The sun supplies energy at a rate of about $1.0$ kilowatt per square meter of surface area ( 1 watt $=1 \mathrm{~J} / \mathrm{s}$ ). The plants in an agricultural field produce the equivalent of $20 . \mathrm{kg}$ sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ per hour per hectare ( $1 \mathrm{ha}=10,000 \mathrm{~m}^{2}$ ). Assuming that sucrose is produced by the reaction
$12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g)$
$\Delta H=5640 \mathrm{~kJ}$
calculate the percentage of sunlight used to produce the sucrosethat is, determine the efficiency of photosynthesis.