1. **Problem statement:** A pump fills a tank of capacity 300 m^3 in 5 hours. The tank is 20 m above the water level. The pump efficiency is 65%. We need to find the power of the engine driving the pump. 2. **Relevant formula:** Power required to lift water is given by $$P = \frac{\rho g h V}{t \times \text{efficiency}}$$ where: - $\rho$ is the density of water (1000 kg/m^3), - $g$ is acceleration due to gravity (9.8 m/s^2), - $h$ is height (20 m), - $V$ is volume (300 m^3), - $t$ is time in seconds (5 hours = $5 \times 3600 = 18000$ s), - efficiency is 0.65. 3. **Calculate power:** $$P = \frac{1000 \times 9.8 \times 20 \times 300}{18000 \times 0.65}$$ 4. **Simplify numerator:** $$1000 \times 9.8 \times 20 \times 300 = 1000 \times 9.8 \times 6000 = 1000 \times 58800 = 58,800,000$$ 5. **Calculate denominator:** $$18000 \times 0.65 = 11700$$ 6. **Divide:** $$P = \frac{58,800,000}{11,700}$$ 7. **Simplify fraction by canceling common factors:** $$P = \frac{\cancel{58,800,000}}{\cancel{11,700}} = 5025.64 \text{ watts}$$ 8. **Convert to kilowatts:** $$P = \frac{5025.64}{1000} = 5.03 \text{ kW}$$ 9. **Check units and efficiency:** The power calculated is the power output of the pump. Since the engine drives the pump, the engine power is the same as calculated here. **Final answer:** The power of the engine is approximately **5.03 kW**. (Note: The given answer 512 kW seems inconsistent with the problem data; the correct calculation yields about 5 kW.)