# How much heat is required to boil 83.0 g of water at its boiling point?

##### 1 Answer

#### Answer:

#### Explanation:

Assuming that pressure is equal to *liquid* at *vapor* at

The amount of heat needed to allow **one mole** of water to undergo this phase change is called the *enthalpy chnge of vaporization*,

For water at

#DeltaH_"vap" = "40.66 kJ/mol"#

This tells you that in order to boil **one mole** of water at its boiling point, you need to provide it with

Your strategy now would be to use water's molar mass to determine how many moles of water you have in

#83.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "4.607 moles H"_2"O"#

So, if you know how many moles of water you have, and how much energy is needed *per mole*, all you have to do is use a simple proportion to get

#4.607color(red)(cancel(color(black)("moles"))) * "44.66 kJ"/(1color(red)(cancel(color(black)("mole")))) = "205.75 kJ"#

Rounded to three sig figs, the number of sig figs you have for the mass of water, the answer will be

#q = color(green)("206 kJ")#