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06-19-2009, 07:50 PM
 DrStiffler Silver Member Join Date: Mar 2009 Posts: 948
The only way to test this...

What does it take to raise 1L of water from 25’C to 100’C (boiling)?

Well 1L = 1,000mL and at STP it takes 1 calorie to raise 1mL of Water 1’C.

So let us assume our 1L of water is at 25’C and we want to get it boiling, which would be 100’C.

Another thing to consider is that our 1L of water is enclosed in an Adiabatic chamber so we can ignore (conduction, convection and radiation).

Thus;

Tboil – Tstart = 100 – 25 = 75

Therefore to raise 1,000 mL 75’C requires the following number of calories.

75 * 1,000 = 75,000 cal.

75,000 cal * 4.187 = 314,000 Joules.

If we want to do this over (1) one hour;

314,000/3,600 = 87.23 J/Sec (Watt Seconds)

Now if we have a so-called gain of 20 we can say 87.23/20 = 4.361 J/Sec

What this says is that if you have an OU of 20 you could Heat 1L of water from 25’C to 100’C in one hour with 4.361W/Hr input.

This is test book perfect ideal… It will never be this way so allow for that.

Unless you test with a calorimeter, forget the results. This looking at a very small average pulse from a coil collapse will not say anything. If you do not believe me, them please?

What is the result of the integration of this so-called super energy pulse? You can ignore complicated math and simple look at it and mentally see that this so called collapsing pulse will not mean any amount of energy worth consideration. You need to do controlled tests in a calorimeter or it is all worthless.
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