Chemistry In Focus
7th Edition
ISBN: 9781337399692
Author: Tro, Nivaldo J.
Publisher: Cengage Learning,
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Chapter 10, Problem 36E
Interpretation Introduction
Interpretation:
Whether an average car has enough surface area to meet its power needs with solar power or not is to be determined.
Concept Introduction:
Power input on the solar cells is the total active area of the solar cells multiplied by solar power. The percent of the input power that is converted to output power by a solar cell is called its efficiency. Percent efficiency of a solar cell is
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Chapter 10 Solutions
Chemistry In Focus
Ch. 10 - Prob. 10.1YTCh. 10 - Prob. 10.2YTCh. 10 - Prob. 1SCCh. 10 - Prob. 1ECh. 10 - Explain why hydroelectric power and wind power are...Ch. 10 - Prob. 3ECh. 10 - Prob. 4ECh. 10 - Prob. 5ECh. 10 - Prob. 6ECh. 10 - Prob. 7E
Ch. 10 - Prob. 8ECh. 10 - Prob. 9ECh. 10 - Prob. 10ECh. 10 - Prob. 11ECh. 10 - Prob. 12ECh. 10 - Prob. 13ECh. 10 - Prob. 14ECh. 10 - Prob. 15ECh. 10 - How does a geothermal power plant generate...Ch. 10 - Prob. 17ECh. 10 - Prob. 18ECh. 10 - Prob. 19ECh. 10 - Prob. 20ECh. 10 - Prob. 21ECh. 10 - Prob. 22ECh. 10 - Prob. 23ECh. 10 - Prob. 24ECh. 10 - Prob. 25ECh. 10 - Prob. 26ECh. 10 - Prob. 27ECh. 10 - Prob. 28ECh. 10 - Prob. 29ECh. 10 - Prob. 30ECh. 10 - Prob. 31ECh. 10 - Prob. 32ECh. 10 - Prob. 33ECh. 10 - Prob. 34ECh. 10 - Prob. 35ECh. 10 - Prob. 36ECh. 10 - Prob. 37ECh. 10 - Prob. 38ECh. 10 - Prob. 42ECh. 10 - Prob. 43E
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- The use of solar power illustrates which of the following green chemistry principles? Design for less waste Design for energy efficiency Atom economy Design for less hazards Design using benign chemicalsarrow_forwardElectrical power is commonly measured in watts (W) or in kilowatts (kW). A commercial solar panel generates about 0.10 watts per square inch of surface area. Based on this, how many kilowatts of power could be generated by a residential solar panel array with a surface area of 621 ft2? amount of power: kWarrow_forwardonsider a cloudless day on which the sun shines down across the United States. If 2751 kJ of energy reaches a square meter m²) of the United States in one hour, how much total solar energy reaches the entire United States per hour? The entire area of the United States is 9,158,960 km². solar energy per hour: kJ/h TOOLS x10 218 PM 9/3/2021 64 F Clear ENGarrow_forward
- The sun supplies about 1 kilowatt(s) of energy for each square meter of surface area (1 kW/m2kW/m2, where a watt = 1 J/sJ/s). Plants produce the equivalent of about 0.19 gg of sucrose (C12H22O11C12H22O11) per hour per square meter.arrow_forwardDefine the terms renewable and nonrenewable as applied to energy resources. Which of the following energy resources are renewable: solar energy, coal, natural gas, geothermal energy, wind power?arrow_forwardAssume that electricity costs 15 cents per kilowatt- hour. Calculate the monthly cost of operating each of the following: a 100 W light bulb, 5 h/day a 600 W refrigerator, 24 h/day a 12,000 W electric range, 1 h/day a 1000 W toaster, 10 min/dayarrow_forward
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- When an electrical appliance whose power usage is X watts is run for Y seconds, it uses X × Y joules of energy. The energy unit used by electrical utilities in their monthly bills is the kilowatt-hour (kWh, that is, 1 kilowatt used for 1 hour). How many joules are there in a kilowatt-hour? If electricity costs $.09 per kilowatt-hour, how much does it cost per megajoule?arrow_forwardThe sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 21 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha = 10,000 m Assuming that sucrose is produced by the reaction m²). 12CO2 (9) + 11H2O(1) → C12H22O11(s) + 120O2(g) AH=5640 kJ calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field. Percent efficiencyarrow_forwardThe sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 23 kg of sucrose (C12H22 O11) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by the reaction 12CO₂(g) + 11H₂O(l) → C12H22 O11 (8) + 1202 (9) AH = 5640 kJ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis in this field. Percent efficiency = %arrow_forward
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