The solar insolation in Sydney in January is I ≈ 5 kilowatt hours/m2/day. This is the energy per unit area per unit time received at ground level.An area of 0.5 m2 is exposed to the solar insolation I for 10 minutes.How much energy is absorbed in kilojoules? Give your answer to one significant figure.
Question
The solar insolation in Sydney in January is I ≈ 5 kilowatt hours/m2/day. This is the energy per unit area per unit time received at ground level.An area of 0.5 m2 is exposed to the solar insolation I for 10 minutes.How much energy is absorbed in kilojoules? Give your answer to one significant figure.
Solution
First, we need to convert the solar insolation from kilowatt hours per square meter per day to kilowatt hours per square meter per minute, since the area is exposed for 10 minutes.
There are 1440 minutes in a day (24 hours x 60 minutes), so the solar insolation in kilowatt hours per square meter per minute is:
I = 5 kWh/m²/day ÷ 1440 minutes/day = 0.00347 kWh/m²/minute.
Next, we calculate the total energy absorbed by the 0.5 m² area in 10 minutes. This is the solar insolation multiplied by the area and the time:
Energy = I x area x time = 0.00347 kWh/m²/minute x 0.5 m² x 10 minutes = 0.01735 kWh.
Finally, we need to convert this energy to kilojoules. 1 kilowatt hour is equivalent to 3,600 kilojoules, so:
Energy = 0.01735 kWh x 3,600 kJ/kWh = 62.46 kJ.
So, the energy absorbed is approximately 60 kJ to one significant figure.
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