Question

**this is a thermodynamics course**

We want to model energy and mass balance issues involved in strenuous exercise, for example the running of a marathon. Suppose a 70 kg marathon runner runs at an ≈8 minute/mile pace, completing the 26.2 mile race distance in 210 minutes, or 3.5 hours.

a.) Physiological measurements indicate that the total power consumed by an average weight runner (70 kg) at a pace of 8 minutes/mile is about 900 kcal/hour. What is the total energy (in kcal) consumed by this runner over the course of the marathon?

b.) Estimates of human mechanical efficiency vary, but there are arguments and studies to suggest that over long periods of time, the efficiency that a well-trained human athlete can convert energy into work is about 0.3 (or 30%). If we assume this value of 30% mechanical efficiency to apply here, how much mechanical work (in Joules) would the runner have produced during the marathon? How much mechanical power (in Joules/sec or Watts), assuming a steady power output for the 3.5 hour race time?

c.) We are left with the question of what happens to the 70% of
the energy consumed that isn't used to produce work. That excess
power must be converted to heat, which the runner's body must get
rid of in order to maintain a constant body temperature. By far,
the most efficient way humans have to remove heat is by evaporation
of water, either through respiration of water vapor or through
perspiration followed by evaporation from the skin. At
physiological temperature, the enthalpy of evaporation of liquid
water is 44 kJ/mol (i.e., the enthalpy of the reaction
H_{2}O (liquid) –> H_{2}O (gas) is ∆H = 44
kJ/mol). Use this value and the density of water (taken to be 1
g/ml) to calculate how much water would need to be evaporated to
carry away the excess heat (70% of the total energy consumed) that
is produced over the course of the marathon, if water evaporation
were the only process the body could use to remove heat.

d.) In order to maintain constant body weight during the marathon, the runner would need to drink as much water as was lost due to evaporation. Your answer to part c will seem like a relatively large amount of water to try to drink over 3.5 hours, though it might be possible. However, a human body has other ways of compensating for inexact energy and mass balances over short periods of time (e.g., a few hours). For example, over the course of a marathon, a healthy runner might lose 2% of their body mass, primarily due to loss of water. For our 70kg runner, how much water volume would a 2% change in body mass correspond to? How does this value compare to that of part c? Would the body's ability to tolerate changes in water content of this amount be able to help maintain body temperature while performing strenuous work over periods of a few hours? Explain your reasoning.

Answer #1

**a) Total Power consumed in 1 hour = 900
KCal/hour**

**Total Power consumed in 210 minutes or 3.5 hour = 900
KCal/hour * 3.5 hour**

**=> 900 KCal * 3.5**

**=> 3150 KCal**

**b)**

**Mechanical Work = 30/100 * Total Work**

**=> 30/100 * 3150**

**=> 945 KCal**

**Mechanical Power = Mechanical Work/Time**

**=> 945 KCal/(210*60)**

**=> 313.8 Joules/sec**

**c)**

**Excess heat = 70/100 * 3150 = 2205 KCal**

**1 Kcal = 4.184 KJ**

**Hence Energy in KJ = 2205 * 4.184 = 9225.72
KJ**

**Number of moles of water required = 9225.72/44 = 209.675
moles**

**Molar mass of Water = 18 gm/mol**

**mass of water required = 209.675 moles * 18 gm/mol =
3774.15 gms or 3.774 Kg of water**

d) Language is quite vague, information is not sufficient to calculate please check and get back to me

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