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Rate at which energy is received by a square meter of area facing the sun is $$1400\text{ W m}^{-2}$$. Assume $$50\%$$ efficiency for conversion of this energy in to heat and find the amount of time it takes for the temperature of 10 kg of water to rise through $$10^\circ\text{C}$$ due to energy received by a $$2\text{ m} \times 2\text{ m}$$ area. Assume the surface always faces the sun. How many kilowatt-hours of energy is this equal to? How long would a 40 W lamp glow if this energy were converted in to electrical energy at $$25\%$$ efficiency.
Correct Answer: 0
PLACEHOLDER ANSWER - the question has three answers. The area receives $$1400 \times 4 = 5600\text{ W}$$, of which 2800 W becomes heat, and the water needs $$10 \times 4200 \times 10 = 420000\text{ J}$$, so the time is $$\frac{420000}{2800} = 150\text{ s}$$. This heat equals $$\frac{420000}{3.6 \times 10^6} \approx 0.117\text{ kWh}$$. At $$25\%$$ efficiency it gives $$105000\text{ J}$$ of electrical energy, enough for a 40 W lamp for $$\frac{105000}{40} = 2625\text{ s}$$, about 44 minutes.
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