The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1J/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 (s) + 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= %

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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The 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 =
%
Transcribed Image Text:The 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 = %
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