The latent heat of vaporizaiton of H2O at body temperature (37.0°C) is 2.42 106 J/kg. To cool the body of a 89 kg jogger (average specific heat capacity = 3500 J/(kg·C°)) by 1.4 C°, how many kilograms of water in the form of sweat have to be evaporated?

Answer 1

masssweat*Lv=massbody*c*deltaTemp

Answer 2

(89)(3500)(1.4)=M(2.42E^6)
436100=M(2.42E^6)
M=.180