# Thermodynamics: Concentric Spherical Shells with Point Source Radiation

## Homework Statement

There is a perfectly absorbing spherical shell with radius R1 suspended in space. Inside is a smaller spherical shell with radius R2. Inside that shell is a ball of radius R3. All three objects are concentric. In the center of the ball is a point source radiation with power W.
Find the temperatures T1, T2, T3 of the outer shell, inner shell, and sphere. The three objects are in thermal equilibrium.

W = σT4(4piR2)

## The Attempt at a Solution

I wondered at first if the temperature of the outer shell
T1 = (W/σ4piR12)$\frac{1}{4}$

But my question is if the sphere in the middle contribute to the total power received by the outer shell? As in, the sphere emits some power, the inner shell absorbs it and emits some power, and the outer shell soaks up both the power from the sphere and the inner shell?
Then what about the inner shell? Does it get the power from the sphere and the outer shell? Or just the sphere?
what about the sphere? Does its temperature depend just on itself? Or is it rather like the greenhouse effect where the sphere resembles the planet and the shells trap radiation and increase the temperature of the sphere itself?

Help appreciated!

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Everybody absorbs heat as well as radiates it at the same time. It is just which of the two has its rates higher. In thermal equilibrium , when all the bodies have definite temperatures, they absorb and radiate same amount of energy , and that is why their temperatures are constant.

I am sorry, I don't understand. Are you saying that all three objects absorb and radiate the heat from each other? So are you saying that they have the same temperature?

No they don't have the same temperature but your first line is true.

I am really confused, can you clarify?
The temperature of the outer shell isn't just T1 = (W/σ4piR^2)^(1/4) because it also absorbs more heat from the inner shell? So how do we find out how much heat the inner shell radiates? Doesn't that also depend on the heat of the outer shell?

Probably too much of a hint:

The problem is steady state. Therefore the energy that is emitted from the ball leaves the outer shell. Since the environment is space, you can write an explicit expression for the temperature of the outer shell, sphere 1, in terms of emissivity, S-B constant, area, and power. This provides an explicit expression for T1.

Once again, since the situation is steady state, all the energy that is radiated from sphere 1 to space must be received from sphere 2. Write an expression for the radiation exchange between sphere 2 and sphere 1. It will contain temperatures, emissivities, shape factors, S-B constant, and areas. But T1 is known; you have an explicit expression now for T2.

Etc, etc.

Probably too much of a hint:

The problem is steady state. Therefore the energy that is emitted from the ball leaves the outer shell. Since the environment is space, you can write an explicit expression for the temperature of the outer shell, sphere 1, in terms of emissivity, S-B constant, area, and power. This provides an explicit expression for T1.

Once again, since the situation is steady state, all the energy that is radiated from sphere 1 to space must be received from sphere 2. Write an expression for the radiation exchange between sphere 2 and sphere 1. It will contain temperatures, emissivities, shape factors, S-B constant, and areas. But T1 is known; you have an explicit expression now for T2.

Etc, etc.
So:
T1 = (W/σ4piR12)1/4
and:
T2 = (W/σ4piR22)1/4 ?

The problem stated that the outer shell is a perfect absorbing medium. That means that it absorbs all radiation that reaches it. None is reflected. But that says nothing about its emissivity. Take for instance white paint. Its absorbtivity is less than 0.3 yet its emissivity is over 0.9. Absorptivity and emissivity are both functions of temperature/wavelength. In this problem I suspect you can assume that absorptivity is unity for all 3 shells. I am not sure what to assume about emissivities. You are the decider of that. If emisivities are unity, the work is less messy.