Calculate the amount of energy it will take to change 20.0 Grams of Water from -3 C to changing it all to vapor at sea level? Please help?

Energy is additive, so you can add up the heat required for the individual steps. ALWAYS outline this type of problem.

#overbrace("20.0 g ice")^(-3^@ "C") stackrel(Heat" ")(->) overbrace("20.0 g ice")^(0^@ "C") stackrel(Mel t" ")(->) overbrace("20.0 g water")^(0^@ "C")#

#stackrel(Heat" ")(->) overbrace("20.0 g water")^(100^@ "C") stackrel(Vapo rize" ")(->) overbrace("20.0 g steam")^(100^@ "C")#

We label these steps as #q_1#, #q_2#, #q_3#, and #q_4#, respectively, so that #q = q_1+q_2+q_3+q_4#.

The total heat flow involved is:

#q = "60.3 kJ"#

Anytime you heat or cool, #DeltaT ne 0#, so #q = mc_sDeltaT# will apply, where #m# is the #"g"# of substance and #c_s# is the specific heat capacity in #"J/g"^@ "C"#.

#q_1 = mc_(s,ice)DeltaT_(ice)#

#= "20.0 g" cdot "2.09 J/g"^@ "C" cdot (0^@ "C" - (-3^@ "C")) = ul"125 J"#

#q_3 = mc_(s,water)DeltaT_(water)#

#= "20.0 g" cdot "4.18 J/g"^@ "C" cdot (100^@ "C" - 0^@ "C") = ul"8360 J"#

Anytime you do a phase transition, you are at constant temperature and pressure.

Constant pressure means #q = nDeltaH#, and constant temperature occurs because all the heat is invested into changing the phase of the substance.

#n# is the mols of the substance, but since you gave values of #DeltaH# in #"J/g"#, we can use #m# instead for mass.

#q_2 = mDeltaH_(fus)#

#= 20.0 cancel"g ice" cdot "334 J"/cancel"g ice" = ul"6680 J"#

#q_4 = mDeltaH_(vap)#

#= 20.0 cancel"g water" cdot "2257 J"/cancel"g water" = ul"45140 J"#

As a result, the total heat involved is

#color(blue)(q) = q_1+q_2+q_3+q_4#

#= "125 J" + "6680 J" + "8360 J" + "45140 J"#

#=# #"60305 J"#

#=# #color(blue)("60.3 kJ")#