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There seems to be different explanations of how microwave ovens work. Here are the two basic forms of explanation I've heard:

A. As microwaves pass through, the molecule dipoles try to align with the electric field, causing rotation and increased kinetic energy (= heating)

B. As microwaves hit the food, it is absorbed by molecules/atoms, the photons converting to kinetic energy (= heating)

In A it makes it sound like the microwaves are not absorbed, but in B they are absorbed.

Understanding microwaves as consisting of photons, and so packets of energy, B makes more sense to me (perhaps with the nuance that it so happens that the Kinetic Energy of B is in the form of dipole alignment). A seems to ignore the idea of photons and quantized energy, and conservation of energy.

However, I'm wondering if the difference is actually because one way explanation is using quantum physics (B), and the other explanation is not (A). Or is there something else I am misunderstanding?

Thanks!

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The two explanations are both correct. Explanation A automatically implies absorption of the microwaves, by conservation of energy. Explanation B is fine, but it doesn't need to invoke photons. We can talk about the microwaves as classical waves that have energy and get absorbed.

We can also, optionally, talk about photons, but this is unnecessary. A microwave oven will heat anything with polar molecules in it. It isn't tuned to a resonance of the water molecule, and in any case water in the liquid phase will not show sharp absorption lines as would water molecules in the gas phase.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Nov 25 '19 at 17:40
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@Ben Crowell answer is correct 👍. The following is only intended to expand upon his points.

A. As microwaves pass through, the molecule dipoles try to align with the electric field, causing rotation and increased kinetic energy (= heating)....In A it makes it sound like the microwaves are not absorbed..

In A the electric field does work in causing the dipole molecules to rotate and acquire rotational kinetic energy. That is correct.

But in this classical wave model energy is transferred from the electric field to the molecule dipoles in the form of work. The field exerts an alternating force on the dipole causing it to align with the field. It is technically a work transfer and not a heat transfer. Heat is energy transfer due solely to temperature difference. As the "temperature" of microwaves (e.g. cosmic background radiation) is only a few degrees Kelvin, there is no heat transfer from microwaves to the food. This is why it is a misnomer to speak of microwave "heating". In the classical wave model the increase in temperature of the food being cooked is not due to heat transfer, but due to work transfer done by an alternating electric field giving the dipoles rotational kinetic energy.

B. As microwaves hit the food, it is absorbed by molecules/atoms, the photons converting to kinetic energy (= heating)...but in B they are absorbed.

In the photon (particle) model, the quantum energy of microwave photons is in the range 0.00001 to 0.001 eV which is in the range of energies separating the quantum states of molecular rotation and torsion. So in this model you can view the transfer of microwave electromagnetic energy in the form of photons absorbed by the molecule dipoles causing them to undergo rotation and torsion, resulting in an increase in the temperature of the food being cooked. But once again it is not a "heating" process.

The end result of both the wave and particle is the same. Whether viewed as a work transfer due to the force exerted by the electric field of the wave and a molecular dipole, or the absorption of a quanta of energy in the molecular quantum state separating molecular rotation and torsion, in either case the molecular dipoles acquire rotational kinetic energy that becomes randomized as an increase in translational kinetic energy, and thus an increase in the temperature of the food.

In concluding, from a thermodynamics perspective, it is important to note that microwave cooking is not a "heating" process, that is, it is not the result of energy transfer due to a temperature difference between the microwave source and the food being cooked.

Thanks for the extra info. Part of my confusion is that sometimes the two explanations seem to be presented as two different ways that can both happen, but I think it is just the same way described from two different models. Is that correct?

Yes, in one case it is in terms of the energy transfer from a wave. In the other in terms of a quanta of energy absorbed by a photon. End result: the same.

Also, isn't it still heating by radiation? Or is not radiation either because it is not a hotter object radiating heat?

It is definitely NOT "heating" by radiation. I cannot emphasize enough that heat is defined as energy transfer solely due to temperature difference. It is not a temperature difference between the microwave energy source and the food that causes energy transfer to the food. As you correctly noted, microwaves are not "hotter objects" radiating heat. As I already noted, the "temperature" of microwave energy is only a few degrees Kelvin.

I'm thinking that IR or Microwave radiation could be working similarly, but then maybe I'm mixing models again...! –

What you are mixing up is that different frequencies of electromagnetic radiation interact with matter in different ways. IR interacts with matter in an entirely different manner than microwaves as does other frequencies of EM waves. For a good overall explanation of the different ways that radiation interacts with matter, look at the Hyperphysics web site.

IR is the primary mode of cooking for broiling in ovens and it is indeed cooking by heat transfer because the temperature of the IR source is way higher than the microwave source. In addition, the photon energy level of an infrared electromagnetic wave is between 0.001 and 1.7 eV, or about three orders of magnitude greater than the microwave photon. It corresponds to the molecular vibration quantum state.

The temperature of the infrared heating element in the oven is typically around 260 - 290 C. Food put under the broiler at room temperature is around 25 C. Because of the temperature difference, heat transfers from the broiling element to the food by thermal radiation. It is an entirely different mode than that for microwaves.

Hope this helps.

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  • $\begingroup$ Hi Bob D. Thanks for the extra info. Part of my confusion is that sometimes the two explanations seem to be presented as two different ways that can both happen, but I think it is just the same way described from two different models. Is that correct? Also, isn't it still heating by radiation? Or is not radiation either because it is not a hotter object radiating heat? I'm thinking that IR or Microwave radiation could be working similarly, but then maybe I'm mixing models again...! $\endgroup$ – cjphysics Nov 25 '19 at 2:03
  • $\begingroup$ @cjphysics I have updated my answer to respond to your follow up questions. Hope it helps. $\endgroup$ – Bob D Nov 25 '19 at 2:42
  • $\begingroup$ Would you also consider burning something with a high-power laser not to be "heating", as the laser light is also coherent and thus has very little "heat" (i.e. entropy)? In both cases, we have electromagnetic radiation being absorbed by matter and transferring energy to it, with that energy being primarily converted to heat, just like in ordinary radiative heating. The coherence (or lack of it) of the radiation being absorbed would seem to make very little difference at the macroscopic level. $\endgroup$ – Ilmari Karonen Nov 25 '19 at 8:42
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    $\begingroup$ I entirely agree with Bob D's insistence that microwaves don't supply heat to the food. The noun 'heat' has a well-established meaning in Physics. The difficulty is that I'm not so sure that the verb 'heat' (or the gerund 'heating') is used so carefully. I might say, "My coffee's gone cold; I'll heat it in the microwave." By *heat it", of course, I'd mean "make it hotter". I think that this usage will be hard to eradicate even among physicists. $\endgroup$ – Philip Wood Nov 25 '19 at 11:34
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    $\begingroup$ @ Ilmari Karonen. Interesting. I think we're alright to say that radiation from the Sun, or from the elements of a toaster, or from a filament lamp conveys heat to an object in its path, because there is a net energy transfer due to a temperature difference. One notes the approximately black body (Planck) spectrum associated with this radiation from a hot source. Suppose we filter out all but a small part of the spectrum leaving approximately monochromatic radiation, no longer identifiable as being from a hot source. Would it still be heat that is being transferred to an object in its path? $\endgroup$ – Philip Wood Nov 25 '19 at 12:52

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