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
Mar 26, 2016

Answer:

#"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.

http://alcheme.tamu.edu/?page_id=2245

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.