Chemistry => Thermochemistry => Enthalpy
We have a material that is heated until it goes from a solid state to a liquid state, that is, it melts.
To calculate the total energy we will take into account two stages:
1. Corresponds to the energy needed to raise the temperature from 20°C to 660°C.
2. The energy needed to melt aluminum at a constant melting temperature (660°C).
We will need the following data that we can find in the literature:
Specific heat of aluminum = 0.896J/g°C
enthalpy of fusion aluminum = 0.322J/g
For stage 1. the energy required will be calculated with the following equation:
![Q=mCp(T_2-T_1)](https://img.qammunity.org/2023/formulas/chemistry/college/63q8bjc1oo6q6er33k10x9kxdycnpf7zud.png)
Where,
Q is the energy required to raise the temperature
m is the mass of the aluminum soda can, 14g
Cp is the specific heat of aluminum, 0.896J/g°C
T2 is the final temperature, 660°C
T1 is the initial temperature, 20°C
We replace the known data:
![\begin{gathered} Q=14g*0.896(J)/(g\degree C)*(660\degree C-20\degree C) \\ \\ Q=8028.16J \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/iv2x5qfc76bzlnmfuq85nowt8o3gnjnxjc.png)
For stage 2 the energy required will be calculated with the following equation:
![Q_m=m\Delta H_m](https://img.qammunity.org/2023/formulas/chemistry/college/ierwul0oynjg4ajcbcfjevsqrpnu8lnmdy.png)
Where
Qm is the energy required to melt the aluminum soda can
m is the mass, 14g
deltaHm is the enthalpy of fusion aluminum,0.322J/g
The energy required will be:
![\begin{gathered} Q_m=14g*0.322(J)/(g) \\ \\ Q_m=4.508J \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/jtject4kvw1hyimn7g50xgbv6semaz9n7n.png)
So, the total energy will be the sum of both energies. So, we will have:
![\begin{gathered} Q_T=Q+Q_m \\ \\ Q_T=8028.16J+4.508J \\ \\ Q_T=8032.668J=8.0kJ \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/thhvcist3lnuiwhd3tf8ta2pdimwslojwe.png)
Answer: It would take 8.0 kJ to melt 1 aluminum soda can