Calculate the mass of glucose metabolized by a 83.1 kg person in climbing a moun

Question

Calculate the mass of glucose metabolized by a 83.1 kg person in climbing a mountain with an elevation gain of 1510 m . Assume that the work performed in the climb is four times that required to simply lift 83.1 kg by 1510 m

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

The metabolism of glucose, C6H1206, yields carbon dioxide, CO2(g), and water, H2O(1), 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 Substance/mol) CO2(g) -393.5 C6 H1206 (s)1273.3 H20() 285.8 O2 (g) 0

Explanation / Answer

the reaction is C6H12O6 (s) + 6O2 (g)---------->6CO2(g)+6H2O (g)

enthalpy of reaction = sum of enthalpy change of products- sum of enthalpy change of reactants

where 6,6, 1 and 6 are coefficients of CO2, H2O, C6H12O6 and O2 respectively

=6*(-393.5)+6*(-285.8)- (-1273.3)= -2802.5 Kj

out of this energy, 63% is got converted to work= 2802.5*0.63 KJ=1765.575 Kj this is the energy released per mole of glucose, moles= mass/molar mass , for 1 mole , mass of glucose= 180 gm

energy required in rising to the height= mass* g(9.8)* height, but work done is 4 times this energy

wrok performed in climbing by 83.1kg person to 1510 m= 4*1510*83.1*9.8 joules=4918855 joules 765.757 Kj= 1765.575*1000 joules

1765.575*1000 joules is produced by 180 gm of glucose

4918885 joules requires 4918885*180/(1765.575*1000)=501.5 gm glucose

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