# Question #09ac6

##### 1 Answer

#### Explanation:

You can find the *latent heat of fusion*, also referred to as the **enthalpy of fusion**,

#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

In your case, gold is said to have an enthalpy of fusion equal to *solid gold* at its melting to *liquid* gold at its melting point, you must provide it with

You're dealing with a

#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)|)))#