Calculate the mass of glucose metabolized by a 60.3 kg person in climbing a moun

Question

Calculate the mass of glucose metabolized by a 60.3 kg person in climbing a mountain with an elevation gain of 1470 m . Assume that the work performed in the climb is four times that required to simply lift 60.3 kg by 1470 m .

Express your answer to three significant figures and include the appropriate units.

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 77.0 % efficiency. Use the data below to answer questions about the metabolism of glucose.

Explanation / Answer

Answer:

W = mgh, Where m = mass of person in kg, g = 9.80 m/s2 , and h = height in meters.

W = 60.3 kg × (9.80 m/s2 ) × (1470 m) = 0.87 * 106 kg.m2/s2 = 0.87 * 106 J

Actual work = 4W = 3.474 x 106 J = 3.47 x 103 kJ

Molar mass (C6H12O6) = 180.16 g/mol

Standard enthalpy of combustion of glucose = 2805 kJ/mol

Moles glucose = (3.47 * 103 kJ) / (2805 kJ/mol) = 1.237 mol

Mass glucose = 1.237 mol * (180.16 g) / (1 mol) = 222.858 gm.

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