To calculate the volume of oxygen consumed in 1 hour by an average man with the given Vo2 max reading, we can follow these steps:
-
Convert the weight of the man to kilograms: \[ \text{Weight in kg} = \text{Weight in lbs} \times 0.453592 \] \[ \text{Weight in kg} = 181 , \text{lbs} \times 0.453592 \approx 82.1 , \text{kg} \]
-
Calculate the total volume of oxygen consumed per minute using the Vo2 max: \[ \text{Vo2 max (mL O2/min)} = \text{Vo2 max (mL O2/kg/min)} \times \text{Weight in kg} \] \[ \text{Vo2 max} = 48.0 , \text{mL O2/kg/min} \times 82.1 , \text{kg} \approx 3946.8 , \text{mL O2/min} \]
-
Calculate the total volume of oxygen consumed in 1 hour: \[ \text{Volume in 1 hour (mL)} = \text{Volume per minute (mL)} \times 60 , \text{minutes} \] \[ \text{Volume in 1 hour} = 3946.8 , \text{mL O2/min} \times 60 , \text{min} \approx 236808 , \text{mL O2} \]
Therefore, the volume of oxygen consumed in 1 hour by an average man weighing 181 pounds and having a Vo2 max of 48.0 mL O2/kg body mass/min is approximately 236,808 mL or 236.8 liters.
1 answer