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## 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/