How much energy is required to vaporize 2 kg of aluminum? Use the table below and this equation: [tex]Q = m L_{\text{vapor}}[/tex].

\begin{tabular}{|c|c|c|c|c|}
\hline
Substance & \begin{tabular}{c}
Latent Heat \\
Fusion \\
(melting) \\
(kJ/kg)
\end{tabular} & \begin{tabular}{c}
Melting \\
Point \\
[tex]$\left(^{\circ}C\right)$[/tex]
\end{tabular} & \begin{tabular}{c}
Latent Heat \\
Vaporization \\
(boiling) [tex]$(kJ/kg)$[/tex]
\end{tabular} & \begin{tabular}{c}
Boiling \\
Point \\
[tex]$\left(^{\circ}C\right)$[/tex]
\end{tabular} \\
\hline
Aluminum & 400 & 660 & 1100 & 2450 \\
\hline
Copper & 207 & 1083 & 4730 & 2566 \\
\hline
Gold & 62.8 & 1063 & 1720 & 2808 \\
\hline
Helium & 5.2 & -270 & 21 & -269 \\
\hline
Lead & 24.5 & 327 & 871 & 1751 \\
\hline
Mercury & 11.4 & -39 & 296 & 357 \\
\hline
Water & 335 & 0 & 2256 & 100 \\
\hline
\end{tabular}

A. 2200 kJ
B. 4520 kJ
C. 800 kJ
D. 1794 kJ



Answer :

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