Cavity and coals: brightness and its consequences

[LCSphysicist]

Homework Statement

Cavities formed by red-hot coals appear brighter than the coals themselves. Is the temperature in such cavities appreciably higher than the surface temperature of an exposed glowing coal?

Relevant Equations

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Cavities formed by red-hot coals appear brighter than the coals themselves. Is the temperature in such cavities appreciably higher than the surface temperature of an exposed glowing coal?

At first i tried to interpret the "brightness" as a synonymum of intensity, anyway, a radiation emitted by the cavities can be more concentrate that the coal as a whole, so the cavities's energy density is greater than the coal as a whole, but the temperature, is equal.

What do you think about?

 

[haruspex]

Cavities formed by red-hot coals appear brighter than the coals themselves. Is the temperature in such cavities appreciably higher than the surface temperature of an exposed glowing coal?

At first i tried to interpret the "brightness" as a synonymum of intensity, anyway, a radiation emitted by the cavities can be more concentrate that the coal as a whole, so the cavities's energy density is greater than the coal as a whole, but the temperature, is equal.

What do you think about?

The question is somewhat deceptive.
If it had asked whether you can deduce from that observation that the temperature is higher then, as you note, you cannot.
Or, had it mentioned the (true) fact that the cavities appear more orange/yellow compared to the red of the surface coals, you could correctly deduce that the temperature is higher.
Yet again, had it omitted any such observation and simply asked whether you would expect the temperature to be higher in a cavity then... what do you think?

So I really don't know what the questioner is looking for here.

 

[Steve4Physics]

Cavities with small openings behave as almost ideal black body radiators (emissivity nearly 1 at the opening).
The coal surface has emissivity rather less than one.
If you assume identical temperatures, what can you deduce?

 

[Keith_McClary]

Cavities with small openings behave as almost ideal black body radiators (emissivity nearly 1 at the opening).
The coal surface has emissivity rather less than one.
If you assume identical temperatures, what can you deduce?

A physics demo: a cubic box (~0.5m edges) has a 4cm diameter hole. The hole looks black, but the prof opens the box to show it is white inside.

 

[LCSphysicist]

Assuming identical temperatures e1/a1=e2/a2 the coal absorb less radiation, the cavity more.
You guys are suggesting that, if the temperature were equal, the cavity would need to be dark?

 

[haruspex]

A physics demo: a cubic box (~0.5m edges) has a 4cm diameter hole. The hole looks black, but the prof opens the box to show it is white inside.

I don't see the relevance. Your example is of a light absorber, not a light emitter.

 

[haruspex]

Assuming identical temperatures e1/a1=e2/a2 the coal absorb less radiation, the cavity more.
You guys are suggesting that, if the temperature were equal, the cavity would need to be dark?

I believe @Steve4Physics is arguing that the cavity will appear brighter even if they are at the same temperature. But that just reinforces my first statement in post #2, that you cannot deduce the cavity is at a higher temperature; it does not rule out the possibility that a cavity would tend to be at a higher temperature.

You can safely ignore the irrelevant post #4.

 

[Steve4Physics]

I don't see the relevance. Your example is of a light absorber, not a light emitter.

Sorry, rather long reply coming...

1. Absorption and emission of ‘black-body’ radiation happen simultaneously, all the time, for any object above absolute zero. If the rates of absorption and emission are equal, the object’s temperature is constant (in the absence of any other energy-transfer processes).

2. If an object is a good absorber, then it is an equally good emitter. If an object is a poor absorber, then it is an equally poor emitter. That’s super-important.

The classic illustrative experiment is:

a) Place a cold black tin can in the sunlight (sunlight is black-body radiation with a temperature of ~6000K). The can quickly heats up (good absorber). Put the hot can in the dark and it quickly cools down (good emitter). (Other heat transfer processes are occurring so the can doesn’t reach 6000K while heating up!)

b) Place a cold shiny tin can in the sunlight. The can slowly heats up (poor absorber). Put the warm can in the dark and it slowly cools down (poor emitter).

3. Consider a small opening to an object with a cavity. If you look at the hole it appears black. The opening acts as a (near perfect) ‘black-body’ absorber. This is true even if the inside of the object is white or even mirrored. That’s the point of post #4.

4. Conversely, if you heat the object (say to a red-hot temperature) the small opening to the cavity acts as a (near perfect) back body emitter. Having a small opening in a furnace is how you can produce black-body radiation for experimental work. This is what’s happening with the coals in the original question.

4. The cavity between the coals will be ‘extra dark’ when the fire is not lit, and ‘extra bright’ when the coals are burning.

 

[haruspex]

Sorry, rather long reply coming...

1. Absorption and emission of ‘black-body’ radiation happen simultaneously, all the time, for any object above absolute zero. If the rates of absorption and emission are equal, the object’s temperature is constant (in the absence of any other energy-transfer processes).

2. If an object is a good absorber, then it is an equally good emitter. If an object is a poor absorber, then it is an equally poor emitter. That’s super-important.

The classic illustrative experiment is:

a) Place a cold black tin can in the sunlight (sunlight is black-body radiation with a temperature of ~6000K). The can quickly heats up (good absorber). Put the hot can in the dark and it quickly cools down (good emitter). (Other heat transfer processes are occurring so the can doesn’t reach 6000K while heating up!)

b) Place a cold shiny tin can in the sunlight. The can slowly heats up (poor absorber). Put the warm can in the dark and it slowly cools down (poor emitter).

3. Consider a small opening to an object with a cavity. If you look at the hole it appears black. The opening acts as a (near perfect) ‘black-body’ absorber. This is true even if the inside of the object is white or even mirrored. That’s the point of post #4.

4. Conversely, if you heat the object (say to a red-hot temperature) the small opening to the cavity acts as a (near perfect) back body emitter. Having a small opening in a furnace is how you can produce black-body radiation for experimental work. This is what’s happening with the coals in the original question.

4. The cavity between the coals will be ‘extra dark’ when the fire is not lit, and ‘extra bright’ when the coals are burning.

So you are saying the absorption case is relevant because it is simply the inverse: the cavity is dark in that case for the same reason it is bright in the coals case.

Then my response is the same as to your post #3, namely, what I wrote in post #2.

 

[Steve4Physics]

... simply the inverse: the cavity is dark in that case for the same reason it is bright in the coals case.

Hi. Yes indeed. Also, as you noted in #2, the original question is very unclear because (for example) the outer surface temperature of the coal could be quite different to the internal cavity temperature. However I've seen this question (better phrased) before so recognised what it was getting at!