i'm not exactly sure which form of Stefan's law you're using. I think that if you use the following form, you should be fine: $$\frac{\mathrm dQ_b}{\mathrm dt}=A\sigma T^4$$ Here, $\mathrm Q_b$ is the radiant energy emitted by a perfectly black body in a small time $\mathrm dt$, so $\frac{\mathrm dQ_b}{\mathrm dt}$ is the rate of emission of energy. $A$ is the surface area of balloon.
Just out of interest, Is $J$ some sort of heat-current-density? It looks like emissive power to me.. –  Manishearth Mar 13 '12 at 5:15
@EricMercer: By first law of thermodynamics, $Q=U+w$. $Q$ is heat, $U$ is internal energy. $w$ is work, but it is zero here because its a vacuum. So here, heat lost=decrease in internal energy. So they're the same here. –  Manishearth Mar 13 '12 at 5:28
I guess what you did for b is correct. $dQ/dt$ is the amount of energy radiated per second. For c, since $Q=U$, their corresponding rates are the same as well (there may be a $\pm$, though). –  Manishearth Mar 13 '12 at 5:29
If it helps, the $Q$ used in the first law of thermoD is $-Q_b$ here (since I took energy emitted) –  Manishearth Mar 13 '12 at 5:31