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Rate at which energy is incident on $$1\text{ m}^2$$ area facing the sun is 1400 W. The average area of the roof top of a moderate home is $$150\text{ m}^2$$. Since the angle of incidence of the energy from sun on the roof top varies during the day, the average rate for a 12 hour duration is half of the maximum rate of incidence at the noon time. Find the energy received in 12 hour duration by the roof of the building. Assuming yield of 200 MeV per reaction in the fission of uranium nucleus, find the number of reactions that produce the same amount of energy as provided by the sun to the roof in the 12 hour duration.
Correct Answer: 0
PLACEHOLDER ANSWER - the question has two answers. The average power on the roof is $$\frac{1}{2} \times 1400 \times 150 = 105000\text{ W}$$, so in 12 hours the energy is $$105000 \times 43200 \approx 4.54 \times 10^9\text{ J}$$. Each fission gives $$200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\text{ J}$$, so the number of reactions is $$\frac{4.536 \times 10^9}{3.2 \times 10^{-11}} \approx 1.42 \times 10^{20}$$.
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