A pump raises 650 tonnes of seawater 55 m in 6 hours. What is the power required to operate the pump if the efficiency is 100%?

To determine the power required to operate the pump under the given conditions, we need to apply the formula for power, which is defined as the work done over time. In this case, the work involves lifting a mass (the seawater) against gravity, which can be calculated using the equation for gravitational potential energy:

Work (W) = mass (m) × gravitational acceleration (g) × height (h).

Given that seawater has a density of approximately 1,025 kg/m³, 650 tonnes of seawater converts to:

[ 650 \text{ tonnes} = 650,000 \text{ kg} ]

The gravitational acceleration is approximately 9.81 m/s², and the height the pump raises the seawater is 55 m. Plugging these values into the equation gives:

[ W = 650,000 \text{ kg} \times 9.81 \text{ m/s}² \times 55 \text{ m} ]

Calculating this yields:

[ W = 650,000 \times 9.81 \times 55 ]

[ W = 3,552,075 \text{ J} ]

Now, to find the power (P),