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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) .

Respuesta :

Answer:

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

[tex]Q = m s\Delta T[/tex]

here we know that

m = 70 kg

s = 3840 J/kg K

[tex]\Delta T = 1.00^o C[/tex]

now we know that

[tex]Q = (70 kg)(3840 J/kg ^oC)(1 ^o C)[/tex]

[tex]Q = 268800 J[/tex]

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

so it is given as

[tex]Q = mL[/tex]

[tex]268800 = m(2.42 \times 10^6)[/tex]

[tex]m = 0.111 kg[/tex]

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