here we go again (yes we did this one on the old forum)
there is a very easy(ish) way to calculate the efficiency of boiling water.
All electrical apparatus have an sticker on the bottom with the power consumption. This is the power they draw from the grid. We shall call it P. However to boil water you need energy (lets call this one E).
The energy is simply a power applied a certain amount of time t:
E = P*t. The unit of the enrgy is J.
If you want to increase the temperature of one cl of water from one °K (or °C) you need I cal or better said 4.18 J.
Now lets say you want to see which way is the more efficient. It’s easy. Take 6 mugs of water (all at the same temperature Ti and around 300 ml). Put 2 in the MWO, 2 in the pan on the hob and the last 2 in the kettle. Now turn them all on for one minute (t=60). Check the new temperature Tf that will be called: Tfm, Tfh, Tfk for the MWO, the hob, the kettle respectively.
You can calculate the real energy you’ve used to warm the water (4.18J per ml per °):
Erm = (Tfm-Ti)/600*4.18
Erh = (Tfh-Ti)/600*4.18
Erk = (Tfk-Ti)/600*4.18 the 600 comes from the volume of water you’ve heated up, and the Er is for real energy.
However you can now compute the energy drawn from the grid simply by using:
Em=Pm*60
Eh=Ph*60
Ek=Pk*60
And you can easily now compute the efficiency (r) of each apparatus being the ration of the real energy over the energy drawn from the grid:
rm=Erm/EM
rh=Erh/Eh
rk=Erk/Ek.
But I am not taking a lot of risks saying that you’ll find your MWO to be the most efficient (At this point it will obviously depend on the MWO indeed but they are the most efficient way to eat up water molecules).