Steam resonator: how fast, and by what, the water heats

With this voltage across this plate cell at this frequency, how fast does the water reach the boil — and is it the dipoles or the current doing it?

The formula

P = (2 π f \cdot ε_{0} \cdot κ + σ) E^{2} \cdot ν

LaTeX

P = \left( 2\pi f \cdot \varepsilon_0 \cdot κ + σ \right) E^{2} \cdot ν

Work it out

Fill from a part the library

A Plate area

cm²

One plate's area facing the other.

d Gap

The water-filled distance between the plates.

V Applied voltage

The peak voltage across the plates.

f Frequency

kHz

The frequency of the field.

w Water

What is between the plates.

σ Water conductivity

µS/cm

What a conductivity meter reads on it.

T Starting temperature

°C

Where the water starts from.

or see a worked example with typical numbers

The volume of water is the plate area times the gap.
The plates as a capacitor, with the field-strength check the plate calculation makes.
The water's loss permittivity at this frequency, from the Cole-Cole model.
The heating, dipoles and ions separately, and the time to 100 °C.

Every assumption of every step. The water between the plates is all the water there is; a vessel with more water around the plates heats slower in proportion.
The conductivity is entered rather than taken from the named water, because it is what decides the answer and the meter reading is the number to use.