A pump lifts 250 kg of water to a height of 12 m in 2 minutes. Calculate the power dissipated by the pump.

Work done ($W$) is equal to the potential energy gained by the water.

$$W = mgh$$ $$W = 250 \text{ kg} \times 10 \frac{\text{m}}{\text{s}^2} \times 12 \text{ m}$$ $$W = 30000 \text{ J}$$

Power ($P$) is work done divided by time.

$$P = \frac{W}{t}$$ $$P = \frac{30000 \text{ J}}{120 \text{ s}}$$ $$P = 250 \text{ W}$$

The power dissipated by the pump is $\text{250 W}$.