The evaporation of sweat is an important mechanism for temperature control in so

Question

The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. Part A What mass of water must evaporate from the skin of a 70.5kg  man to cool his body 1.00?C? The heat of vaporization of water at body temperature (37.0?C) is 2.42 The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. Part A What mass of water must evaporate from the skin of a 70.5kg  man to cool his body 1.00?C? The heat of vaporization of water at body temperature (37.0?C) is 2.42 Part A What mass of water must evaporate from the skin of a 70.5kg  man to cool his body 1.00?C? The heat of vaporization of water at body temperature (37.0?C) is 2.42 Part A What mass of water must evaporate from the skin of a 70.5kg  man to cool his body 1.00?C? The heat of vaporization of water at body temperature (37.0?C) is 2.42 Part A What mass of water must evaporate from the skin of a 70.5kg  man to cool his body 1.00?C? The heat of vaporization of water at body temperature (37.0?C) is 2.42 Part A What mass of water must evaporate from the skin of a 70.5kg  man to cool his body 1.00?C? The heat of vaporization of water at body temperature (37.0?C) is 2.42

Explanation / Answer

Th amount of energy (heat) needed to reduce the temp of a mass:
Q = m*c*dT
where m is the mass, c is the specifi heat and dT is the temperatue change.
In your cae it would be:
Q = 70*4.19x10^3*1 (J)
This heat should be equal to the evaporation heat of water (or sweat):
70*4.19*1000*1 = mw*2420x10^3
Then you can find mw (which is water mass).

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