Question

on a day when air temperature is 34°c (normal skin temperature) a cyclist is maintaining a speed of 15 km/hr. how many grams of water must this cyclist evaporate each minute to get rid of the body heat produced by his activity?

Answer 1

$Fv=(1-\eta)P,$

$Mgv=\frac{(1-\eta)(Lm+cm(100°-t°))}{t},$

$m=\frac{Mgvt}{(1-\eta)(L+c(100°-t°))},$

$m=\frac{82\cdot 9.8\cdot 15/3.6\cdot 60}{(1-0.2)\cdot(2260000+4200\cdot (100-34))}=100~g.$