How is thermal radiation transformed into KE in a body (atom/lattice)? Suppose a piece of metal at 0 C° is exposed to thermal radiation (350 °K or thereabouts, EMR = 7 x 10^12 Hz)),  its temp  will raise , they say, because the KE of its atoms (or of the lattice as a whole) will increase. But how is the incoming radiation trnsformed into KE?
The only way I can think of is scattering, but there are a couple of obscure points I hope you can clarify:


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*1) since the energy of the incoming photon is small, an electron (and even less a proton, but I learned [Does a nucleus in a lattice emit thermal radiation? that nuclei play no part in this process) will get only a tiny fraction of its energy and could never reach the above temp (350 °K). Moreover I suppose that only free electrons can interact, since orbiting or bonded electrons cannot be perturbed, is that right?

*2) in the linked answer some comments by a distinguished member specify that atoms do not react individually and the whole lattice and particularly phonons are involved. Now, can you explain how/ according to what formula/ rule/principle incoming thermal radiation interacts with these or other elements and produces an increase of KE and eventually of the body's temperature?
 A: If you imagine starting with a perfect crystal at absolute zero then we have an array of regularly spaced nuclei surrounded by a sea of electrons. The innermost electrons in the atoms are strongly localised to that atom, but the outermost electrons are delocalised so in effect they spread out over the whole crystal. These electrons aren't necessarily free in the sense that electrons in the conduction band of a metal are free, but they are spread out.
Now bring our perfect crystal up to room temperature or $0$ºC in your example. As we introduce thermal energy the nuclei start to vibrate. I say the nuclei vibrate rather than the atoms vibrate because most of the mass of the atoms is in the nuclei.
Since the geometry of the crystal changes as the nuclei vibrate, and since the electrons respond to the motion of the nuclei, that means the electron density in the sea of delocalised electrons must vary as well. But this means that the electrons must be accelerating to and fro in response to the oscillations, and we know that accelerated charges emit EM radiation. So:


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*the thermal energy makes the nuclei oscillate around their mean position

*the electrons oscillate in response

*the accelerating electrons emit EM radiation
And this is basically what black body radiation is. It's a broad band of frequencies emitted at any temperature above absolute zero due to the thermal motion in the material. Note that it isn't associated with any specific atomic excitations. The oscillations of the sea electrons have a large range of frequencies so they emit radiation over a large range of frequencies.
A minor diversion: the oscillations of the nuclei would look completely random if you watched individual nuclei. However we can Fourier transform them to describe the motion as a sum of oscillatory modes with well defined frequencies. These modes are our phonons. Each phonon is a plane wave with a well defined frequency and many phonons moving in different directions with different frequencies sum to give the random thermal oscillations.
By now you're wondering when I'm going to actually answer your question. Well it's just the above process in reverse. An EM wave has an oscillating electric field. This field exerts a force on the sea electrons in the material and makes them oscillate, and the nuclei oscillate in response. Energy is transferred from the incoming photon to create a phonon with the same frequency as the incoming photon. However that phonon is quickly scattered and the energy gets spread out over all the oscillatory modes of the crystal lattice. And that's how photons heat the material.
