1. **Problem statement:** Calculate the amount of heat energy (in kJ) delivered by an electric boiler with a power rating of 7.5 kW over a period of one hour. 2. **Formula used:** Power ($P$) is the rate of energy transfer, so energy ($Q$) delivered over time ($t$) is given by: $$Q = P \times t$$ where - $Q$ is energy in kilojoules (kJ), - $P$ is power in kilowatts (kW), - $t$ is time in hours (h). 3. **Important rules:** - 1 kilowatt (kW) = 1000 watts (W), but since power is given in kW and time in hours, the energy will be in kilowatt-hours (kWh). - To convert kWh to kJ, use the conversion factor: $$1\text{ kWh} = 3600\text{ kJ}$$ 4. **Calculation:** Given: $$P = 7.5\text{ kW}, \quad t = 1\text{ hour}$$ Calculate energy in kWh: $$Q = 7.5 \times 1 = 7.5\text{ kWh}$$ Convert to kJ: $$Q = 7.5 \times 3600 = 27000\text{ kJ}$$ 5. **Answer:** The electric boiler delivers **27000 kJ** of heat energy in one hour.