Magnet Myths and Misconceptions

[MarkE]

MH says read books-look on the net
Quote: http://www.solarchoice.net.au/blog/power-and-energy/
Power, on the other hand, is a measure of how much we CAN produce. It is not a measure of how much energy there actually is, but a way of describing how much could be produced.

Quote: What is a kilowatt hour? As a unit of measurment, it is actually the same thing as a joule, it is just a way of measuring energy.
It is very hard to use terms and measurements you guys like,when what we are being told is different depending on who is telling us.

It is unfortunate that the www has lots of misinformation.  Fortunately, one can check more than one reference to see if it other references support or contradict it.  If you read a bit further in the reference it states correctly:

1 Watt = 1 Joule per second

[TinselKoala]

I think where many people go wrong is in the "units" of the values being discussed. I see many workings where units are left out, so that it is easy to get confused. The "units" must be respected and treated like any other algebraic value in equations. If the units don't work out all the way through, you've gone wrong somewhere. So it's a good idea to include the units in every calculation, not just the numbers themselves.

For example, the "units" of Power, the Watt, are in Joules per second, and "per" almost always indicates a division operation algebraically. 

1 W = 1 J/s  therefore
1 WattSecond = 1 Joule  (algebraically multiplying both sides by the unit "second")

So to convert from kiloWattHours to Joules properly, we need to do some algebra with the units, as well as with the numbers.

1 kWH = 1000 WattHours x 60 min/hour x 60 seconds/minute = 3 600 000 WattSeconds (or Joules).    (note the correct cancelling of the _units_ of time (hours, minutes) in the algebra.)

The kWH is a large unit of _energy_. This is what you pay for on your electric bill. You don't pay for how _fast_ you use energy, you pay for how _much_ you use.  The kW, or smaller Watt, is a _rate_ of energy usage.  If you use 1 kWH (3,600,000 Joules)  in one hour, you could express this as 1 kWH/H (one kiloWattHour per Hour)  and the "hours" units cancel algebraically and you are left with 1 kW, which is the _rate_ at which you have used, or dissipated, that amount of Joules of energy during that hour: 3,600,000 Joules per hour.

A slight confusion arises because the kiloWattHour and other such multiplications are commonly represented with a dash:  kW-H    that looks like a "minus" sign. But it is really a multiplication operation algebraically. I have avoided the use of this convention in the above remark.

[tinman]

That's not actually how the question read.
And it is impossible to answer that question literally without a given amount of time the power source was turned ON.

If you were simply looking for an answer to question if the total energy in your system is conserved, then the numbers presented in the diagram are irrelevant and no computation is required. But I am curious to see how you yourself would take that 60W source and determine the answer to your question.

The 60 watts is just what the cell uses when in operation V*I. No time is given because that is irrelevant to the question. The question remains,and applies for all devices-->will all the output energies(wether disipated or stored)equal the input energy?.The law of the conservation of energy says it will,and i was asking if the same applies to the electrolisis system i gave a diagram of.

If so,then that very same law says it is also possable to have an added energy output from an open system outside the closed system of the cell.
Like i said,if we are able to account for all the energy that is going into the cell,and that cell can switch another system from being pulled upon by gravity,to being pushed upon by becomeing buoyant,then any energy that can be gained by the later two is above that of the already accounted for energy used to create that change.

But enough of this on this thread,as it has been sidetracked too much already.

[MarkE]

The 60 watts is just what the cell uses when in operation V*I. No time is given because that is irrelevant to the question. The question remains,and applies for all devices-->will all the output energies(wether disipated or stored)equal the input energy?.The law of the conservation of energy says it will,and i was asking if the same applies to the electrolisis system i gave a diagram of.

Energy is conserved in your electrolysis example.  If you wish to challenge that, then you are welcome to perform an energy balance and show a discrepancy.

If so,then that very same law says it is also possable to have an added energy output from an open system outside the closed system of the cell.

If you do not place a boundary around what you are evaluating then evaluation is basically meaningless. 

By example:  One could point to some vessel like a swimming pool and note its capacity:  Say 20,000 gallons and note that it is full.  Then one could point to a drain pipe delivering 1000 gph, and ask:  "When will the pipe run dry?"  The question only has a definite answer if the rate at which any input to pool is defined.  One way to define that is to prohibit any:  Close the system everywhere except the drain pipe output. 

Like i said,if we are able to account for all the energy that is going into the cell,and that cell can switch another system from being pulled upon by gravity,to being pushed upon by becomeing buoyant,then any energy that can be gained by the later two is above that of the already accounted for energy used to create that change.

An energy balance like any other balance calculation must account for all the credits and debits.  Selectively ignoring either is just accounting error.

But enough of this on this thread,as it has been sidetracked too much already.

[tinman]

Energy is conserved in your electrolysis example.  If you wish to challenge that, then you are welcome to perform an energy balance and show a discrepancy.If you do not place a boundary around what you are evaluating then evaluation is basically meaningless. 

I do not wish to challenge that at all,as it was the answer i needed in order to prove that extra energy can be produced via this system,when all energy has been accounted for within the electrolysis system.