Respuesta :
Explanation:
Moles of sucrose = [tex]\frac{\text{mass of sucrose}}{\text{molecular weight}}[/tex]
= [tex]\frac{24 kg \times 1000 g/kg}{342.3 g/mol}[/tex]
= [tex]\frac{70.114 mol}{hr-ha}} \times \frac{1 ha}{10000 m^{2}} \times \frac{1 hr}{3600 s}[/tex]
= [tex]1.948 \times 10^{-6} mol/(m^{2}s)[/tex]
Energy required for photosynthesis = 5640 kJ/mol
Energy needed to produce 24 kg sucrose
= Energy required for photosynthesis x Moles of sucrose
= [tex](5640 kJ/mol) \times [1.948 x 10^{-6} mol/(m^{2}s)][/tex]
= [tex]0.01098 kJ/(m^{2}s)[/tex]
Energy supplied from sun = [tex]1 \frac{kW}{m^{2}}[/tex]
= 1 [tex]kJ/(m^{2}s)[/tex]
Efficiency for photosynthesis = [tex]\frac{\text{Energy needed to produce 24 kg sucrose}}{\text{Energy supplied from sun}}[/tex]
= [tex]\frac{0.01098 kJ/(m^{2}s)}{1 kJ/(m^{2}s)}[/tex]
= [tex]0.01098 \times 100[/tex]
= 1.098%
Thus, we can conclude that efficiency of photosynthesis for a plant that is being considered as a source of biofuels is 1.098 %.
