# 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?

##### 1 Answer

#### Explanation:

The **enthalpy of fusion**, sometimes referred to as *latent heat of fusion*, tells you how much heat is required in order to convert

In your case, you know that water has an enthalpy of fusion,

This tells you that in order to melt

So, if **every gram** of ice at

#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

#"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**.