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When you place a water or food in a microwave oven, it heats. Which process commits more energy to that: dielectric heating, or ion drag i.e. resistive heating?

AFAIK, in distilled water (which is a dielectric) dielectric heating is close to 100%.

What about regular (non-distilled) water? Mineral water? Sea water? Salty & wet food?

Is dielectric heating still gives more energy to the water/food then the resistive heating i.e. ion drag?

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Is resonance absorption called dielectric heating? –  Ron Maimon Mar 31 '12 at 2:51
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1 Answer

Common microwaves always operate on the principle of dielectric heating; resistive heating is not a factor. In general, microwaves are designed to be non-ionizing, meaning that the individual photons don't have enough energy to knock off electrons. A typical microwave operating at $2.45$ $\text {GHz}$ has photons with an energy $E = hf = 1.01\times10^{-5}\text {eV}$.

Typical electron orbital energies are on the order of $\text {eV}$, so this is far too small to dislodge an electron and generate a current, and thus resistive heating. Thus, the fact that something is salty and thus has a reduced resistivity doesn't help it transmit a current here because there are no free electrons. For metals, however, the ionization energy is much lower and ionization is possible, which is why one doesn't put tin foil in the microwave.

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The waves don't have to ionize anything. Electrolytes (such as sea water) already have free ions. Google for "Electromagnetic Waves in Conducting Media" - the EM wave will create the electric current in the media. –  Soonts Feb 17 '13 at 5:25
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