Chapter 5: Problem 110
The Sun supplies about 1.0 kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{~kW} / \mathrm{m}^{2},\right.\) where a watt \(\left.=1 \mathrm{~J} / \mathrm{s}\right)\) Plants produce the equivalent of about \(0.20 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H=5645 \mathrm{~kJ} \end{aligned} $$
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