How much perspiration must vaporize per hour to dissipate extra energy

[Jtwa]

Homework Statement

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.

Homework Equations

Q=mcΔT
Q=mHv

The Attempt at a Solution

Thermal energy to be dissipated in 1.00h is
U=(30J/s)(3600s/hr)=108KJ/hr

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

I'm not sure if I'm doing it correctly. I never used 37 degrees/

 

[mfb]

I'm not sure if I'm doing it correctly. I never used 37 degrees/

Did you have body temperatures in previous problems? Maybe the players start at lower temperatures?

 

[NascentOxygen]

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

I'm not sure if I'm doing it correctly. I never used 37 degrees/

You use 37° when reading from tables to determine what value of ∆H_vap to use because it is not a fixed value across all temperatures, ∆H_vap falls slightly as temperature rises. http://www.thermexcel.com/english/tables/vap_eau.htm