Metabolic activity in the human body releases approximately 1.0 x 104kJ of heat per day. Assuming the body is 50 kg of water, how much would the body temperature rise if it were an isolated system? How much water must the body eliminate as perspiration to maintain the normal body temperature (98.6 0F)? Does the result make sense? Explain. The heat of vaporization of water may be taken as 2.41 kJ/g.

This makes sense because firstly the value obtained is positive then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature

Explanation:

Considering the relationship (between the heat released and the mass of the object) as shown below

q = msΔT

where q is the heat released per day = [tex]1.0 * 10^{4} kJ[/tex]

m is the mass of the body = 50 kg

ΔT is the temperature rise = ?

s is the specific heat of water = [tex]4,18kJ/kg^{o} C[/tex]

substituting values we have

[tex]1.0 * 10^{4} kJ[/tex] = [tex](50kg) (4.18kJ/kg^{o}C )[/tex] ΔT

ΔT = [tex]\frac{1.0 * 10^{4}kg}{(50kg)(4.18kJ/kg^{o} C)}[/tex]

= 47.85°C

To maintain the normal body temperature (98.6F = 37°C) the amount of heat released by metabolism activity must be utilized for evaporation of some amount of water

Hence

[tex]Amount of water that must be evaporated =\frac{heat released per day}{heat of vaporization of water}[/tex]

[tex]= \frac{1.0* 10^{4}kJ}{2.41kJ/g} \\= 4149.38g\\=4.15kg[/tex]

Note (1 kg = 1000 g)

This makes sense because firstly the value obtained is positive then secondly it is a normal occurrence in the real world that in a place where the temperature is high the body usually produce sweat in order to balance its internal temperature