Evaporation of Water on a Jogger

[Boozehound]

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 I am pluggin in some wrong numbers somewhere. any help is appreciated

 

[G01]

I think you may have a calculator error, for Q=cmt, I get Q=(3550)(71.9)(1.10)=28769.5J. This may be your problem. See what you get as an answer using this value for Q.

 

[Boozehound]

yes that was it...i used 3350 instead of 3550 on accident. thank you.

 

[G01]

No problem!