Question

It takes 15kWh(kilowatt-hours) of electrical energy to produce 1.0 kgof aluminum metal from aluminum oxide by the Hall–Heroult process. Compare this to the energy necessary to melt 1.0 kgof aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is$\mathbf{10}\mathbf{.}\mathbf{7}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{(}\mathbf{1}\mathbf{W}\mathbf{=}\mathbf{1}\mathbf{J}\mathbf{/}\mathbf{s}\mathbf{)}$.]

Short answer

The amount of energy produced by the Hall-Heroult process is$5.4\times {10}^7\mathrm{J}/\mathrm{kg}$ and the amount of energy produced by melting is $3.96\times {10}^5\mathrm{J}/\mathrm{kg}$.

The melting of aluminum is feasible as it requires less energy for the same amount of aluminum.

Step-by-step solution

Concept Introduction

The overall heat content in a system is enthalpy, equal to the system's internal energy plus the product of volume and pressure.

Hall-Heroult Process

The Hall-Heroult process is the major industrial process for extracting aluminum from its oxide, alumina.

The amount of electrical energy used to produce 1.0 kg of aluminum is 15kWh.

The enthalpy of fusion for aluminum is $10.7\raisebox{1ex}{$\mathrm{KJ}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$.

Converting a given amount of electrical energy from KWh to Ws–

$15\mathrm{kWh}\times \frac{1000\mathrm{W}}{1\mathrm{kW}}\times \frac{3600\mathrm{s}}{1\mathrm{h}}=5.4\times {10}^7\mathrm{Ws}$

So, the amount of electrical energy used to produce 1.0 kg of aluminum is equal to 15kWh, which is equal to $5.4\times {10}^7\mathrm{Ws}$.

$5.4\times {10}^7\mathrm{Ws}\times \frac{1\mathrm{J}}{{\mathrm{W}}_{\mathrm{s}}}\times \frac{1}{\mathrm{kg}}=5.4\times {10}^7\mathrm{J}/\mathrm{kg}$

Energy needed to melt the metal

Now, the energy required to melt 1.0 kg of aluminum is obtained by dividing the enthalpy of the fusion of aluminum by the molar mass of aluminum. To make it easier to compare with the previously obtained value, it is converted into appropriate units of measurement.

$\frac{10.7\mathrm{kJ}}{1\mathrm{mol}}\times \frac{1\mathrm{mol}}{27.0\mathrm{g}}\times \frac{1000\mathrm{g}}{1.0\mathrm{kg}}\times \frac{1000\mathrm{J}}{1\mathrm{kJ}}=3.96\times {10}^5\mathrm{J}/\mathrm{kg}$

The energy required for the Hall-Heroult process is greater than the energy required to melt the same amount of aluminum, so it is feasible to recycle aluminum by melting it.

Therefore, the Hall-Heroult method produces$5.4\times {10}^7\mathrm{J}/\mathrm{kg}$ of energy, whereas melting produces$3.96\times {10}^5\mathrm{J}/\mathrm{kg}$ of energy. Aluminum melting is practicable because it requires less energy to melt the same amount of aluminum.