Apparently the concept of a thought experiment is beyond you.

]]>Brad: One surface vs two can be tricky. Imagine cutting a 2x2x2 cube into eight 1x1x1 cubes. The surface area doubles and so does the total emission. Where does the extra power come from? The temperature is still the same.

The extra power comes from twice as much internal energy being converted to radiation – its temperature falls twice as fast …. but only in intrastellar space. Everywhere else, there is a local temperature, local thermal radiation field and twice as much surface area to absorb it. So the temperature may not fall. Twice as much power could be exchanged by radiation, but not necessarily any “heat”.

Now consider reassembling the eight cubes into one big one. Then consider the difference between a solid sphere (with one surface) and a hollow sphere with two). In all cases, you need to consider both outgoing and incoming radiation at each surface.

This works well until you slice objects so thin they are semi-transparent and no longer behave like blackbodies. This happens when you consider layers of atmosphere. “Optically thick” layers by definition emit blackbody radiation, but our atmosphere isn’t optically thick at all wavelengths and eventually becomes optically thin at all wavelengths if you go high enough. “Optically thin” layers modify the radiation passing through them. Those modifications add up. That is why we use the Schwarzschild eqn for radiative transfer through the atmosphere, but the S-B equation elsewhere. The S-B equation is only appropriate for radiation in equilibrium (via repeated absorption and emission) with the medium it is passing through. In classical experiments, a black cavity with a pinhole was used to create such an equilibrium. When radiation is not in equilibrium with its surroundings (semi-transparency, for example), you need the Schwarzschild equation.

]]>Ah, yes. The temperature would just be equivalent to the emission from one side.

As you said DeWitt, the incoming flux would be double the emitted flux in the green plate experiment. So my incorrect thinking was that the flux arriving to the inside of the shell needed to support emission from both sides. My reasoning was that I was interpreting sb to be emission from a surface, regardless the direction the surface is facing. Sb though is for emission from a body. Incoming flux to the shell wouldn’t need to be double, as it was on the green plate experiment, since only the emission on the outside shell is “from the body”.

I think I need to stick to reading through that MIT textbook for now.

]]>Brad,

It’s the surface area that doubles when you consider both sides, not the radiant flux density in W/m². If only one side is illuminated, then the input flux density has to be twice the emission flux density at steady state.

]]>In the green plate effect on Eliās site the plates had to have a temperature equivalent to double the emission from a single side

What? No! They have a temperature equivalent to the emission of **one** side. Not double.

I had this realization today:

“The StefanāBoltzmann law describes the power radiated **from a black body** in terms of its temperature.” …emphasis added

I’ve read this plenty of times and have failed to fully grasp the …from a body….. I’ve been thinking of it as emission away from a surface.

In the green plate effect on Eli’s site the plates had to have a temperature equivalent to double the emission from a single side. That’s obviously a different case since the emission away from a plane can’t be absorbed by the same object. I realize everyone that has responded understands this very clearly, just sharing in case it is a stumbling block for someone else that passes through the blog.

]]>Brad wrote: “my mind it simply wasnāt making sense why emission from shell wall to shell wall could conceivably count as power into the shell.”

My comment was held up in moderation and so missed the back and forth.

If you go back to the problem posed by this post, the 3 m of rubber provides a lot of insulation. The outside temperature is 133 K and the inside 423 K. That is roughly a 100-fold difference in T^4. So the internal transfer of energy from one location on the inner surface of the sphere to another is very large – but no heat transfer is involved since the net flux is zero (by symmetry).

SOD cleverly didn’t give his internal heat source a definite size and therefore didn’t need to get bogged down in details of viewing angles, the “shadow” cast by the central power source, nor energy returning to the central power source (:)) It is an elegant teaching example as written, but one that is much more challenging (and therefore less clear) if the central power source has a defined size. The smaller the power source gets, the hotter it must be to radiate 30,000 W.

Many commenters above were outraged about how much power was traveling inside the sphere. Where did it come from? If you imagine starting the system up from 0 K, negligible energy is escaping through the rubber sphere for a long time. Insulation keeps that energy inside – it builds up because it can’t escape until there is a large temperature gradient. Power * time is energy – the energy that is going nowhere when being radiated from spot on the inner surface to another.

]]>Cancel that last question. I see now that is really the initial point of my misunderstandings.

]]>DeWitt,

From your example:

“That means that the shell is emitting 25W/mĀ² and has a temperature of 144.9K. So we plug that into the heat transfer equation

100 = 0.0000000567(T1^4 -144.9^4)

T1 = (100/0.0000000567 + 144.9^4)^0.25

T1 = 216.68K for an emission of 125W/mĀ² not 200 W/mĀ²

I agree that emission on the outside of the sphere must equal 25w/m^2 for the given radii ratio of .5. But that would mean the total emission of the shell is 50, correct. Solving that I got a shell temp of 172 and surface temp of 227.

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