Heating water power calculation

[Rheinhardt]

Dear Reader,

I know this is a chemical engineering forum and thought this might be my best bet, I seem do hav displaced all my physics lectures from varsity ..

I warm up 1m ³ water from 12 °C to 16°C within 15 min…

Does anybody have the formula to work out the required power in kilo-watts (kw)

I would appreciate any help on this

Kind Regards

Rheinhardt.


--Off all the things i've lost , i miss my mind the most--

 

[25362]

Take it from 1 kW = 1 kJ/s = 3600 kJ/h = 860 kcal/h

 

[Rheinhardt]

I have done the following....
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C == 4186 j/kg °C

Q=c*m*deltaT

c = 4186 j/kg°C
m = 1000kg = 1000l = 1m³ of water
deltaT = 60°C-12°C >> I have an error in first post it is 60°C deltaT=48°C

Q=200.928 Million Joules
Since a kilowatt-hour is 3.6 million Joules, this energy amounts to about 55.81 kw/h

Now 15 min of an hour is 4 times quicker

Result is 223.24 kW not taking into account thermal losses

Regards

Rheinhardt

--Off all the things i've lost , i miss my mind the most--

 

[25362]

Rheinhardt, excellent thinking. One thing you didn't lose is your intellectual power.