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## Main Question or Discussion Point

Hi Guys

I'm new the PF and hope I'm posting I the correct place, apologies to admin if not ;-). I'm working on a project at home and maths is not my strong suit. I have an equation that I need to complete and involves converting thermal heat capacity in water into electric power. From what I could find with the following parameters this is the calculation.

Max temp 90degree C

Min temp 22degree C

Water heat retention factor of 4.19KJ/kg.k

Volume of water is 10kg/10L

To get the power or kj energy stored in the water I used this formula

Q= (4.19kj/kg.k)(10kg)((90)-(22))

Q= 2849.2kj of energy

Now using a converter that amount of energy is equal to 791.44 Watt hours of power

Here is the tricky part does that mean to get electrical power I use ohms law.

So 791.44W / 12v = 65.95A current. Is that correct and is that the same as saying it would take 65.95A of current at 12v to heat 10L of water from 22 degrees to 90 degrees ? Excluding efficiently losses ?

Any help would be great !

I'm new the PF and hope I'm posting I the correct place, apologies to admin if not ;-). I'm working on a project at home and maths is not my strong suit. I have an equation that I need to complete and involves converting thermal heat capacity in water into electric power. From what I could find with the following parameters this is the calculation.

Max temp 90degree C

Min temp 22degree C

Water heat retention factor of 4.19KJ/kg.k

Volume of water is 10kg/10L

To get the power or kj energy stored in the water I used this formula

Q= (4.19kj/kg.k)(10kg)((90)-(22))

Q= 2849.2kj of energy

Now using a converter that amount of energy is equal to 791.44 Watt hours of power

Here is the tricky part does that mean to get electrical power I use ohms law.

So 791.44W / 12v = 65.95A current. Is that correct and is that the same as saying it would take 65.95A of current at 12v to heat 10L of water from 22 degrees to 90 degrees ? Excluding efficiently losses ?

Any help would be great !