Question #2ca27

1 Answer
Jun 14, 2017

Answer:

#"2257 J"#

Explanation:

In order to be able to answer this question, you need to know the value of the enthalpy of vaporization of water

#DeltaH_"vap" = "2257 J g"^(-1)#

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

The enthalpy of vaporization tells you the energy needed to convert #'1 g"# of a given substance from liquid at its boiling point to vapor at its boiling point.

In this case, you know that in order to convert #"1 g"# of water at its normal boiling point of #100^@"C"# to #"1 g"# of vapor at #100^@"C"#, you need to provide it with #"2257 J"#.

You can use the molar mass of water to determine the molar enthalpy of vaporization, which represents the energy needed to convert #1# mole of water from liquid at #100^@"C"# to vapor at #100^@"C"#.

#18.015color(white)(.)color(red)(cancel(color(black)("g")))/"mol" * "2257 J"/(1color(red)(cancel(color(black)("g")))) = "40,660 J mol"^(-1)#

This means that in order to convert #"18.015 g"# of water, the equivalent of #1# mole of water, at its normal boiling point to vapor at #100^@"C"#, you need to provide

#"40,660 J " = " 40.66 kJ"#