1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaporation of Water on a Jogger

  1. May 24, 2007 #1
    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
     
  2. jcsd
  3. May 24, 2007 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. May 24, 2007 #3
    yes that was it...i used 3350 instead of 3550 on accident. thank you.
     
  5. May 24, 2007 #4

    G01

    User Avatar
    Homework Helper
    Gold Member

    No problem!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Evaporation of Water on a Jogger
  1. The Overheated Jogger (Replies: 2)

  2. Evaporation of water (Replies: 3)

Loading...