To determine the energy required to vaporize 2 kg of aluminum, you can use the equation:
[tex]\[ Q = m \cdot L_{\text{vapor}} \][/tex]
where:
- [tex]\( Q \)[/tex] is the energy required (in kJ),
- [tex]\( m \)[/tex] is the mass of the substance (in kg),
- [tex]\( L_{\text{vapor}} \)[/tex] is the latent heat of vaporization (in kJ/kg).
From the given table, we see that the latent heat of vaporization for aluminum is 1100 kJ/kg.
Given:
- Mass of aluminum, [tex]\( m = 2 \)[/tex] kg,
- Latent heat of vaporization for aluminum, [tex]\( L_{\text{vapor}} = 1100 \)[/tex] kJ/kg.
Plug these values into the equation:
[tex]\[ Q = 2 \, \text{kg} \times 1100 \, \text{kJ/kg} \][/tex]
[tex]\[ Q = 2200 \, \text{kJ} \][/tex]
Therefore, the energy required to vaporize 2 kg of aluminum is:
A. 2200 kJ
