The basal rate of consumption of O_2 by a 70. kg human is about 17 mol of O_2 pe

Question

The basal rate of consumption of O_2 by a 70. kg human is about 17 mol of O_2 per day. This O_2 oxidizes food and is reduced to H_2O, providing energy for the organism. O_2 + 4 H^+ + 4 e^- 2 H_2O to what current (in amperes = C/s) does this respiration rate correspond? (Current is defined by the flow of electrons from food to O_2.) 7.6E1 A Compare your answer in part (a) with the current drawn by a refrigerator using 5.20 times 10^2 W at 119 V. Remember that power (in watts) = work/s = E middot I. 4.37 times more current than the refrigerator If the electrons flow from nicotinamide adenine dinucleotide (NADH) to O_2, they experience a potential drop of 1.1 V. What is the power output (in watts) of our human friend? 8.3E1 W

Explanation / Answer

For option b

I= 5.20*10^2/119

=>I= 4.69A

Hence the required answer is 4.69 A

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