# How would you calculate the enthalpy change, delta H, for the process in which 33.3 g of water is converted from liquid at 4.6 C to vapor at 25.0 C? For water: H = 44.0 kJ/mol at 25.0 C and s = 4.18J/g C for H2O(l).

Jul 31, 2017

I got $\text{84 kJ}$.

This is interesting... you want to vaporize at room temperature, which will require more heat than at ${100.0}^{\circ} \text{C}$.

Heat flow is a path function, so we can separate this into specific steps. All of this is done at constant pressure, so the heat flow is equal to the enthalpy, a state function, and the order of these steps doesn't matter.

We set a "boiling point" at ${25.0}^{\circ} \text{C}$, and thus have two steps:

1. Heat from ${4}^{\circ} \text{C}$ to ${25.0}^{\circ} \text{C}$.
2. Vaporize at ${25.0}^{\circ} \text{C}$.

HEATING AT CONSTANT PRESSURE

Step 1 would be a simple heating at constant pressure:

${q}_{1} = m {C}_{P} \Delta T$,

where $m$ is mass in $\text{g}$, ${C}_{P}$ is the specific heat capacity at constant atmospheric pressure in $\text{J/g"^@ "C}$, and $\Delta T$ is the change in temperature until the phase change temperature.

In this case we specify ${T}_{f} = {25.0}^{\circ} \text{C}$:

q_1 = 33.3 cancel"g" xx "4.184 J/"cancel"g"cancel(""^@ "C") xx (25.0cancel(""^@ "C") - 4.6cancel(""^@ "C"))

$=$ $\underline{\text{2842.27 J}}$

VAPORIZATION AT CONSTANT TEMPERATURE AND PRESSURE

Step 2 would be vaporization at constant pressure again, but at a CONSTANT, lower temperature than usual by ${75}^{\circ} \text{C}$, requiring about $\text{3.33 kJ/mol}$ more energy than usual.

Recall that $q = \Delta H$ at constant pressure (provided they have the same units). Thus, we just have:

${q}_{2} = {n}_{w} \Delta {H}_{v a p} \left({25}^{\circ} \text{C}\right)$

= (33.3 cancel("g H"_2"O") xx (cancel"1 mol")/(18.015 cancel("g H"_2"O")))("44.0 kJ/"cancel"mol")

$=$ $\underline{\text{81.33 kJ}}$

Thus, the total heat required to obtain water vapor at ${25}^{\circ} \text{C}$ from water liquid at ${4}^{\circ} \text{C}$ (at least transiently, before it condenses), is:

$\textcolor{b l u e}{{q}_{t o t}} = {q}_{1} + {q}_{2} = \text{2.84227 kJ" + "81.33 kJ}$

$=$ $\underline{\textcolor{b l u e}{{84.}_{17} \text{ kJ}}}$

or $\text{84 kJ}$ to two sig figs, the number of sig figs you have allowed yourself.