We're told that the photon energy of microwave frequency radiation ($\sim 10^{-5}~\text{eV}$) is not high enough to break hydrogen bonds. But if that's true, how does dielectric heating of water work? Liquid water is a network of polar molecules held together by H bonds so that they CAN'T rotate, in concert with the microwave beam or anything else ….
Seems like this is a problem of the two ways of looking at radiation — classical wave vs photon stream — being incompatible.
Best Answer
In a solid or liquid we have collective vibrations of the whole system. It can be useful to think of these as quasiparticles called phonons, that is when we add vibrational energy to the system as a whole it generates a phonon, or conversely a phonon can decay and emit energy.
Black body radiation is (mostly) the emission of photons from the collective vibrations i.e. the decay of phonons to emit photons, and dielectric heating is the reverse process i.e. the absorption of photons to create phonons. This is what happens in your microwave oven. The heating is due to the excitation of the collective vibrations, not to the interaction of photons with hydrogen bonds. The quanta of these collective excitations (i.e. their phonons) are generally very small so they can absorb photons of even very low energies.
In real materials the collective vibrational modes are anharmonic oscillators so they all interact with each other and the vibrational energy is distributed between them in accordance with the Boltzmann distribution. That means the vibration energy from the absorbed microwave photons is quickly equilibrated with higher energy modes such as the vibrational excitations of hydrogen bonds, and it can break those bonds. This means energy can be absorbed in small units from microwave photons and still build up sufficiently to break the much higher energy hydrogen bonds.