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The surface area of a concentrator type ...

The surface area of a concentrator type solar cooker heater is `5 m^2`. It reflects `80 %` of the radiation incident on it. Calculate the enegry concentrated by the heater in `1` hour if the solar energy were delivered to it at the rate of `0.66 kW//m^2`.

Text Solution

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The correct Answer is:
`9504 kJ`
Energy concentrated `= (80//100) (0.66 kW//m^2)(5 m^2)(1 xx 60 xx 60 s) = 9504 kJ`.
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