Suppose a bag full of ice (450 g) at 0.0 °C sits on the counter and begins to melt to liquid water. How much energy must be absorbed by the ice if 2/3 of it melted?

1 Answer
Dec 23, 2015

Answer:

#q = 1.0 * 10^2"kJ"#

Explanation:

All you have to do here is figure out how much heat is required to convert ice at #0^@"C"# to liquid water at #0^@"C"#, i.e. how much heat is required for that much ice to undergo a solid #-># liquid phase change.

As you know, phase changes take place at constant temperature and depend on the given substance's enthalpy of fusion, which in water's case is equal to about

#DeltaH_f = 334"J"/"g"#

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

What that means is that in order to melt #"1 g"# of ice at #0^@"C"# to liquid water at #0^@"C"#, you need to provide it with #"334 J"# of heat.

In your case, you need to know how much heat is required to melt #2/3# of that #"450-g"# sample of ice. Well, if #"1 g"# requires #"334 J"# of heat, it follows that you have

#2/3 * 450 color(red)(cancel(color(black)("g"))) * "334 J"/(1color(red)(cancel(color(black)("g")))) = "100,200 J"#

Rounded to two sig figs and expressed in kilojoules, the answer will be

#q = color(green)(1.0 * 10^2"kJ")#