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 72.5 kg man to cool his body 1.30 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).

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

Explanation / Answer

Mass of man = 72.5 kg

decrease in temperature = 1.3 degree C

specific heat capacity of body = 3480 J/Kg-k

So heat lost = 72.5 * 1.3 * 3480 = 327990 J

Let the mass of water to be evaporated be m

heat gain by water to evaporate = mass * heat of vaporization = 2.42*106 m J

So by heat gained = heat lost

2.42*106 m = 327990

=> m = 0.1355 Kg = 135.5 gram

So mass of water to be evaporated = 0.1355 Kg = 135.5 gram

B) So the volume of water a man need to drink to replenish it = 135.5 gram = 135.5/355 = 0.38 soft-drink can

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