Estimate the mass of water that must evaporate from the skin to cool the body by 0.45 ∘c. assume a body mass of 95 kg and assume that the specific heat capacity of the body is 4.0 j/g⋅∘c.

0.0757 kg

Explanation:

m = mass of the body = 95 kg = 95000 g

ΔT = change in the temperature of the body = 0.45 °C

c = specific heat capacity of the body = 4.0 J/(g °C)

M = mass of water evaporated

L = latent heat of vaporization of water = 2260 J/g

Using conservation of heat

Heat gained by water = heat lost by body

ML = m c ΔT

M (2260) = (95000) (4.0) (0.45)

M = 75.7 g

M = 0.0757 kg

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