Question

In a calorimeter, 10.0 g of ice melts at 0°C. The enthalpy of fusion of the ice is 334 J/g. How much heat was absorbed?

Answer 1

`"3.34 kJ"`

Explanation:

The enthalpy of fusion, sometimes referred to as latent heat of fusion, tells you how much heat is required in order to convert `"1 g"` of a given substance from solid at its melting point to liquid at its melting point.

In your case, you know that water has an enthalpy of fusion, `DeltaH_"fus"`, equal to `"334 J g"^(-1)`.

This tells you that in order to melt `"1 g"` of ice at `0^@"C"` to `"1 g"` of liquid water at `0^@"C"`, you need to supply it with `"334 J"` of heat.

So, if every gram of ice at `0^@"C"` requires `"334 J"` of heat in order to become liquid water at `0^@"C"`, it follows that `"10.0 g"` of ice will require

`10.0 color(red)(cancel(color(black)("g"))) * overbrace("334 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)(DeltaH_"fus")) = color(green)(|bar(ul(color(white)(a/a)"3,340 J"color(white)(a/a)|)))`

If you want, you can express the answer in kilojoules by using the fact that `"1 kJ" = 10^3"J"`

`"heat needed" = color(green)(|bar(ul(color(white)(a/a)"3.34 kJ"color(white)(a/a)|)))`

The answer is rounded to three sig figs.