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).

I got #"84 kJ"#.

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.