Reference no: EM13883412
A human being (75 kg ) consumes about 6000 Kj of food per day. Assume that the food is all glucose and that overall reaction is- C6H12O6 +6O2 ------à 6CO2 +6H20 , Δ H= -2816 KJ
Find man's metabolic rate ( the rate of living ,loving and laughing ) in terms of moles of oxygen used per m3 of person per second.
Solution -
Ro2= -ro2 = - 1/V*(Dno2/dt ) = mol O2 used/m3 of person -s
Let us evaluate the two terms in this equation. First of all, from our life experience , we estimate the density of man to be-
Ρ= 1000 kg/m3
So, for the person in the question-
Vperson = 75/1000 =0.075 m3
Next , noting that each mole of glucose consumed uses 6 moles of oxygen and releases 2816 kj of energy,
DNO2/dt= dno2/dt =6000/2816 *1/6 =12.8 mol O2/day -ro2 '' = 1/0.075*12.8 mol o2 used/day *1/24*3600 =0.002 mol -O2/M3-S
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