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20 days ago

9

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) .

1 answer:

inna [2.7K]20 days ago

7 0

Answer:

m = 0.111 kg

Explanation:

The heat required for the body to drop in temperature by 1 degree Celsius is determined by

$Q = m s\Delta T$

where we have the known values

m = 70 kg

s = 3840 J/kg K

$\Delta T = 1.00^o C$

Now considering the same heat being required to vaporize water from the body

so it can be expressed as

$Q = mL$

$268800 = m(2.42 \times 10^6)$

$m = 0.111 kg$

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