# Thermal energy to be dissipated in an hour

1. Mar 25, 2012

### dani123

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. Mar 25, 2012

### Staff: Mentor

You don't seem to have taken into account the starting temperature of the perspiration.

3. Mar 25, 2012

### technician

your calculation for the energy needed to vaporize that amount of water is correct.

4. Mar 25, 2012

### 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.