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Thermal energy to be dissipated in an hour

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data

    During the game, the metabolism of basketball players often increases by as much as 30.0 W. How much perspiration must a player vaporize per hour to dissipate this extra thermal energy? Assume that perspiration is simply pure water and that perspiration starts at temperature of 37 degrees celcius.

    2. Relevant equations

    Q=mcΔT
    Q=mHv

    3. The attempt at a solution

    Thermal energy to be dissipated in 1.00h is
    U=(30J/s)(3600s/h)=1.08x10^5J

    The amount of water this energy transmittes as heat would vaporize is,
    m=(1.08x10^5J)/(2.26X10^6J/kg)= 4.78x10^-2kg

    ----> Just looking for someone to double check my reasoning/calculations and to make sure the number of significant figures has been respected. Thank you so much in advance!
     
  2. jcsd
  3. Mar 25, 2012 #2

    gneill

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    Staff: Mentor

    You don't seem to have taken into account the starting temperature of the perspiration.
     
  4. Mar 25, 2012 #3
    your calculation for the energy needed to vaporize that amount of water is correct.
     
  5. Mar 25, 2012 #4

    gneill

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    Staff: Mentor

    Hmm. The heat of vaporization for water varies with temperature. At the boiling point (100C) it's 2.257 x 106 J/kg, but at 37C it's more like 2.40 x 106 J/kg.
     
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