Question #e02ea

1 Answer
Jan 8, 2016

Answer:

#"350 J"#

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 #0^@"C"#. This means that a solid #-># liquid phase change takes place when you're going from solid at #0^@"C"# to liquid at #0^@"C"#.

Notice that your sample of ice starts at a temperature of #-1.1^@"C"#. This means that before you can melt it, you must first heat it from solid at #-1.1^@"C"# to solid at #0^@"C"#.

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/latent-heat-melting-solids-d_96.html

Two equations will come in handy here

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat absorbed / lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

#color(blue)(q = m * DeltaH_"fus")" "#, where

#q# - heat absorbed / lost
#m# - the mass of the sample
#DeltaH_"fus"# - the enthalpy of fusion of the substance

So, to get #"3.5 g"# of ice to go from solid at #-1.1^@"C"# to solid at #0^@"C"#, you need to provide it with

#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 #0^@"C"# to liquid at #0^@"C"#, you'll need

#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 #-1.1^@"C"# to liquid water at #0^@"C"# will thus be

#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")#