Question

A.) What mass of water must evaporate from the skin of a 71.5 kg man to cool his body 1.15 ∘C? The heat of vaporization of water at body temperature (37.0 ∘C) is 2.42×10^6 J/kg. The specific heat of a typical human body is 3480 J/(kg⋅K). Express your answer in kilograms to three significant figures.

B.) What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is 355 cm^3. Express your answer in cubic centimeters to three significant figures.

Answer 1

[A] ans

step;1

Given that

mass m1=71.5 kg

temperature T1=1.15+273= 274.15 K

temperature T2=37+273=310 k

heat of vaporization L=2.42*10^6 J/kg

specific heat of human body S=3480 J/kg.k

step;2

now we find the heat

heat =ms[T2-T1]=71.5*3480[310-274.15]=8920197 J

step;2

now we find the mass

mass =8920197/2420000=3.7 kg

Question B;

step;1

Given that

volume of soft drink V2=355^3

density of soft drink d2=2.1 g/cm^3

density of water d1=1 g/cm^3

step;2

now we find the volume of water

v1d1=v2d2

V1*1=355*2.1=

V1=745.5 cm^3