How much heat is required to heat ethanol from #25^@ "C"# to a gas at #78.24^@ "C"#? #C_P = "2.460 J/g"^@ "C"#, #DeltaH_(vap)^@ = "38.56 kJ/mol"#, #T_b = 78.24^@ "C"#

I got #"44.04 kJ"# total.

Well, ethanol starts as a liquid, with the following:

• #DeltaH_"vap"^@ = "38.56 kJ/mol"#
• #T_b = 78.24^@ "C"#

• #C_P = "2.460 J/g"^@ "C"#

• Heat it from room temperature to boiling point.

• Keep temperature constant at boiling point and put heat in to vaporize.

If we start at room temperature, #T = 25^@ "C"#, and we assume #DeltaH_"vap"^@# and #C_P# vary negligibly within the temperature range of #25^@ "C" -> 78.24^@ "C"#, then under natural conditions (constant pressure and varied temperature):

#color(green)(q_"heating") = mC_PDeltaT#

#= ("45.5 g")("2.460 J/g"^@ "C")(78.24^@ "C" - 25^@ "C")#

#=# #"5959.15 J"#

#=# #color(green)("5.959 kJ")#

That accounts for the temperature change up to the boiling point. Then, while boiling, the temperature is held constant, so that at the boiling point:

#DeltaH_"vap"^@ = q_"vap"/(n_"EtOH")#

#=> color(green)(q_"vap") = n_"EtOH"DeltaH_"vap"^@#

#= (("45.5 g EtOH")/("46.0684 g/mol"))("38.56 kJ/mol")#

#=# #color(green)("38.08 kJ")#

Thus, to vaporize #"45.5 g"# of ethanol starting at #25^@ "C"#, heating it up through to the boiling point of #78.24^@ "C"#, and turning it into the vapor phase, is:

#color(blue)(q_"tot") = q_"heating" + q_"vap"#

#= "5.959 kJ" + "38.08 kJ" =# #color(blue)("44.04 kJ")#