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) .

m = 0.111 kg

Explanation:

Heat required to release from the body of the person when his temperature cool down by 1 degree C is given as

here we know that

m = 70 kg

s = 3840 J/kg K

now we know that

now the same heat is used to vaporize water of the body

so it is given as

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