14.36 Part A The evaporation of sweat is an important mechanism for temperature

Question

14.36

Part A

The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals.

What mass of water must evaporate from the skin of a 70.5 kg man to cool his body 1.00 C? The heat of vaporization of water at body temperature (37.0 C) is 2.42×106J/kg. The specific heat capacity of a typical human body is 3480 J/(kgK).

m=_______kg

Part 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 cm3

v=__________ cm^3

Explanation / Answer

Part A:

Q=mc(delta)T

Then mass of water must evaporate from the skin= (70.5kg)(3480J/kg*K)(-1.00C) / (2.42*10^6J/Kg) =0.10138 kg

Part B

use that 0.10138 kg/1000kg/m^3 and you get 1.0138*10^-4m^3

which is the same as 101cm^3

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