The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. a. What mass of water must evaporate from the skin of a 66.0 kg man to cool his body 1.30 °C? The heat of vaporization of water at body temperature (37.0 ∘C) is 2.42×10^6J/kg. The specific heat capacity of a typical human body is 3480 J/(kg⋅K).
b. What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is 355 cm^3

a) m = 26.03 kg

b) V = 26000 cm³  

Explanation:

a) The mass of water that should evaporate from the skin can be calculated using the following equation:

Where:                                

m: is the mass of water =?

Q: is the heat of vaporization of water = 2.42x10⁶ J/kg      

C: is the specific heat capacity = 3480 J/(kg*K)

ΔT: is the temperature difference = 1.30 °C = 274.3 K  

M: is the mass of the man = 66.0 kg

Hence, 26.03 kg of the mass of water must evaporate from the skin.  

b) The volume (V) of water is:    

Where:

d: is the density of water = 997 kg/m³          

Compared to a soft drink can of 355 cm³, the man should drink approximately 73 cans of soft drink to compensate for the evaporated water.  

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