Question #1c1a3

Well, you'll have parts of the "curve" that are flat from phase changes, and parts of the curve that increase with the amount of heat added simply to change the temperature.

If you want actual data points, here they are. [You don't need them. Why?]

#(0,T_0) = ("0 kJ",-50^@ "C")#

#(q_1,T_1) = ("21.080 kJ",0^@ "C")#

#(q_1+q_2,T_2) = ("87.913 kJ",0^@ "C")#

#(q_1+q_2+q_3,T_3) = ("171.593 kJ",100^@ "C")#

#(q_1+q_2+q_3+q_4,T_4) = ("623.106 kJ",100^@ "C")#

#(q_1+q_2+q_3+q_4+q_5,T_5) = ("643.066 kJ",150^@ "C")#

For simplicity, dots + straight lines = this graph, since we assumed the specific heat capacities were constant over the temperature range. If you want you can add some curves at the corners of phase changes. It should look kind of like this:

First, outline it...

#"heat" stackrel(-50^@ "C" -> 0^@ "C"" ")(->)"melt"stackrel(0^@ "C"" ")(->)"heat"stackrel(0^@ "C" -> 100^@ "C"" ")(->)"boil"stackrel(100^@ "C"" ")(->)"heat"stackrel(100^@ "C"->150^@ "C"" ")(->)"done"#

So there should be five distinct sections on the graph for a constant mass. You'll then need to switch between heating and phase change equations.

#(1)# #q_1 = mc_"ice"DeltaT#, #" "" "DeltaT = 0 - (-50^@ "C")#

where #c_"ice" = "2.108 J/g"^@ "C"# is the specific heat capacity of ice, assumed constant in the temperature range. #DeltaT# is change in temperature, and #m# is mass in #"g"#.

#q_1 = 200 cancel"g" cdot "2.108 J/"cancel"g"cancel(""^@ "C") cdot 50cancel(""^@ "C") = "21080 J"#

#=# #ul"21.080 kJ"#

#(2)# #q_2 = nDeltabarH_"fus"#

where #n# is the mols of water and #DeltabarH_"fus" = "6.02 kJ/mol"# is the enthalpy of fusion of water.

#q_2 = 200 cancel"g" xx (cancel"1 mol")/(18.015 cancel"g water") xx "6.02 kJ"/cancel"mol"#

#=# #ul"66.833 kJ"#

#(3)# #q_3 = mc_wDeltaT#, #" "" "DeltaT = 100 - 0^@ "C"#

where #c_w = "4.184 J/g"^@ "C"# is the specific heat capacity of water, assumed constant in the temperature range.

#q_3 = 200 cancel"g" cdot "4.184 J/"cancel"g"cancel(""^@ "C") cdot 100cancel(""^@ "C") = "83680 J"#

#=# #ul"83.680 kJ"#

#(4)# #q_4 = nDeltabarH_"vap"#

where #DeltabarH_"vap" = "40.67 kJ/mol"# is the enthalpy of vaporization of water.

#q_4 = 200 cancel"g" xx (cancel"1 mol")/(18.015 cancel"g water") xx "40.67 kJ"/cancel"mol"#

#=# #ul"451.513 kJ"#

#(5)# #q_5 = mc_gDeltaT#, #" "" "DeltaT = 150 - 100^@ "C"#

where #c_g = "1.996 J/g"^@ "C"# is the specific heat capacity of steam, assumed constant in the temperature range.

#q_5 = 200 cancel"g" cdot "1.996 J/"cancel"g"cancel(""^@ "C") cdot 50cancel(""^@ "C") = "19960 J"#

#=# #ul"19.960 kJ"#

Now you have enough data to make your plot. Take #q# to be on the #x# axis and #T# to be on the #y# axis. Then you have 5 sections and 6 data points:

#(0,T_0) = ("0 kJ",-50^@ "C")#

#(q_1,T_1) = ("21.080 kJ",0^@ "C")#

#(q_1+q_2,T_2) = ("87.913 kJ",0^@ "C")#

#(q_1+q_2+q_3,T_3) = ("171.593 kJ",100^@ "C")#

#(q_1+q_2+q_3+q_4,T_4) = ("623.106 kJ",100^@ "C")#

#(q_1+q_2+q_3+q_4+q_5,T_5) = ("643.066 kJ",150^@ "C")#

I'll let you make the graph. Just put dots on it and connect them with straight lines.