Heating water with a field: dielectric or conduction

At this field and frequency, how fast does the water between the plates heat — and is it the dipoles or the ions doing it?

E = \frac{V}{d} \qquad D = 2\pi f \cdot \varepsilon_0 \cdot κ \cdot E^{2} \cdot ν \qquad J = σ \cdot E^{2} \cdot ν \qquad P = D + J

Method

1. The field is the voltage over the gap, in volts per metre.
2. Dielectric heating per unit volume is ω·ε₀·ε″·E², the standard loss in a dielectric with imaginary permittivity ε″ — the same expression that heats food in a microwave oven, where ε″ is thirty; at ten kilohertz it is a thousandth.
3. Conduction heating per unit volume is σ·E²: the current density σE times the field.
4. Both times the volume of water between the plates give watts. Their sum over the water's heat capacity (4186 J/kg·K) times its mass times the degrees to 100 °C gives the time to reach the boil.

Assumptions

• The field is uniform and the given voltage is a sinusoidal peak; both loss formulae are for the RMS field, so a factor of two hides here for a square wave and a factor of √2 for a sine. The order of magnitude is the point, and the ratio between the two mechanisms does not depend on it.
• No heat is lost to the plates, the vessel or the air, and none of the current goes into electrolysis rather than heat. Both make the real time longer.
• The conductivity and the loss permittivity are the cold water's. Conductivity rises about 2 % per degree, so an electrolyte heats faster as it goes.
• The archive takes no view on what the steam resonator was for. It records that at any frequency Meyer's circuits produced, this arithmetic says the water is heated by the current through it, which is the ordinary electrode boiler.