Question

Question: Mountain climbers carry bottled oxygen when at very high altitudes.

(a) Assuming that a mountain climber uses oxygen at twice the rate for climbing 116 stairs per minute (because of low air temperature and winds), calculate how many liters of oxygen a climber would need for 10.0 h of climbing. (These are liters at sea level.) Note that only 40% of the inhaled oxygen is utilized; the rest is exhaled.

(b) How much useful work does the climber do if he and his equipment have a mass of 90.0 kg and he gains 1000 m of altitude?

(c) What is his efficiency for the 10.0-h climb?

Short answer

(a) The total liters of oxygen needed is \(5880{\rm{ L}}\).

(b) The useful work done is \(882000{\rm{ J}}\).

(c) The efficiency of his climb is \(1.79\% \).

Step-by-step solution

Energy consumption

Since the digestive process is basically one of oxidizing food, energy consumption is directly proportional to the rate of consumption of oxygen.

Amount of oxygen needed

(a)

The required amount of oxygen for climber is,

\({V_{req}} = \frac{{{V_{stair}}}}{\eta } \times t\)

Here, \({V_{stair}}\) is the oxygen consumption rate for climbing stairs \(\left( {{V_{stair}} = 1.96{\rm{ L}}/{\rm{min}}} \right)\), \(\eta \) is the amount of inhaled oxygen is utilized \(\left( {\eta = 40\% } \right)\), and t is the time of climbing \(\left( {t = 10.0{\rm{ h}}} \right)\).

Putting all known values,

\(\begin{aligned} {V_{req}} &= \frac{{\left( {1.96{\rm{ L}}/{\rm{min}}} \right)}}{{40\% }} \times \left( {10.0{\rm{ h}}} \right)\\ &= \frac{{\left( {1.96{\rm{ L}}/{\rm{min}}} \right)}}{{40 \times \frac{1}{{100}}}} \times \left( {10.0{\rm{ h}}} \right) \times \left( {\frac{{60{\rm{ min}}}}{{1{\rm{ h}}}}} \right)\\ &= 2940{\rm{ L}}\end{aligned}\)

At high altitude the climber uses oxygen at twice rate. Therefore, total amount of oxygen required,

\(V = 2 \times {V_{req}}\)

Putting all known values,

Therefore, the required total liters of oxygen needed is \(5880{\rm{ L}}\).

Amount of useful work done by the climber

(b)

Work done in climbing is given as,

\(W = mgh\)

Here, m is the mass of the climber \(\left( {m = 90{\rm{ kg}}} \right)\), g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and h is the altitude \(\left( {h = 1000{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned} W &= \left( {90{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {1000{\rm{ m}}} \right)\\ &= 882000{\rm{ J}}\end{aligned}\)

Therefore, the useful work done is \(882000{\rm{ J}}\).

Efficiency

(c)

The power needed to climbing stairs is,

\({P_{stair}} = 685{\rm{ W}}\)

At high altitude the climber uses oxygen at twice rate. Therefore, total power input is,

\({P_{in}} = 2{P_{stair}}\)

Putting all known values,

\(\begin{aligned} {P_{in}} &= 2 \times \left( {685{\rm{ W}}} \right)\\ &= 1370{\rm{ W}}\end{aligned}\)

The work input is,

\({W_{in}} = {P_{in}}t\)

Here, \({P_{in}}\) is the total input power \(\left( {{P_{in}} = 1370{\rm{ W}}} \right)\), and t is the time of climbing \(\left( {t = 10.0{\rm{ h}}} \right)\).

Putting all known values,

\(\begin{aligned} {W_{in}} &= \left( {1370{\rm{ W}}} \right) \times \left( {10{\rm{ hr}}} \right)\\ &= \left( {1370{\rm{ W}}} \right) \times \left( {10{\rm{ hr}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hr}}}}} \right)\\ &= 49320000{\rm{ J}}\end{aligned}\)

The efficiency of the work done is,

\(\eta = \frac{W}{{{W_{in}}}} \times 100\% \)

Here, W is the work done in climbing \(\left( {W = 882000{\rm{ J}}} \right)\), \({W_{in}}\) is the work input \(\left( {{W_{in}} = 49320000{\rm{ J}}} \right)\).

Putting all known values,

\(\begin{aligned} \eta &= \frac{{882000{\rm{ J}}}}{{49320000{\rm{ J}}}}\\ &= 1.79\% \end{aligned}\)

Therefore, the required efficiency of his climb is \(1.79\% \).