Question

Question 4 of 10

How much energy is required to vaporize 2 kg of gold? Use
the table below and this equation: Q = mLvapor
Substance
Aluminum
Copper
Gold
Helium
Lead
Mercury
Water
Latent Heat
Fusion
(melting)
(kJ/kg)
400
207
62.8
5.2
24.5
11.4
335
Melting
Point
(°C)
660
1083
1063
-270
327
-39
0
Latent Heat
Vaporization
(boiling) (kJ/kg)
1100
4730
1720
21
871
296
2256
Boiling
Point
(°C)
2450
2566
2808
-269
1751
357
100

Answer 1

To vaporize 2 kg of gold, 3440 kJ of energy is required. This calculation uses the latent heat of vaporization value, which is 1720 kJ/kg for gold.

Step-by-step explanation:

The amount of energy required to vaporize 2 kg of gold can be calculated using the latent heat of vaporization. According to the provided table, the latent heat of vaporization for gold is 1720 kJ/kg. Using the formula Q = mLvapor, where Q represents the energy required, m is the mass of the substance, and Lvapor is the latent heat of vaporization, we find that the energy required to vaporize 2 kg of gold is:

Q = (2 kg) × (1720 kJ/kg) = 3440 kJ.

Therefore, 3440 kJ of energy is needed to vaporize 2 kg of gold.