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

\begin{tabular}{|c|c|c|c|c|}
\hline
Substance & \begin{tabular}{c}
Latent Heat \\ Fusion \\ (melting) \\ [tex]$(kJ/kg)$[/tex]
\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}



Answer :

To determine how much energy is required to melt 2 kg of gold, you can use the formula:

[tex]\[ Q = m \cdot L_{\text{fusion}} \][/tex]

where [tex]\( Q \)[/tex] is the energy required, [tex]\( m \)[/tex] is the mass of the substance, and [tex]\( L_{\text{fusion}} \)[/tex] is the latent heat of fusion of the substance.

From the table provided, we can see that the latent heat of fusion for gold ([tex]\( L_{\text{fusion}} \)[/tex]) is [tex]\( 62.8 \, \text{kJ/kg} \)[/tex].

Given:
- Mass of gold, [tex]\( m = 2 \, \text{kg} \)[/tex]
- Latent heat of fusion of gold, [tex]\( L_{\text{fusion}} = 62.8 \, \text{kJ/kg} \)[/tex]

Plug these values into the formula:

[tex]\[ Q = 2 \, \text{kg} \times 62.8 \, \text{kJ/kg} \][/tex]

Multiplying these values gives:

[tex]\[ Q = 125.6 \, \text{kJ} \][/tex]

Therefore, the energy required to melt 2 kg of gold is [tex]\( 125.6 \)[/tex] kJ.