Hello, Sharon,
I'll work through this problem for you, but it actually can't be answered without knowing the amount of water in the tank. For the sake of doing the problem, let's assume a 50 gallon tank, filled with 50 gal of water. I like metric, so I'll convert this using the following conversion factors:
1 gal = 3.75 liter
1 liter = 1000 ml
density of water is 1 g/cm^3, or 1 g/ml.
1 liter H2O = 1kg H2O
Therefore, 50 gallons of water = 50 gal*(3.75 liter/gal)*(1 kg H2O/liter), or 187.5 kg of water. Note that the units cancel to leave only kg water.
Next we need the specific heat of water, a value that tells us how much energy is required to heat an certain amount of water for each unit of temperature change. I'm vague on the words "amount and temperature," since the same number can be expressed in a variety of units. For this example I will use the specific heat (Cp) of water using the following value:
Cp = 4.2 kJ/kg/C
You can literally read this as "it takes 4,200 Joule to raise 1 kilogram of water 1 degree C. So if you have 1 kg of water and raise it from 24 to 25 degrees C, it will require 4.2 kJ. Twice that, you raise it 2 degrees C. Be careful to check the units on the Cp you are using. They all are correct, but only for the conditions shown. Some will be written with moles, degrees K or F, or even calories instead of Joules.
In this example, we want to heat 187.5kg of water, with Cp = 4.2kJ/kg/C, by 25 degrees C.
Looking at the units, it becomes clear:
187.5kg * 4.2kJ/kg/C * 25C = 19,613 kJ total energy required in Joules.
But the question wants the answer expressed in kw-hr. So we need to covert the kJ into kw-hr. The conversion factor I find is:
1J = 2.777778⋅10-7kWh
This seems incredibly low to me, so please double check. Perhaps I'm just accustomed to inefficient water heaters. If it is correct, then the total kwhr needed is = 2.777778⋅10-7kWh/J * 19,613 kJ = 5.45 kwhr.
I hope this helps. Feel free to ask questions,
Bob
