the heat capacity of water is 4190 Joules/kg deg C
1 meter^3 of water is about 1000 kg
so Joules to heat = 4190 J/kg C*1000 *volume in m^3 * change in temp
a Joule is a watt second
If the price of electrical energy is $0.10 per kilo-watt*hour, what is the cost of using electrical energy to heat the water in a swimming pool (12.0m*9.00m*1.5m) from 15 to 27 degree C?
Can anyone please give me some hints to do it? Thanks a lot!
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1 answer