Question

It takes 15 kWh (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is 10.7 \(\mathrm{kJ} / \mathrm{mol}(1 \text { watt }=1 \mathrm{J} / \mathrm{s}) . ]\)

Short answer

The energy required to melt 1.0 kg of aluminum metal is significantly less than the 15 kWh of energy needed to produce it from aluminum oxide. This difference in energy consumption makes recycling aluminum cans economically feasible, as less energy is expended in the recycling process compared to producing aluminum from raw materials.

Step-by-step solution

Determine the number of moles in 1 kg of aluminum metal

To find the number of moles in 1 kg of aluminum, we will use the atomic weight of aluminum (26.98g/mol). 1 kg of aluminum = 1000 g of aluminum (as kg needs to be converted to grams) Number of moles = \( \frac{1000 \ \mathrm{g}}{26.98\ \mathrm{g/mol}} \)

Calculate the energy required to melt 1 kg of aluminum metal

Using the enthalpy of fusion for aluminum metal that is 10.7 kJ/mol, we'll now calculate the energy required to melt 1.0 kg of aluminum metal. Energy required to melt 1.0 kg aluminum = Number of moles × Enthalpy of fusion Energy required to melt 1.0 kg aluminum = \( \frac{1000 \ \mathrm{g}}{26.98\ \mathrm{g/mol}} \times 10.7\ \mathrm{kJ/mol} \)

Compare the energy required for melting aluminum with the energy required to produce it from aluminum oxide

Now that we have calculated the energy required to melt aluminum, we will convert it to kilowatt-hours (kWh) and compare it with the 15 kWh energy required to produce aluminum from aluminum oxide. 1 watt = 1 J/s, therefore 1 kilowatt = 1000 J/s. 1 kilowatt-hour = 1 kW × 1 hour = 1000 J/s × 3600 s = 3.6 × 10^6 J Now, convert the energy required to melt 1.0 kg aluminum from kJ to kWh: Energy_required(kWh) = \( \frac{Energy_required(kJ) \times 10^3}{3.6 \times 10^6} \) Energy_required(kWh) = \( \frac{\left( \frac{1000 \ \mathrm{g}}{26.98\ \mathrm{g/mol}} \times 10.7\ \mathrm{kJ/mol} \right) \times 10^3}{3.6 \times 10^6} \)

Discuss the economic feasibility of recycling aluminum cans

Comparing the energy required to melt aluminum metal with the energy required to produce it from aluminum oxide, we can see that: Energy_required_for_melting_aluminum(kWh) << 15 kWh The energy required for melting aluminum is significantly less than the energy needed to produce aluminum from aluminum oxide. This shows that recycling aluminum cans saves a substantial amount of energy, making it economically feasible to recycle them.

Key concepts

Hall-Heroult process

The Hall-Heroult process is the principal method used to produce aluminum from aluminum oxide, also known as alumina. This procedure is a cornerstone in the aluminum industry due to its effectiveness. In essence, alumina is dissolved in molten cryolite and then subjected to electrolysis.
This process requires a significant amount of electrical energy, approximately 15 kWh for the production of just 1 kg of aluminum metal. The high energy requirement is primarily due to the breaking of the strong oxygen-aluminum bonds in alumina, necessitating substantial electrical input to separate oxygen and aluminum.

• Involves dissolving alumina in cryolite
• Electrolysis separates aluminum from oxygen
• Consumes about 15 kWh per kg of aluminum

The Hall-Heroult process made aluminum production economically feasible, drastically reducing the cost of aluminum and making it a prominent metal for industrial applications. However, its energy consumption remains a significant concern, especially in sustainability discussions.

enthalpy of fusion

Enthalpy of fusion, often referred to as the heat of fusion, is the amount of energy required to change a substance from a solid to a liquid at its melting point. For aluminum, the enthalpy of fusion is 10.7 kJ/mol. This means every mole of aluminum requires 10.7 kJ of energy to transition from solid to liquid.
To calculate the energy needed to melt 1 kg of aluminum, we first determine the number of moles in 1 kg of aluminum. The atomic weight of aluminum is 26.98 g/mol. Thus,

• 1 kg of aluminum = 1000 g
• Number of moles = \( \frac{1000\ \mathrm{g}}{26.98\ \mathrm{g/mol}} \)

Using this molar quantity, we calculate the energy:

• Energy Required = Number of Moles × 10.7 kJ/mol

This energy is considerably less than what is required for aluminum production via the Hall-Heroult process.

energy comparison

Comparing the energy required to produce aluminum via the Hall-Heroult process with the energy needed to melt aluminum reveals crucial insights. Production uses about 15 kWh per 1 kg of aluminum, while the melting process requires far less energy.
To compare, we convert the energy required to melt 1 kg of aluminum from kilojoules to kilowatt-hours:

• Use relation: 1 watt = 1 J/s and 1 kilowatt-hour = 3.6 × 10^6 J
• Energy required to melt aluminum (kWh) = \( \frac{\text{Energy (kJ)} \times 10^3}{3.6 \times 10^6} \)

When you do the calculations, you'll find that the energy needed to melt aluminum is significantly less than the energy used to produce it from alumina. This substantial difference highlights the economic and environmental benefits of recycling aluminum. Indeed, recycling saves a large amount of energy, making it a vital practice to meet sustainability goals.