Chapter 5: Problem 116
The Sun supplies about 1.0 kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{kW} / \mathrm{m}^{2},\) where a watt \(=1 \mathrm{J} / \mathrm{s}\right)\) 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. $$\begin{array}{c}{12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\Delta H=5645 \mathrm{kJ}}\end{array}$$
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