8kw heat conversion to centigrade

8kw heat conversion to centigrade
Reply to
simoncouter
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Interesting concept. It reminds me of the thorny problem of converting third order differential equations to women.

Reply to
The Natural Philosopher

You're not serious, right? It is like asking to convert mph to kilos.

Reply to
DrTeeth

This is meaningless. What are you talking about?

Reply to
Tim Streater

An 8kW electric pottery kiln can run at 1300°C. Is that helpful?

Reply to
Steve Firth

But you are confusing things by using Celsius... :-)

Reply to
polygonum

Hmmm, I was taught that women can be integrated, but that triple integration was likely to lead to problems with maintaining the necessary path.

Reply to
unknown

Really, did you get repeatable results on that one? Brian

Reply to
Brian Gaff

Hmm, well I suspectthere was a key part missing in the title and as both message bodies were blank. I suspect these whould not have been sent.

Reply to
Brian Gaff

No. It proved impossible to reproduce the singularities effectively.

Without introducing infinity.

Reply to
The Natural Philosopher

42
Reply to
Chris Hogg

Sounds like something for which you found a brilliant solution, in a dream just before you woke up!

Reply to
Windmill

The AA cell in my bedside clock (output about 0.5 watts) can heat a tungsten filament to about 2000°C, so the relationship is clearly an inverse one.

Reply to
WetClay

A problem faced by randy but socially-inept mathematicians everywhere.

Owain

Reply to
Owain

Precisely so.

an endemic problem with anyone who deals with advanced theories or technologies...

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Reply to
The Natural Philosopher

Reply to
Steve Firth

42
Reply to
Peter Parry

I think more commonly they attempt the inverse problem, perhaps thinking of themselves as strange attractors.

Reply to
Gib Bogle

You will need to tell us what you are heating or at least its heat capacity (Joules per Kelvin).

If the matter being heated will go through a phase change(s) e.g. ice to water to steam, we will need to know the starting temperature and the latent heat for each phase change. Again Joules per Kelvin.

You will need to tell us how much stuff you are heating (in KG please).

You will need to tell us how long it is being heated for.

Any answer we give will only be an approximation as we will not be able to account for heat conducted, convected and radiated way during the heating process.

Also I'm not sure anyone will be bothered enough to work it out for you.

Philip

Reply to
philipuk

Actually brain fart. Joules per Kg for latent heat and specific heat capacity (Joules per KG per Kelvin) for heat capacity.

Philip

Reply to
philipuk

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