# How much energy does it take to boil a gallon of water?

This question cannot be answered accurately unless you define a few more parameters. You need to know what the beginning temperature of the water is and the elevation. See the website below for a good explanation as to why:

http://whatscookingamerica.net/boilpoint.htm

If you say you had a gallon of water at a temperature of 68 degrees (room temperature) and wanted to raise the temperature to 208 degrees Fahrenheit it would require you to raise the temperature 140 degrees. TO FIGURE OUT A PROBLEM LIKE THIS YOU NEED TO KNOW THE SPECIFIC HEAT OF WATER!

Come on, wiki’ don’t fail me now!

From the page titled “Specific Heat Capacity” :::

For example, the heat energy required to raise waterâ€™s temperature one kelvin (equal to one degree Celsius) is 4186joules per kilogram[1].

So we just have to do a few convesions and we’ll have it… let’s get gallons to cubic centimeters and we can use a density of water as 1 gram per cubic center D = M / V    so converter (we need you!)  — thanks calculateme.com   : “

# Convert Gallons to Cubic Centimeters

1 Gallon = 3785.41178 Cubic Centimeters

OK so we then have 3785.41178 cubic centimeters which will weigh that much in grams — because 1 cubic centimeter of water (with a density of 1) has a mass of 1 gram — 3785.4 grams x (1 kg / 1000 grams) = 3.7854 kg of water.

We’re almost there:

We need to now take the figure 140 degrees and the fact that for each degree we raise 1 kg it will cost us 4186 joules of energy. So we will multiply these two terms to get the total that it would take to raise the temperature of 1 kg of water and that is (dashboard…) 586,040 joules. But remember we have 3.7854 kg of water, not just 1 kg of water — so if you assume it will all heat directly at the same time (it wont, but there will be some serious convection as the temperature starts to rise…) so we have to multiply one last time by 3.7854 and we get a total of 2,218,395.8 joules of energy.

But the thing is, you can never achieve perfect efficiency in heat transfer, there will always be some loss to the surrounding environment (the atmosphere around the stove top will be directly heated as you attempt to boil the spaghetti pot full of a gallon of water—and it requires energy to heat the metal pot up itself first too).

So this figure we have is a bear MINIMUM amount of energy it would require. It might be a safe estimate to say that it might actually require 3 million (or even much more) joules of energy. Which doesn’t actually equate to that much in the end… 3 million joules is only .83333333 kW/h…. but that’s still a considerable amount of energy.