A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[Floor]

                                          OHMIIC AND NON OHMIC RESISTANCE

The two most common types of electric resistors are carbon resistors and wire
wound resistors. The amount of current that will flow through these types of
resistors, is in proportion to the voltage applied to them.  They function in accord
with ohms law. (E = I x R, R = E / I and  I = E / R) with the exception that their
electrical resistance increases when their temperature is increased.

E is electromotive force and is measured in units of volts (V).
I is electric current FLOW and is measured in amperes (A).
R is resistance to electric current flow and is measured in units of ohms (Ω).

An electric current flowing through these common types of resistors causes heating
of the resistor.

The heating of these common types of resistors is in proportion to the magnitude of
the current flowing through the resistor.
    But ......
Their resistance increases when their temperature increases.

If the heat generated by the electric current flow, is allowed to dissipate into the
surrounding environment, the resistor may reach an equilibrium between the heat
produced and the heat dissipated.  The magnitude of its electrical resistance is then
stabilized.

Common resistors are rated in terms of the power as heat energy they can dissipate
(their wattage).  Generally the larger the physical size of a resistor, the more heat it
can rapidly dissipate (the more wattage it can handle).  This is because a larger surface
area is in contact with a greater amount of the surrounding air.  Heat transfer is more
rapid when that contact area is greater.

A common electric resistor component AT A GIVEN TEMPERATURE, behaves in accord
with ohm's equations ( E = I x R,  R = E / I  and  I = E / R ). The amount of electric
current that flows through a given resistor, is then in dependent upon the applied
voltage. Its resistance is "ohmic".

Other wise it is "non ohmic", in that Ohm's law does not account for change in its
resistance due to change in temperature.

Example:

A 40 watt, filament type (incandescent) light bulb passes a current of  around 0.33 A)
                     WHEN IT IS AT ITS OPERATING TEMPERATURE.
                       ( P power = E x I), (40 watts / 120 volts equals about 0.33 A)

Its resistance is about  363.6  Ω
                     WHEN IT IS AT ITS OPERATING TEMPERATURE.
                                  (E / I = R), (120 V / 0.33 A = 363 Ω) .

Its resistance when MEASURED at room temperature is around 0.03 Ω.

120 V / 0.03 Ω  =  4000 A. This would result in a (very brief) 480,000 watt power
dissipation (120 V x 4000 A = 480,000 watts), if the the supply system could deliver
that  peak power draw before the filament reaches its operating temperature and
AT WHICH TIME its resistance has increased to 363.6 Ω 

When operated at very low voltage and power input, a 40 watt incandescent lamp
behaves in accord with ohm's equations.  When operated at high voltages and power
input, it is said to behave in a "non ohmic" manner.  The current flowing through it
at a given voltage, has greatly decreased because of its high operating temperature.
Its resistance increased.

Electrolytes behave in accord with ohmic equations at a given temperature except that
there are also chemical reactions which affect and / or affected by temperature and resistance.
                    But also, also .....
the electrical resistance DOES change with simple temperature change.

Electric current through the electrolyte will decrease as the electrolytes resistance
has increased due to increase in its temperature.  But the chemical activity of the
electrolyte will be affected by temperature as well.  Many kinds of chemical reactions
occur faster at higher temperatures than they would occur at lower temperatures.

It seems complex, but to get past a need for temperature stabilization of an electrolyte
during experimentation, one may simply measure power as electric wattage input into
the system, while also accounting for the duration, in time, of the electrolysis.

Measure...
Power input (wattage). Use a watt meter.

Time duration of the electrolysis. Use a timer.

Temperature of the electrolyte before electrolysis.  Use a thermometer.

Volume of the electrolyte. Use a measuring cup.

Temperature of the electrolyte at the end of the electrolysis Use a thermometer.

Volume of electrolyte remaining after electrolysis.  Use a measuring cup.

Temperature of the evolved gases (including water vapor) at the finish time of the
electrolysis. Use a thermometer.

Volume of the H and O (including water vapor) while under normal atmospheric pressure
and also while at the same temperature as the electrolyte was at, at the end of the
electrolysis.

Remove the water vapor. Pass the gas mixture through a freezing pipe.

How much water was removed from the electrolyte as water vapor. Thaw the ice and
measure its volume with a measuring cup 

Consider this thawed volume of water, to have been heated along with the rest of the
electrolyte and add these calories gained, to those gained from during the rest of
electrolyte's heating during the electrolysis.

Volume of the H and O at regular atmospheric pressure, with out the water vapor, but
also while knowing the temperature of the H and O..

Calculate the calories present within the gas which were gained by the gas during
its temperature increase during the electrolysis.

Calculate the Temperature of the high heat combustion of the H and O produced and
the calories that combustion will produce.

    OR...
look all these things up in a reference source.

[George1]

To Floor.
===================
I know that you аre a touchy man and that you get angry very easily. But risking to insult you seriously again I would repeat again: YOU ARE NOT READING MY POSTS AT ALL! AND THIS IS NOT A DIALOGUE AND THIS IS NOT AN HONEST DISCUSSION! THIS IS YOUR MONOLOGUE!
===================
In our post of January 03, 2021, 03:48:09 PM (this post's text has been published many times before in this topic) it is CLEARLY WRITTEN THAT: CONSTANT PURE WATER AND COOLING AGENT SUPPLY COULD KEEP CONSTANT THE ELECTROLYTE'S TEMPERATURE, HEAT EXCHANGE, MASS AND OHMIC RESISTANCE, RESPECTIVELY.
===================
AM I CLEAR NOW? READ CAREFULLY MY POSTS!!!!!

[Floor]

Yes it is obvious that you do not understand that it is not necessary
to cool the electrolyte, in order that one may conduct the experiment.

In fact cooling the electrolyte would complicate the measurements done
in the experiment.

I understand that you have stated that you do not intend to do experiment.
So your whole topic is without merit anyway, except for the input
from some posters other than your self.

[George1]

To Floor.
======================
======================
FIRSTLY.
Prof. S. L. Srivastava's solution is given below.
Prof. S. L. Srivastava's solution consists of two lines only.
LINE 1. Current through the electrolyte is given by I = (m)/(Z x t).
LINE 2. Power consumed = (I) x (I) x (R) =  ((m)/(Z x t)) x ((m)/(Z x t)) x (R) = 31.86 W.
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I am asking you (PERSONALLY!) my question for the 25th time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!
-----------------------------------
All members of this forum are waiting for your PERSONAL(!) answer for the 24th time. Only one word -- either "yes" or "no"!
======================
======================
SECONDLY.
The three equalities V = I x R (1), V x I x t = I x I x R x t (2) and H = Z x I x t x HHV (3) (that is, Ohm's law, Joule's law of heating and Faraday's laws of electrolysis, respectively) have been validated experimentally millions of times within a period of 200 years. And you reject this simple obvious truth!

[Floor]

Only on word, yes or no ?
Dream on.