At a certain location, the solar power per unit area reaching Earth's surface is 200 W/m^2, averaged over a 24-hour day.If the average power requirement in your home is 2.8 kW and you can convert solar power to electric power with 8.0 % efficiency, how large a collector area will you need to meet all your household energy requirements from solar energy?

To determine the size of a collector area needed to meet your household energy requirements from solar energy, we need to calculate the amount of solar power that can be converted to electric power and compare it to your average power requirement.

First, we need to convert the average power requirement from kilowatts to watts:
2.8 kW * 1000 = 2800 watts.

Next, we need to calculate the amount of solar power that can be converted to electric power. The efficiency of the conversion is given as 8.0%, which means only 8.0% of the solar power can be converted to electric power.

Solar power available = 200 W/m^2 (provided in the question).

To find the electric power that can be generated:
Electric power = Solar power available * Efficiency
Electric power = 200 W/m^2 * 8.0% = 16 W/m^2.

Now, we can calculate the collector area needed to meet the average power requirement:
Collector area = Average power requirement / Electric power per unit area
Collector area = 2800 watts / 16 W/m^2.

Dividing the total watts by the watts per square meter cancels out the units, leaving us with square meters as the unit for collector area.

Collector area = 2800 W / 16 W/m^2 ≈ 175 square meters.

Therefore, you will need a collector area of approximately 175 square meters to meet all your household energy requirements from solar energy.