Question 1 An excersing subject inspires air at a rate of 05

Question 1.
An excersing subject inspires air at a rate of 0.5mol/min, which is on a water-resistant basis. The expired air containing 0.02 mole fraction CO2 and 0.18 mole fraction O2.How much glucose does the subject use in units of g/min? How does the value of glucose consumption compare to the basal metabolic rate?

Question 2.
Assuming that 0.11 kcal/step is the energy consumption for stair climbing, how many steps 0.15m in height woukd have to be climbed for 1 g of fat to be metabolized?

Solution

Expiration rate = 0.5 mole/min

CO₂ in expired air = 0.02 mole fraction

O₂ in expired air = 0.18 mole fraction

To calculate basal metabolic rate we can use the equation:

nO₂ = N_E (0.265(1- f_{E ,} CO₂ )- 1.265 f_{E ,} O₂ )

Putting the values in the equation we get;

nO₂ =0.5 (0.265(1- 0.02) - 1.265 * 0.18)

            = 0.016 mol O₂/min

Oxygen consumption corresponds to 0.016 mol O₂/min * 180 g glucose / 686kcal = 2.88 g / min

Basal metabolic rate in terms of glucose consumption is 72 kcal / 60 min * 180 g glucose / 686kcal = 0.31 g glucose/ min

This is about 10 times greater than the basal metabolic rate.

Basal metabolic rate: BMR is the amount of energy expended while at rest in a neutrally temperate environment.

Energy consumption for stair climbing = 0.11 kcal per step

We now that the energy content of fat is 3500 kcal/lb

therefore, after converting the unit to kcal /g we get;

= 3500 / 454 = 7.71 kcal/g

Using this value to get the number of steps;

= 0.11 kcal step^-1 * X = 7.71

Then, X = 70 steps