What is the enthalpy change when #"5 g"# of #Y# are dissolved in #"100 g"# of water to decrease its temperature from #22^@ "C"# to #17^@ "C"#? #C_P = "4.184 J/g"^@ "C"# for pure water near #25^@ "C"#.

#DeltaH = q_P = -"2196.6 J"#.

It doesn't really matter what #Y# is, since we are only finding #DeltaH# in #"J"#, not #"J/mol"#. We thus don't need the molar mass of #Y#.

At constant pressure, by definition, the heat flow #q_P# is given by

#q_P = DeltaH = mC_PDeltaT#,

where

• #DeltaH# is the change in enthalpy of the solution.
• #m# is the mass of the solution.
• #C_P# is the specific heat capacity of the solution in #"J/g"^@ "C"#.
• #DeltaT# is the change in temperature of the solution (in guess what units? #"K"#? Sure. But we'll use #""^@ "C"#).

As a result, there is not much of an extra step here. It's just regular heat flow calculations, with a different label for what #q# is at constant pressure.

(If it were constant volume, we would have given you an empirically-obtained calorimeter heat capacity in #"kJ/"^@ "C"#, and asked for #DeltaE# instead.)

We assume:

• the same specific heat capacity of water as usual, despite the solution not being pure water
• that #C_P# does not change within the temperature range.

#color(blue)(q_P) = ("5 g + 100 g")("4.184 J/g"^@ "C")(17.0^@ "C" - 22.0^@ "C")#

#=# #color(blue)(-"2196.6 J")#