An athlete in the weight room lifts a 50 kg mass through a vertical distance of 2.0 m. The mass is a — Thermodynamics and Thermochemistry Chemistry Question
Question
An athlete in the weight room lifts a 50 kg mass through a vertical distance of 2.0 m. The mass is allowed to fall through the 2.0 m distance while coupled to an electrical generator. The electrical generator produces an equal amount of electrical work, which is used to produce aluminium by Hall electrolytic process. $\text{Al}_2\text{O}_3\text{(solution)} + 3\text{C(graphite)} \to 2\text{Al(l)} + 3\text{CO(g)}; \Delta G^\circ = 600\text{ kJ}$. How many times must the athlete lift the 50 kg mass to provide sufficient Gibbs energy to produce 27 g Al? ($g = 10\text{ m/s}^2$)
💡 Solution & Explanation
Work per lift = $mgh = 50 \times 10 \times 2.0 = 1000\text{ J}$. The reaction produces 2 moles of Al ($54\text{ g}$) for $600\text{ kJ}$. To produce $27\text{ g}$ (1 mole) of Al requires $600 / 2 = 300\text{ kJ} = 300,000\text{ J}$ of energy. Number of lifts needed = $300,000 / 1000 = 300$. Format as 0300.