How much heat is required to convert 5.88 g of ice at 12.0°C to water at 25.0°C, if the heat capacity of ice is 2.09 J/°C?
1 Answer
Answer:
Explanation:
Your strategy here will be to determine

how much heat is required to convert that sample of ice from solid at
#12.0^@"C"# to solid at#0^@"C"# 
determine how much heat is required to convert that sample of ice from solid at
#0^@"C"# to liquid at#0^@"C"# 
finally, determine how much heat is required to heat that sample of now liquid water from
#0^@"C"# to#25.0^@"C"#
In order to be able to calculate all this, you will need to know the values for the specific heat of ice and that of water, and the value of the enthalpy of fusion,
#c_"ice" = 2.09"J"/("g" ""^@"C")" "# and#" "c_"water" = 4.18"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 your
#q_1 = 5.88 color(red)(cancel(color(black)("g"))) * 2.09"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [0  (12.0)]color(red)(cancel(color(black)(""^@"C")))#
#q_1 = "147.47 J"#
To get you sample to undergo a solid
#q_2 = 5.88 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g")))#
#q_2 = "1963.9 J"#
Finally, to heat the sample from
#q_3 = 5.88 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25.0  0)color(red)(cancel(color(black)(""^@"C")))#
#q_3 = "614.46 J"#
The total heat* required to get your sample from ice at
#q_"total" = q_1 + q_2 + q_3#
#q_"total" = "147.47 J" + "1963.9 J" + "614.46 J"#
#q_"total" = "2725.83 J"#
Rounded to three sig figs and expressed in kilojoules, the answer will be
#q_"total" = color(green)("2.73 kJ")#