Answer to Question #116840 in Molecular Physics | Thermodynamics for Rose

Question

The maximum rate at which high performance athletes can do work is often limited by the rate at which they can get rid of heat from their body.

Assume that it is a warm day and an athletes only means of getting rid of body heat is the evaporation of sweat (water) from their skin. Assume the certain athlete does mechanical work with an efficiency of 15% and is capable of rehydrating at a rate of 1.5 litres of water per hour.

How much work can this athlete can do in one hour without dehydrating and while maintaining a constant body temperature? (to 2 s.f and in MJ)

[Lvap-water = 2560×103 J kg−1 at body temperature, for water 1 litre = 1 kg, assume all sweat evaporates]

Answer 1

$Q=\lambda \times m=2560\times 10^{3}\times1.5=3840\times10^3J;\\3840\times10^3-85\%\\x-15\%\\x\times 85=15\times 3840\times10^3;\\A=\frac{15\times 3840\times10^3}{85}=677647.1J;\\Answer:A=677.647kJ$