Question

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).

Answer 1

I got `"84 kJ"`.

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This is interesting... you want to vaporize at room temperature, which will require more heat than at `100.0^@ "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^@ "C"`, and thus have two steps:

1. Heat from `4^@ "C"` to `25.0^@ "C"`.
2. Vaporize at `25.0^@ "C"`.

HEATING AT CONSTANT PRESSURE

Step 1 would be a simple heating at constant pressure:

`q_1 = mC_PDeltaT`,

where `m` is mass in `"g"`, `C_P` is the specific heat capacity at constant atmospheric pressure in `"J/g"^@ "C"`, and `DeltaT` is the change in temperature until the phase change temperature.

In this case we specify `T_f = 25.0^@ "C"`:

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

`=` `ul("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^@ "C"`, requiring about `"3.33 kJ/mol"` more energy than usual.

Recall that `q = DeltaH` at constant pressure (provided they have the same units). Thus, we just have:

`q_2 = n_wDeltaH_(vap)(25^@ "C")`

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

`=` `ul("81.33 kJ")`

Thus, the total heat required to obtain water vapor at `25^@ "C"` from water liquid at `4^@ "C"` (at least transiently, before it condenses), is:

`color(blue)(q_(t ot)) = q_1 + q_2 = "2.84227 kJ" + "81.33 kJ"`

`=` `ul(color(blue)(84._(17)" kJ"))`

or `"84 kJ"` to two sig figs, the number of sig figs you have allowed yourself.