# Radiation, Heat and Matter

1. Jun 2, 2009

### treddie

Hello, and thank you for taking the time to read my question.

I know this is a posting with multiple questions, but I would be perfectly fine if instead of answering all of the questions, someone could steer me to a book or online sources of information that cover these questions. So far, I have been unable to find any online sources that do this.

Suppose that a hollow box is made of perfectly insulating sides so that no heat can escape. Inside the box is a gas at some pressure and say 0deg C. Also in the box is a heating element shaped like a 2-dimensional square, where only one side emits radiation. I have two questions about this setup. If I say that the heating element is radiating energy at 100 deg C, as measured by a temperature sensing "gun", then does this mean that every particle of the heating element is radiating at 100 deg, or does it mean that OVERALL, the heating element is at 100 deg? In other words, if I were to take the heating element and cut it into 10 smaller squares, would they each radiate 10 deg of heat, or each 100 deg of heat?

My second question is that since the heater is emitting radiation constantly at 100 deg, and once all of the atoms of the gas in the box all reach equilibrium at 100 deg, does this mean that the heat radiated from the heater can no longer raise the temperature of the gas above 100 deg? In other words, are the photons of the heat radiation simply reflected off of the atoms of the gas, (not absorbed/re-emitted). If not where do the photons go when the box is perfectly insulated? They are increasing in number, are they not?

Thank you very much,
Tom Reddie

2. Jun 2, 2009

### Andy Resnick

It sounds like you are essentially describing a blackbody cavity within the cavity of another blackbody.

For your first question, if the heating element is radiating power equivalent to that of a 100 °C blackbody, then by subdividing the heating element into smaller units, each unit will continue to radiate in a manner consistent with a 100 °C blackbody.

I don't exactly understand your second question. It sounds like you are wondering what happens as equilibrium is reached. In equilibrium, the interior object will no longer be able to transfer energy to the surroundings, essentially becoming invisible.

3. Jun 2, 2009

### treddie

Your answers were perfect. And yes, a perfect blackbox. I actually "knew" the answer to the first question, but when I started thinking about the consequences, things wouldn't add up. Your response to the second question, partly clarified matters. I guess, simply put, the heater no longer becomes a heater when equilibrium is reached. But this is where it gets confusing. If it is an electric heater, then electrical resistance produces heat. But if the heater and everything in the entire box are in thermal equilibrium, should not the heater still be radiating energy in excess of what is already inside the box? And if there were a second heating element in the box that was turned on when the box temperature was 100deg, and producing 200 deg of heat, then before thermal equilibrium is reached, should not the original heater become instead a heat sink, until equilibrium is reached? If so, how can the first heater, which is still turned on, not be radiating energy?

4. Jun 3, 2009

### Mapes

There are different types of heaters; some easily maintain a constant temperature under changing external conditions, and some do not.

A heater that pumps 100°C water through coils, for example, will bring the insulated box to 100°C and keep it there in equilibrium. Until the box heats to 100°C, the water exiting the coils will be cooler; afterward, the exit temperature will be 100°C because no energy was removed.

A heater that dissipates constant electrical power, however, will not stay at 100°C as the temperature of the environment changes. As long as current flows, it will be hotter than the rest of the box. For a perfectly insulated box, the heater temperature will climb towards infinity. (In reality, of course, the box would leak some energy and would eventually reach equilibrium at some large temperature >>100°C.)

On the other hand, if you regulate the heater's electrical current through a control system to keep its temperature at 100°C, then the box will reach equilibrium at 100°C as in the first case, and you'll find that the necessary input current will decrease to zero.

5. Jun 3, 2009

### treddie

I see your points, but now as I try to look at this whole thing at an atomic level, things get strange.

First off, I want to correct myself when earlier I confirmed that the box was a blackbody. It would actually be a whitebody...a perfect reflector...zero absorption.

I guess my confusion is over the concepts of conduction, radiation and vibration. For instance it seems that conduction is nothing more than one atom colliding with another and transferring momentum in the process. In this way, the particles of the gas in the box move faster at higher temperatures and collide at a higher rate than when at a lower temperature. But heat does not seem to be synonymous with linear speed. For instance, heat is often described as the vibration of the particles in a system. Also, a particle cannot know how fast it is moving, therefore linear speed is not the heat in the system...it is merely a consequence of the heat. However, in a collision, I can imagine that not only would there be a transfer of momentum, but that the particles in the collision would ring like tiny bells as a result of the collision, and that THIS is perhaps, the heat. If so, then in order to conserve energy, part of the collision energy is transferred into linear speed, and the rest as vibration in the particles (heat).

But I think the vibration analogy is only partially applicable for two reasons:
1. A physical bell (or any macroscopic object for that matter), can vibrate at an infinite number of different modes simultaneously. But an atom is restricted, because,
2. The "vibration" per say would just be the time it takes for the affected electrons to return to their ground state orbitals, correct?

If this is correct, then any atom in isolation and in collision with a fixed number of photons, can not radiate heat any longer than one cycle of vibration for each one of the electrons that absorbed a photon and re-emitted it, and the "frequency" of any one cycle would depend on how high the electron was kicked out of its ground state. Therefore, the time it would therefore take an isolated atom to completely lose its heat would be the frequency of the lowest energy photon involved. I am assuming that the time it takes an electron to be kicked up and then settle down equals the frequency of the photon .

But the gas in the box is not an isolated atom...it is an incredibly HUGE amount of atoms all colliding with each other and constantly transferring momentum and heat, and therefore constantly having their electrons kicked out of their ground states and back again, over and over. So it seems that conduction is not as efficient as pure radiation in transferring heat because during conduction, part of the energy of the collision is "wasted" as transferred linear momentum. But also it seems that no conductive transfer can EVER be without radiative heat transfer as well...each heated atom is in the process of trying to return to its ground state and therefore is constantly radiating energy away as released photons.

So I offer another hypothetical situation: Suppose that the heater in the system is a pipe of flowing, heated water. But imagine that the pipe itself acts like a perfect check valve or diode...it only allows heat to pass one way...into the box. In this situation, would not the temperature in the box slowly rise to infinity, since radiation of the heat of the water molecules is constantly being added to the interior of the box? This part bugs me. If the energy of the radiated photons is equivalent to 100deg of heat per photon, then these photons cannot cause any of the atoms of the gas to vibrate (go from ground state to higher state and back again) any faster than the frequency of these photons. Hence, once all of the atoms of the gas have reached 100deg, all of the interactions are still only 100deg interactions and the heat should stay constant at 100deg. I can only assume then that the growing amount of energy in the box is transferred into linear momentum (the linear speed of all the atoms of the gas).

But maybe this is the solution.

Since linear momentum is constantly increasing, a healthy portion of this overall increasing momentum is being transferred into higher heat values with each new collision. In other words, higher collision energies means part of that collision energy will be portioned out as higher linear speed, and the rest as higher energy photons (higher frequency photons, meaning deeper penetrations into the electron shells of the atoms by these photons, meaning that the deeper electrons will be kicked up out of THEIR ground states into higher orbitals and drop back down again. And since the energy of these deeper ground states is higher, then like stronger rubber-bands, these electrons will snap back quicker than ground state electrons in higher orbitals. Walla, higher heat).

Or as best as I can surmise!

6. Jun 4, 2009

### treddie

So can anyone tell me if I am at least REMOTELY close to the truth?