Question

How much heat input in `"kJ"` is required to produce `"450 g"` of water vapor from liquid water?

Answer 1

You are referring to vaporization, a situation where heat flow `q` and enthalpy `DeltaH` are related at constant pressure AND constant temperature.

So, this is asking for the enthalpy of vaporization, `DeltaH_"vap"^("H"_2"O")`, with no change in temperature. No need to use `q = mcDeltaT`.

When a substance vaporizes, its vapor pressure is constantly equal to the atmospheric pressure, and we have the following relationship between the enthalpy and the heat flow:

`color(green)(DeltaH = q_p)`

Naturally, if we divide by the `"mol"`s of water, we get:

`(DeltaH_"vap")/(n_("H"_2"O")) = DeltabarH_"vap"^("H"_2"O") = q_p/(n_("H"_2"O"))`

The molar enthalpy `\mathbf(DeltabarH_"vap"^("H"_2"O"))` for the vaporization of water is known to be about `"40.7 kJ/mol"`. You should have it somewhere in your book, or you can look it up.

`\mathbf(q_p = n_("H"_2"O")DeltabarH_"vap"^("H"_2"O"))`

At this point, all we need to do is figure out the `"mol"`s of water:

`"mols H"_2"O" = "450" cancel("g H"_2"O") xx ("mol H"_2"O")/("18.015" cancel("g H"_2"O"))`

`= 24.979` `"mols"`

and now we can solve for the heat flow `q_p`:

`color(blue)(q_p) = "24.979" cancel("mols H"_2"O") xx "40.7 kJ"/cancel("mol H"_2"O")`

`=` `color(blue)("1016.65 kJ")`