The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?

1 Answer
Nov 16, 2016

Answer:

#"6.48 kJ"#

Explanation:

The molar heat of vaporization, #DeltaH_"vap"#, sometimes called the molar enthalpy of vaporization, tells you how much energy is needed in order to boil #1# mole of a given substance at its boiling point.

In water's case, a molar heat of vaporization of #"40.66 kJ mol"^(-1)# means that you need to supply #"40.66 kJ"# of heat in order to boil #1# mole of water at its normal boiling point, i.e. at #100^@"C"#.

#DeltaH_"vap" = color(blue)("40.66 kJ") color(white)(.)color(red)("mol"^(-1)) #

You need #color(blue)("40.66 kJ")# of heat to boil #color(red)("1 mole")# of water at its normal boiling point.

Now, the first thing to do here is to convert the mass of water to moles by using its molar mass

#2.87 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.1593 moles H"_2"O"#

You can now use the molar heat of vaporization as a conversion factor to determine how much heat would be needed to boil #0.1593# moles of water at its boiling point

#0.1593 color(red)(cancel(color(black)("moles H"_2"O"))) * "40.66 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("6.48 kJ")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of the sample.