Timeframes are required to calculate this.
Here's some ideas tho. I used 55 gallons FULL, going from 100*F to 155*F, with a timeframe of 30 minutes. The following calculations take into account NO losses from water to air, kettle to air, ect. This is obviously not possible, but it gives you a starting point.
We need to calulate the number of joules required by the following formula.
Q=cm(dT)
where c = 4.186 J/g (Pure Water)
m = mass (in grams)
dT = Change in Temp (in degrees C)
In your case,
Q=(4.187 J/g)*(207,900g)*(30*C)
Q= 26,108,082 J
Then since 1 watt = 1 J/second,
Watts = 26,108,082 / (30minutes*60Seconds)
Watts = 14,504
I entered the formulas into Excel, so if you want to change the variables, just let me know.
