Question #09ac6

1 Answer
May 29, 2016

Answer:

#1.0 * 10^2"kJ"#

Explanation:

You can find the latent heat of fusion, also referred to as the enthalpy of fusion, #DeltaH_"fus"#, for gold listed here

#DeltaH_"fus" = "63 kJ kg"^(-1)#

http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

So, a substance's enthalpy of fusion tells you how much heat is required in order to melt #"1 kg"# of that substance.

In your case, gold is said to have an enthalpy of fusion equal to #"67 kJ kg"^(-1)#. This tells you that in order to melt #"1 kg"# of gold, i.e. go from solid gold at its melting to liquid gold at its melting point, you must provide it with #"67 kJ"# of heat.

You're dealing with a #"1.5-kg"# sample of gold, which means that you'll need

#1.5 color(red)(cancel(color(black)("kg"))) * overbrace("67 kJ"/(1color(red)(cancel(color(black)("kg")))))^(color(blue)(DeltaH_"fus")) = "100.5 kJ"#

Rounded to two sig figs, the number of sig figs you have for the mass of gold, the answer will be

#"energy required" = color(green)(|bar(ul(color(white)(a/a)color(black)(1.0 * 10^(2)"kJ")color(white)(a/a)|)))#