# How much heat is needed to vaporize "25.0 mL" of water at its normal boiling point? DeltaH_(vap) = "40.65 kJ/mol".

May 20, 2017

$\text{54.1 kJ}$ into the water.

Vaporizing at constant pressure (and of course, temperature) allows us to say:

$\boldsymbol{{q}_{v a p} = n \Delta {H}_{v a p}}$,

where $\Delta {H}_{v a p}$ is the change in enthalpy of vaporization, $n$ is the mols of water, and ${q}_{v a p}$ is the heat flow through the water. For water, $\Delta {H}_{v a p} = \text{40.65 kJ/mol}$.

Knowing that the density of water at ${100}^{\circ} \text{C}$ is $\text{0.9583665 g/mL}$:

$25.0 \cancel{\text{mL" xx "0.9583665 g"/cancel"mL}} = {23.9}_{591625}$ $\text{g water}$

where the subscripts indicate digits past the last significant digit.

This amount of water corresponds to

23.9_(591625) cancel"g water" xx "1 mol water"/(18.015 cancel"g water")

$= \underline{{1.32}_{9956} \text{ mols of water}}$.

So, the heat used to vaporize it was:

$\textcolor{b l u e}{{q}_{v a p}} = {1.32}_{9956}$ $\cancel{\text{mols water" xx "40.65 kJ"/cancel"mol water}}$

$=$ $\textcolor{b l u e}{\text{54.1 kJ}}$, INTO the water,

to three sig figs.