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The latent heat of vaporization of H2O at body temperature (37.0°C) is 2.38E+6 J/kg. To cool the body of a 71.9 kg jogger [average specific heat capacity = 3550 J/(kg*°C)] by 1.10°C, how many kilograms of water in the form of sweat have to be evaporated?
Q=cmt Q=mL
Q=3550(71.9)1.10=264951.5J
264951.5J=m(2.38E6) m=.111kg
the answer is wrong and i think im pluggin in some wrong numbers somewhere. any help is appreciated
Q=cmt Q=mL
Q=3550(71.9)1.10=264951.5J
264951.5J=m(2.38E6) m=.111kg
the answer is wrong and i think im pluggin in some wrong numbers somewhere. any help is appreciated