How Much Water Evaporates to Remove Heat from Basal Metabolism?

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To calculate the amount of water needed to evaporate for cooling from basal metabolism, the heat generated is 60 kcal/hr for a 65 kg person. The formula Q = mL, where L is the heat of vaporization (539 kcal/kg), is used to determine the mass of water required. Setting up the equation yields m = 60 kcal/hr divided by 539 kcal/kg, resulting in approximately 0.111 kg/min. This translates to about 6.679 kg/min, or nearly seven liters of water evaporating per minute to remove the heat generated by basal metabolism. The calculations highlight the significant amount of water required for effective thermoregulation through evaporation.
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I can't seem to get this problem to work out...any suggestions?

-HOw much water would have to evaporate from the skin/min to take away all of the heat generated by basl metabolism (60kcal/hr) for a 65Kg person.
 
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The heat generated by this process needs to change the water from its liquid to its vapor state. The amount of heat, Q, required to alter a mass m of water to its vapor state is given by
Q=mL where L is the heat of vaporization of water which is 539\ kcal/kg.
 
So would I set the problem up like this:

60 Kcal/hr (60 min/hr)= m (539 Kcal/kg)

m=6.679Kg/min
 
That is about seven liters of water per minute!
 
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