How Much Water Evaporates to Remove Heat from Basal Metabolism?

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Homework Help Overview

The problem involves calculating the amount of water that must evaporate to remove heat generated by basal metabolism, specifically for a person weighing 65 kg. The context is rooted in thermodynamics and the heat of vaporization of water.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between heat generation and the heat of vaporization, with one participant attempting to set up the problem using the formula Q = mL. Questions arise regarding the setup and calculations involved.

Discussion Status

The discussion is ongoing, with participants exploring the calculations necessary to determine the mass of water required for evaporation. There is no explicit consensus, but attempts to clarify the problem setup and calculations are evident.

Contextual Notes

Participants are working under the assumption that the entire heat generated by basal metabolism must be accounted for through water evaporation, and they reference specific values for heat generation and heat of vaporization.

that1grrl
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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, [itex]Q[/itex], required to alter a mass [itex]m[/itex] of water to its vapor state is given by
[tex]Q=mL[/tex] where [itex]L[/itex] is the heat of vaporization of water which is [itex]539\ kcal/kg[/itex].
 
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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