How much heat is needed to heat #"25.0 g"# of ice from #-25^@ "C"# into steam at #115^@ "C"#?

I got #"77.3 kJ"#. But you should go through this. It's an extensive exercise (get it?).

Well, you'll need a bunch of constants... Here are the specific heat capacities (which we assume are constant in the temperature range), and the enthalpies of phase transitions:

• #C_(ice) = "2.09 J/g"^@ "C"#
• #C_(w) = "4.184 J/g"^@ "C"#
• #C_(g) = "2.01 J/g"^@ "C"#
• #DeltaH_(fus) = "6.02 kJ/mol"#
• #DeltaH_(vap) = "40.67 kJ/mol"#

We'll need to distinguish between two processes.

1. Heating with a change in temperature, #q = mcDeltaT#
   i.e. mass times specific heat capacity times the change in temperature

2. Heating to melt or boil, constant temperature, #q = nDeltaH_(trs)#
   i.e. moles times the enthalpy of phase transition.

I would always outline the problem. Separate this into 5 regions:

#"heat ice" stackrel(-25^@ "C" -> 0^@ "C"" ")(->) "melt ice" stackrel(0^@ "C"" ")(->) "heat water" stackrel(0^@ "C" -> 100^@ "C"" ")(->) "boil water" stackrel(100^@ "C"" ")(->) "heat steam" stackrel(100^@ "C" -> 115^@ "C"" ")(->) "hot steam"#

Call these regions 1, 2, 3, 4, 5, with heat flow #q_1, q_2, q_3, q_4, q_5#. Then:

#q_1 = mC_(ice)DeltaT_(ice) = 25.0 cancel"g" xx "2.09 J/"cancel"g"cancel(""^@ "C") xx (0cancel(""^@ "C") - (-25cancel(""^@ "C")))#

#= "1306.25 J" = ul"1.306 kJ"# to heat the ice

#q_2 = nDeltaH_(fus) = 25.0 cancel"g" xx cancel("1 mol")/(18.015 cancel"g water") xx "6.02 kJ/"cancel"mol"#

#=# #ul"8.354 kJ"# to melt the ice

#q_3 = mC_(w)DeltaT_(w) = 25.0 cancel"g" xx "4.184 J/"cancel"g"cancel(""^@ "C") xx (100cancel(""^@ "C") - 0cancel(""^@ "C"))#

#= "10460 J" = ul"10.46 kJ"# to heat the water

#q_4 = nDeltaH_(vap) = 25.0 cancel"g" xx cancel("1 mol")/(18.015 cancel"g water") xx "40.67 kJ/"cancel"mol"#

#=# #ul"56.44 kJ"# to vaporize the water

#q_5 = mC_(g)DeltaT_(g) = 25.0 cancel"g" xx "2.01 J/"cancel"g"cancel(""^@ "C") xx (115cancel(""^@ "C") - 100cancel(""^@ "C"))#

#= "753.75 J" = ul"0.7538 kJ"# to heat the steam

And all of these are additive (heat is extensive).

#color(blue)(q_(t ot)) = q_1 + q_2 + q_3 + q_4 + q_5 + . . .#

#= "1.306 kJ" + "8.354 kJ" + "10.46 kJ" + "56.44 kJ" + "0.7538 kJ"#

#=# #color(blue)("77.3 kJ")#

to three significant figures from the mass.