What mass of water must evaporate from the skin of a 70.0 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⋅k). the specific heat of a typical human body is 3480 j/(kg⋅k) . what volume of water must the man drink to replenish the evaporated water?

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

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