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If the water is not already at boiling temperature, then you will need equations 1 and 2. If the water is already at boiling temperature, you will only need equation 2.

1. Q = m X C X ΔT

This equation is used to calculate how much energy is required to change the temperature of a given object, of given mass, by a given number of degrees.

Q = the total amount of energy required, in joules

m = mass, in grams, of the object being heated (in this case, the water)

C = the specific heat of the object (for water, 4.186)

ΔT = the total change in temperature

2. Q = 2.257 joules X m

The 2.257 in this equation is the heat of vaporization of water: that is, the amount of energy, required per gram of water, to boil water: 2.257 joules per gram. If you were using this equation for a different substance, you would have to look up its heat of vaporization, and substitute it in this equation.

Q = the amount of energy required, in joules

m = mass, in grams, of the water

For example, suppose you were asked to calculate how much energy it would take to boil 256 grams of water which is currently at 40 degrees Celsius. We know that the boiling temperature of water is 100 degrees Celsius; therefore the change in temperature, ΔT, is 100 - 40, which equals 60. Calculate as follows:

Q = 256 grams X 4.186 X 60°C

Q = 64296.96 joules

This is how much energy it will take to raise the temperature of the water from 40°C to 100°C. Now calculate how much energy it will take to boil the water once it reaches 100°C:

Q = 2.257 joules X 256 grams

Q = 577.792 joules

We now take the energy required to raise the temperature of the water from 40°C to 100°C and add it to the energy required to boil the water:

64296.96 joules + 577.792 joules = 64874.752 joules

Convert to kilojoules:

64875.752 joules / 1000 = 64.875752 kilojoules

Round to 64.88 kilojoules.

If you are required to express your answer in scientific notation, then express it as

6.488 x 103 kilojoules.

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