Discussion
A solar pond has electricity generating capacity of 600 MWe. If the efficiency of solar energy to electric generation process was 2% and solar energy supply rate was 300 W/m\(^{2}\), what is the area of solar pond ?
Answer: A
The thermal energy needed is the electric power divided by the efficiency, and the area is that power divided by the solar supply rate.
\[ P_{solar} = \frac{600 \text{ MW}}{0.02} = 30000 \text{ MW} \]
\[ A = \frac{3 \times 10^{10}}{300} = 10^{8} \text{ m}^2 \]
\[ A = 100 \text{ km}^2 \]
So the answer is 100 km\(^{2}\), option A. The other areas do not follow from the data.
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