Answer :
Answer:
The body temperature would rise by 47.85 °C
The amount of water the body evaporates is 4.15 kg.
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