Question

Use the following data to determine the maximum rate at which a standard man can climb a mountain: Blood contains 15.7 wt% hemoglobin (with molecular weight 65,000 g/mol). Each hemoglobin molecule can carry four oxygen molecules. The heart pumps 107 cm³/s blood of density 1.06 g/cm³. Each oxygen molecule can oxidize one sugar unit (the chemical formula per sugar unit is CH₂O, which is an organic alcohol group) to CO₂ and H₂O; the oxidation of 1 g sugar yields about 17 kJ of energy, of which 25% can be used to do muscle work. (Assume the climber has a mass of 60 kg.)

Answer 1

given

Blood contains 15.7 wt% hemoglobin

molecular weight 65,000 g/mol

heart pumps 107 cm³/s

blood of density 1.06 g/cm³

the oxidation of 1 g sugar

yields about 17 kJ of energy

which 25% can be used to do muscle work

the climber has a mass of 60 kg

107 cm³/sec = 113.42 g/sec

here 15.7 wt% hemoglobin

= 15.7 x 113.42 / 100

= 17.806 g/sec

= 17.806 / 65000

= 2.7393 x 10^-4 moles/sec

to calculate the O₂ rate

= 2.7393 x 10^-4 x 4

= 1.0957 x 10^-3 moles/sec

CH₂O + O₂ ---------> CO₂ + H₂O

= 30 gm / 1 mole O₂

= 30 x 1.0957 x 10^-3 g/sec

= 32.872 mg/sec

the output energy = 32.872 x 10^-3 x 17 x 10³

= 558.8344 J/sec

the energy is used by the muscle = 25 % of 558.8344

= 0.25 x 558.8344

= 139.7086 J/m

the energy is required per meter is = m g h

= 60 x 9.8 x 1

= 588 J/m

so the rate of climbing is = 139.7086 / 588

= 0.237599 m/sec

the rate of climbing is = 23.7599 cm/sec