Question

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?

Answer 1

`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")`