# Question 09ac6

May 29, 2016

$1.0 \cdot {10}^{2} \text{kJ}$

#### Explanation:

You can find the latent heat of fusion, also referred to as the enthalpy of fusion, $\Delta {H}_{\text{fus}}$, for gold listed here

$\Delta {H}_{\text{fus" = "63 kJ kg}}^{- 1}$

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 $\text{1 kg}$ of that substance.

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

You're dealing with a $\text{1.5-kg}$ 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)|)))#