What mass of water must evaporate from the skin of a 70.5 kg man to cool his bod

Question

What mass of water must evaporate from the skin of a 70.5 kg man to cool his body 1.10 C? The heat of vaporization of water at body temperature (37.0 C) is 2.42×106J/kg. The specific heat of a typical human body is 3480 J/(kgK).

Part A = .112 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

I cant figure out part B. I keep getting .315 but it is incorrect. I need answer in three sig figs and in cm^3.

Explanation / Answer

density of water = 1000 kg / m^3

density = mass / volume

1000 = 0.112 / volume

volume = 1.12 * 10^-4 m^3 or 112 cm^3

volume of water man must drink = 112 cm^3

compare to 355 cm^3 of softdrink can its = 112 / 355 or 0.31549 times

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