Question #e02ea
1 Answer
Answer:
Explanation:
Before doing anything else, make sure that you understand what happens when a solid melts.
The first thing to take note of here is that since melting is a phase change, it must take place at constant temperature.
As you know, pure water melts and freezes at
Notice that your sample of ice starts at a temperature of
This means that you will have to know two things

the specific heat of ice, which tells you how much heat is needed to increase the temperature of
#"1 g"# of ice by#1^@"C"# 
the enthalpy of fusion for water, which tells you how much heat is needed in order to convert
#"1 g"# of ice from solid to liquid at its melting point
You can find these values here
#c_"ice" = 2.09"J"/("g" ""^@"C")" "#
#DeltaH_"fus" = 334"J"/"g"#
http://www.engineeringtoolbox.com/latentheatmeltingsolidsd_96.html
Two equations will come in handy here
#color(blue)(q = m * c * DeltaT)" "# , where
#color(blue)(q = m * DeltaH_"fus")" "# , where
So, to get
#q_1 = 3.5 color(red)(cancel(color(black)("g"))) * 2.09"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [(0  (1.1)]color(red)(cancel(color(black)(""^@"C"))) = "8.05 J"#
To get the solid ice from solid at
#q_2 = 3.5 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g"))) = "337.5 J"#
The total heat needed to go from ice at
#q_"total" = q_1 + q_2#
#q_"total" = "8.05 J" + "337.5 J"#
#q_"total" = "345.55 J"#
Rounded to two sig figs, the answer will be
#q_"total" = color(green)("350 J")#
To get the answer in kilojoules, use the conversion factor
#"1 kJ" = 10^3"J"#
#q_"total" = 350 * 10^(3)"kJ" = color(green)("0.35 kJ")#