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

Feb 8, 2016

You are referring to vaporization, a situation where heat flow $q$ and enthalpy $\Delta H$ 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 $q = m c \Delta T$.

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:

$\textcolor{g r e e n}{\Delta H = {q}_{p}}$

Naturally, if we divide by the $\text{mol}$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 $\text{40.7 kJ/mol}$. You should have it somewhere in your book, or you can look it up.

$\setminus m a t h b f \left({q}_{p} = {n}_{\text{H"_2"O")DeltabarH_"vap"^("H"_2"O}}\right)$

At this point, all we need to do is figure out the $\text{mol}$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"))

$= 24.979$ $\text{mols}$

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")

$=$ $\textcolor{b l u e}{\text{1016.65 kJ}}$