another small breakthrough on our NERD technology.

[evolvingape]

Rosemary....  let's say I have 100 grams of water at 20 degrees C, and I raise that water to 80 degrees C over a timespan of 60 minutes. How much energy have I put into the water to heat it up? Assume a nice insulated container for the water.
What is the _average_ power that I must use?

Please show your work.

We can all use an online calculator, which I presume you did as you show no working, and 6.97 Watts is only a valid answer if attempting to hide the working, which would cause embarrassment to some. It is also an incorrect answer, the online calculator you used rounded down without telling you, or you rounded to two decimal places without telling us.

The answer is Average Power of 6.973 recurring Watts.


The specific heat capacity of water is 4.184 J/g C


Q = mc dt
Q = 100 x 4.184 x (80 â€" 20 )
Q = 100 x 4.184 x 60
Q = 25104 Joules


A Joule is a unit of energy and is not a Watt.


A Watt is a unit of Power and therefore requires a time component because it is a rate at which energy is consumed.


1 Watt = 1 Joule per Second


Therefore 25104 J / 60 minutes / 60 seconds = 6.9733333333333333333333333333333333333333333333 Watts


6.973 Watts = 6.973 Joules per Second (to 3 decimal places)



I smell a RAT


RM

[Rosemary Ainslie]

I'm delighted to see that everyone's answering TK's question.  I'm not sure I'd dare.  And IF that was a question by Glen Lettenmaier related to batteries - then here's the thing.  We were donated those beautiful Raylites.  But they have no ratings detailed on the battery itself.  We've tried to determine this and were advised that they're 40 ampere hour AND subsequently - that they're 60 ampere hour.  Don't know for sure and there's no way that we can find this out definitively.

We erred on the side of caution and have used the 60 AH rating for our paper.  Here's the 'extract' from our paper.

Some mention must be made of those aspects of the tests that have not been thoroughly explored. The first relates to the batteries’ rated capacity. The batteries used in these experiments have been used on a regular basis now, for over 18 months. They have been dissipating an average wattage conservatively assessed at 12 watts for five hours of each working day, during that period, continually subjected as they were, to both light and heavy use. Notwithstanding this extensive use, they have never shown any evidence of any loss of voltage at all. Nor have they been recharged except for two batteries that caught fire. Bearing in mind that the batteries’ rating is is not more than 60AH, there is evidence of out performance related to that rating.  However there has not been a close analysis of the electrolytic condition of the batteries, before, during or even after their use. This would require a detailed analysis of the supply’s electrolytic properties that is outside the scope of this presentation and expertise. Results therefore were confined to classical measurement protocols with the distinction that the energy dissipated at the resistor element was established empirically and as it related to the heat dissipated on that resistor.

So TK - YOU do the math.  And when you do this - factor in the continual use of 6 batteries only - as 2 were taken out of circulation some time back.  Or better still - average it at 7 batteries.  And then factor in that we've had possession of those batteries since late Jan early feb of 2010. Which means that its usage has been FAR more extensive than the conservatively assessed 18 months of continual use.  The usage has NOW actually spanned closer to 26 months.  And then try and explain why there is apparently absolutely NO LOSS OF VOLTAGE OVER ANY OF THE BATTERIES SINCE THE DAY WE TOOK POSSESSION. 

Kindest regards,
Rosemary

[Rosemary Ainslie]

And MileHigh

I was given your absolute assurance that you'd 'given up' on this subject.  Clearly you're undertakings mean nothing.  But let me advise you - if you think that I'm depending on that battery data - then it's belied by the text in the paper.  What a strange man you are.  You look at the written word and assume that it's not written.  I only mention the evidence that the batteries are still showing the same voltage as when we first took possession which implies - if anything - that they've outperformed their watt hour rating.  But that's taking the assumption that we use 20 working days per week - when our most extensive testing is at the weekends.  And in terms of my own calculations - which are suspect at BEST - but at their absolutely nominal average - then I THINK we should have discharged not less than 90% of each battery's capacity. 

But read the text.  The battery performance is absolutely NOT the claim.  It is just a passing reference.  I do however, confidently predict that our circuit would outperform a control.  And I'd be very happy to test that - as mentioned - with some attention to a small caveat that we've also mentioned - all over the place.

Kindest regards,
Rosie Pose.

And MileHigh - as a word of caution.  If you want to garner any credibility in this continual background 'muttering' you need to learn the trick of posting over the ENTIRE post - else we'll all start thinking that you're applying some kind of editorial EMPHASIS - that is intended to mitigate against our claim.  I fondly believe that one day your offspring (if you have any) will thank the likes us poor experimentalists into this over unity reach.  And they'll hopefully be able to reference your own diligence in trying to prevent this knowledge.  Which may confuse them somewhat.  It certainly confuses me.  On the whole I'm hoping that this information will be progressed.  On the whole you're hoping that this information won't be progressed.  I'm not sure that we're on the same side - somehow.

Again,
as ever,
Rosie Pose,
   

[eatenbyagrue]

6.973 Watts = 6.973 Joules per Second (to 3 decimal places)

I smell a RAT

What?  Are you such a pedant that you are seriously trying to put my answer down as wrong?  Ever heard of rounding?


And I did not use any online calculator.  There is no Google function for 100 grams of water raised by 60 degrees for 60 minutes.  I used physics formulas.  This is not rocket science, dude.

[fuzzytomcat]

We can all use an online calculator, which I presume you did as you show no working, and 6.97 Watts is only a valid answer if attempting to hide the working, which would cause embarrassment to some. It is also an incorrect answer, the online calculator you used rounded down without telling you, or you rounded to two decimal places without telling us.

The answer is Average Power of 6.973 recurring Watts.


The specific heat capacity of water is 4.184 J/g C


Q = mc dt
Q = 100 x 4.184 x (80 â€" 20 )
Q = 100 x 4.184 x 60
Q = 25104 Joules


A Joule is a unit of energy and is not a Watt.


A Watt is a unit of Power and therefore requires a time component because it is a rate at which energy is consumed.


1 Watt = 1 Joule per Second


Therefore 25104 J / 60 minutes / 60 seconds = 6.9733333333333333333333333333333333333333333333 Watts


6.973 Watts = 6.973 Joules per Second (to 3 decimal places)



I smell a RAT


RM

Well actually the question was for Rosemary .....

A) Multiply the mass of water by the temperature, to calculate the required energy in calories. For example, to raise the temperature of 100 grams of water by 60 degrees, you need to calculate: 100 x 60 = 6,000 calories.


B) Multiply the energy in calories by 4.186 to convert it to joules (J). In this example, the energy is 6,000 x 4.186 = 25,116 J.


C) Multiply the time in minutes by 60 to convert it to seconds. For example, if you need to raise the temperature in 60 minutes -- then 60 x 60 = 3600 seconds.


D) Divide the energy by the time, to calculate the electrical power needed. In this example, the required power is 25,116 / 3600 = 6.9766666 Watts.


Answer - 6.9766666 Watts


FuzzyTomCat