The latent heat of vaporization of H2O at body temperature (37.0°C) is 2.39E+6 J/kg. To cool the body of a 74.?
The latent heat of vaporization of H2O at body temperature (37.0°C) is 2.39E+6 J/kg. To cool the body of a 74.4 kg jogger [average specific heat capacity = 3550 J/(kg*°C)] by 1.30°C, how many kilograms of water in the form of sweat have to be evaporated?
[74.4(3550)(37-1.3)]/( 2.39E+6) =3.459
my home work says i am wrong
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The problem I see immediately is the change in temp you used should be 1.30°C. So 74.4 x 3550 x 1.3 = 343,356 J.
Now 343,356 / 2.39E6 = 0.144 kg or 144 g of water in the form of sweat.