Answer:
The heat of vaporization of the gold sample is 1577.397 joules per gram.
Explanation:
The latent heat of vaporization of gold ([tex]h_{v}[/tex]), in kilojoules per gram, is the heat required by a unit mass of gold to transform the material from liquid to gas:
[tex]h_{v} = \frac{E}{m}[/tex] (1)
Where:
[tex]E[/tex] - Energy required for vaporization, in joules.
[tex]m[/tex] - Mass, in grams.
If we know that [tex]E = 92120\,J[/tex] and [tex]m = 58.40\,g[/tex], then the heat of vaporization of the gold sample is:
[tex]h_{v} = \frac{E}{m}[/tex]
[tex]h_{v} = 1577.397\,\frac{J}{g}[/tex]
The heat of vaporization of the gold sample is 1577.397 joules per gram.
