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Evaporation of Water on a Jogger

  • Thread starter Boozehound
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  • #1
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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
 

Answers and Replies

  • #2
G01
Homework Helper
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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.
 
  • #3
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yes that was it...i used 3350 instead of 3550 on accident. thank you.
 
  • #4
G01
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No problem!
 

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