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 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha = the reaction 10,000 m²). Assuming that sucrose is produced by 12CO2 (g) + 11H20(1) · → C12 H22 O11 (s) + 1202(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 efficiency =

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[References]
1 J/s). The plants in an agricultural field
The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt =
produce the equivalent of 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by
the reaction
12CO2 (g) + 11H,O(1) → C12 H22 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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Transcribed Image Text:[References] 1 J/s). The plants in an agricultural field The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = produce the equivalent of 20. kg of sucrose (C12H22O11) per hour per hectare (1 ha = 10,000 m²). Assuming that sucrose is produced by the reaction 12CO2 (g) + 11H,O(1) → C12 H22 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 % Submit Answer Try Another Version 3 item attempts remaining
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