A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[AlienGrey]

https://youtu.be/hlrhW33xk7U
Well, here's an alternating current electrolysis.No diodes, no rectifiers, just a transformer.

That video idea looks a bit dangerous is all it is at what I can see is a container
full of a liquid possible plain water fed by an isolation transformer or auto transformer.
What hapens wen all the water evaporates or some one picks it up while it's live ?

[kolbacict]

The circuit will break and there will be no current.

[George1]

To kolbacict.
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Hi kolbacict,
Thank you for your reply. And my respect to your enthusiasm for seeking the truth despite of the difficulties in your life!
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May be it would not be necessary to use a calorimeter. Please read the text below, if you like. (The text below repeats some considerations from our previous posts.)
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1) It is perfectly valid for the copper wire circuit (which is hidden inside a black box 1) that (a) the voltmeter registers a voltage of 4VDC, (b) the watt-meter registers a power of 16 Watts, (c) the ammeter registers a current of 4 A, (d) the ohmmeter registers an Ohmic resistance of 1 Ohm and (e) the clock registers a period of 1000 seconds, within which a current of 4 A flows through the copper wire.
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2) It is perfectly valid for the PEM electrolyzer circuit (which is hidden inside a black box 2) that (a) the voltmeter registers a voltage of 4VDC, (b) the watt-meter registers a power of 16 Watts, (c) the ammeter registers a current of 4 A, (d) the ohmmeter registers an Ohmic resistance of 1 Ohm and (e) the clock registers a period of 1000 seconds, within which a current of 4 A flows through the PEM electrolyzer.
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3) Let us assume that an electric engineer must measure the Joule's heat generated by the load, which is hidden inside black box 1. Having in mind the readings of the measuring devices (voltmeter, ammeter, ohmmeter, watt-meter and clock) he/she would inevitably conclude that the Joule's heat, generated by the hidden load, is just equal to 16000 J.
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4) Let us assume that an electric engineer must measure the Joule's heat generated by the load, which is hidden inside black box 2. Having in mind the readings of the measuring devices (voltmeter, ammeter, ohmmeter, watt-meter and clock) he/she would inevitably conclude that the Joule's heat, generated by the hidden load, is just equal to 16000 J.
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5) In one word, any electric engineer in the world, without being interested what loads are hidden inside black boxes 1 and 2, would inevitably conclude that the Joule's heat, generated by any of the two hidden loads, is just equal to 16000 J. Any electric engineer in the world would only look at the readins of the measuring devices and after that he/she would only make some simple calculations leading to 16000 J of Joule's heat generated by any of the two hidden loads.
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What is your opinion? Two black boxes and two unknown loads hidden inside these two black boxes. For the any of the two cases you have only the readings of the measuring devices (voltmeter, ammeter, ohmmeter, watt-meter and clock). How to get the Joule's heat generated by any of the two hidden and unknown loads?
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Looking forward to your answer.
Best regards,
George   

[kolbacict]

Hi.
the resistance of the copper wire is constant and the power is constant for a long time.the resistance of the electrolytic cell will not be constant.there will be chemical reactions, electrodes will dissolve, changing the composition of the electrolyte.Although, within 1000 seconds it will be quite small.
Already will introduce an error.

[George1]

Hi kolbacict,
Thank you for your reply.
Good answer indeed! Yes, an error could really appear just as you mention in your last post. This error however can be easily
eliminated by two parallel (simultaneous) ways. Firstly, you have to add regularly distilled water in the PEM eletrolyzer and
secondly, you have to cool down regularly that same PEM electrolyzer thus keeping constant its temperature and its Ohmic
resistance, respectively. It's simple.
(Please look also at our post of March 26, 2019, 10:39:21 AM, almost two years ago. It is written there that:"Constant pure/distilled water and cooling agent supply could keep constant the electrolyte's/electrolyzer's temperature, heat exchange, mass and ohmic resistance, respectively.)
Looking forward to your answer.
Regards,
George