Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
15
The Laws of Thermodynamics
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15-1 and 15-2: First Law of Thermodynamics
15-3: Human Metabolism
15-5: Heat Engines
15-6: Refrigerators, Air Conditioners, Heat Pumps
15-7: Entropy
15-10: Statistical Interpretation
15-11: Energy Resources

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 53
Q

Solar cells (Fig. 15–26) can produce about 40 W of electricity per square meter of surface area if directly facing the Sun. How large an area is required to supply the needs of a house that requires 24 kWh/day? Would this fit on the roof of an average house? (Assume the Sun shines about 9 h/day.)

Problem 53.
Figure 15-26.
A
72 m2 Yes, the panel will fit on the roof72 \textrm{ m}^2 \textrm{ Yes, the panel will fit on the roof}
Giancoli 7th Edition, Chapter 15, Problem 53 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli answers with Mr. Dychko. One square meter of solar panel can produce 40 watts of power. And times that by 24 times ten to the three watt hours per day of the requirements for a typical house. And then times by one day for every nine hours and you see that the units work out here nicely to get a square meters and hours cancel there and the days cancel there and the watts cancel there and we're left with 66.67 square meters or about 70 square meters of area required for solar panels. A typical house, typical house might have a thousand square feet area. And let's assume it has a flat roof to make our calculations easy. And… and and times it by one meter for every 3.28 feet squared to convert it into 92.95 square meters. And so yes the panels will fit because the total roof area of 93 is greater than the area required for the solar panels of 70.

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