How much electrical energy (in kWh) is needed to heat the water in a well-insulated electric hot water heater of capacity 40 gal from 208C to 508C (688F to 1228F)?
40 gallons x 3.785 L/gallon = 151.4 L = 151,400 grams if we assume the water has a density of 1.00 g/mL.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q = 151,400 g x 4.184 J/g*C x (508-208) = ? J
Then convert J to kwh. 1 kwh = 3.6E6 Joules.
Post your work if you get stuck.