# Step-by-Step Solution Given: - Average daily energy use: \(83,835\) kJ (per day) --- ## a. How many 100W lightbulbs could you run for 24 hours with this energy? ### Step 1: Convert energy to kilowatt-hours (kWh) 1 kWh = 3,600 kJ \[ \text{Daily energy (kWh)} = \frac{83,835\, \text{kJ}}{3,600\, \text{kJ/kWh}} \approx 23.29\, \text{kWh} \] ### Step 2: Energy used by one 100W bulb in 24 hours - Power = 100 W = .1 kW - Time = 24 hours \[ \text{Energy per bulb per day} = .1\, \text{kW} \times 24\, \text{h} = 2.4\, \text{kWh} \] ### Step 3: Maximum number of bulbs \[ \text{Number of bulbs} = \frac{23.29\, \text{kWh}}{2.4\, \text{kWh/bulb}} \approx 9.7 \] **Final answer (a):** You could run about **9.7** (or **9 full**) 100W lightbulbs for 24 hours. --- ## b. If this energy is used for a 1,000 W microwave, how many hours can it run? ### Step 1: Power and energy - Microwave power = 1,000 W = 1 kW - Total energy available: 23.29 kWh ### Step 2: Time it can operate \[ \text{Time} = \frac{\text{Total energy}}{\text{Power}} = \frac{23.29\, \text{kWh}}{1\, \text{kW}} = 23.29\, \text{hours} \] **Final answer (b):** The microwave would run for **23.3 hours** before the energy runs out. --- ## c. Annual energy use: Why does 84,000 kJ/day ≈ 8,500 kWh/year? ### Step 1: Daily to annual energy (in kJ) \[ \text{Annual energy (kJ)} = 84,000\, \text{kJ/day} \times 365\, \text{days/year} = 30,660,000\, \text{kJ/year} \] ### Step 2: Convert to kWh \[ \text{Annual energy (kWh)} = \frac{30,660,000\, \text{kJ}}{3,600\, \text{kJ/kWh}} \approx 8,517\, \text{kWh} \] ### Step 3: Explanation - 8,500 kWh/year is a rounded value from the above calculation. - Actual daily energy can vary, but the average is around this value. **Final answer (c):** A home using **84,000 kJ/day** on average uses about **8,500 kWh/year** because: \[ 84,000\, \text{kJ/day} \times 365\, \text{days} \div 3,600\, \text{kJ/kWh} \approx 8,500\, \text{kWh} \]