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

1 Answer
Nov 25, 2015

Answer:

#"206 kJ"#

Explanation:

Assuming that pressure is equal to #"1 atm"#, boiling water at its boiling point implies providing it with enough heat to turn it from liquid at #100^@"C"# to vapor at #100^@"C"#.

The amount of heat needed to allow one mole of water to undergo this phase change is called the enthalpy chnge of vaporization, #DeltaH_"vap"#.

For water at #100^@"C"#, the enthalpy change of vaporization is equal to

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

http://www2.bren.ucsb.edu/~dturney/WebResources_13/WaterSteamIceProperties/EnthalpyOfVaporizationH2O.pdf

This tells you that in order to boil one mole of water at its boiling point, you need to provide it with #"40.66 kJ"# of heat.

Your strategy now would be to use water's molar mass to determine how many moles of water you have in #"83.0 g"#

#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")#