Question

The metabolism of glucose, C6H12O6, yields carbon dioxide, CO2(g), and water, H2O(l), as products. Energy released in this metabolic process is converted to useful work, w, with about 63.0 % efficiency. Use the data below to answer questions about the metabolism of glucose. Calculate the mass of glucose metabolized by a 75.5 kg person in climbing a mountain with an elevation gain of 1870 m . Assume that the work performed in the climb is four times that required to simply lift 75.5 kg by 1870 m .

Answer 1

Potential energy of lift= 75.5* 1870* 9.8 joules=1383613 joules

Work performed =4 times the work required on lift= 4*1383613=5534452 joules

Efficiency of work =63%

Work to be performed = 5534452/0.63=8784844 joules

Energy released during glucose metabolism =1350 kcal/mole=1350*1000 cal/mole= 135000*4.2 joules/mole=567000 joules/mole

Moles of glucose required= 8784844/567000=15.49 moles

Molecular weight of glucose ( C6H12O6)= 72+12+96=180

Mass of glucose required = 15.49*180 gms =2788.2 gms =2.782 Kgs