To find the amount of energy required to vaporize 2 kg of aluminum, we can use the equation \( Q = m L_{\text {vapor}} \), where:
- \( Q \) is the energy required,
- \( m \) is the mass,
- \( L_{\text {vapor}} \) is the latent heat of vaporization.
From the provided table, we can find that the latent heat of vaporization for aluminum is \( 1100 \, \text{kJ/kg} \).
Given:
- Mass of aluminum, \( m = 2 \, \text{kg} \),
- Latent heat of vaporization of aluminum, \( L_{\text {vapor}} = 1100 \, \text{kJ/kg} \).
Substituting these values into the equation \( Q = m L_{\text {vapor}} \):
[tex]\[ Q = 2 \, \text{kg} \times 1100 \, \text{kJ/kg} \][/tex]
[tex]\[ Q = 2200 \, \text{kJ} \][/tex]
Therefore, the amount of energy required to vaporize 2 kg of aluminum is \( 2200 \, \text{kJ} \).
The correct answer is:
D. [tex]\( 2200 \, \text{kJ} \)[/tex]
