09-29-2003, 09:33 AM
I've tried to look for an alternative source of data, but every single site has taken their data from the same source, which lists the volume as 1.640 KM^3. That's no use. You need the mass of the water in kilogrammes to work out how much energy is needed
ah... found a convertor
1 cubic KM = 1 x 10^12 L
= 1 x 10^12 Kg
1 x 10^12 x 1.640 = 1.64 x 10^12 Kg
change in energy = Specific heat capacity x mass of substance x temperature change in Kelvin
SHC of water = 4200 J Kg^-1 K^-1
E = 4200 x 1.64 x 10^12 x 1
E = 6.888 x 10 ^15 J
or 6.888 PJ
in comparison, you average electric lightbulb (100W) uses 360,000 J per hour, or 8,640,000 J per day
so it would take 797,222,222 days for an electric lightbulb to use that much energy, or 2,182,675.5 years. Unless they have a way of pumping over 2 million lightbulb years of energy into the lake in a short time, I don't think we can worry about the temerature change much.
Of course, all the above doesn't take into account the loss of temperature to evaporation, the air and the land around.
-Bob
ah... found a convertor
1 cubic KM = 1 x 10^12 L
= 1 x 10^12 Kg
1 x 10^12 x 1.640 = 1.64 x 10^12 Kg
change in energy = Specific heat capacity x mass of substance x temperature change in Kelvin
SHC of water = 4200 J Kg^-1 K^-1
E = 4200 x 1.64 x 10^12 x 1
E = 6.888 x 10 ^15 J
or 6.888 PJ
in comparison, you average electric lightbulb (100W) uses 360,000 J per hour, or 8,640,000 J per day
so it would take 797,222,222 days for an electric lightbulb to use that much energy, or 2,182,675.5 years. Unless they have a way of pumping over 2 million lightbulb years of energy into the lake in a short time, I don't think we can worry about the temerature change much.
Of course, all the above doesn't take into account the loss of temperature to evaporation, the air and the land around.
-Bob