Question

Question #2ca27

Answer 1

`"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"`