Respuesta :
Answer:
0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.
It will take 17,447.41 days long to gain 1 lb of fat by this person.
Explanation:
Calorie intake of an adult in a day =[tex]2.2\times 10^3 calorie[/tex]
Calorie burnt by an adult in a day = [tex]2.0\times 10^3 calorie[/tex]
Excess nutritional energy in day=
[tex](2.2\times 10^3 calorie)-(2.0\times 10^3 calorie)[/tex]
[tex]=2.0\times 10^2 calorie[/tex]
1 kcal = 4.184 kJ
So , 1000 cal = 4.184 kJ
[tex]1 cal = 4.184\times 10^{-3} kJ[/tex]
[tex]2.0\times 10^2 calorie=2.0\times 10^2\times 4.184\times 10^{-3} kJ[/tex]
[tex]=0.8368 kJ[/tex]
0.8368 kiloJoules of an excess nutritional energy is consumed by an adult in a day.
[tex]14.6\times 10^3[/tex] kilo joules of excess nutritional energy = 1 lb fat
The 1 kilo joule of excess nutritional energy = [tex]\frac{1}{14.6\times 10^3}[/tex] lb fat
Excessive nutritional energy of an adult per day = 0.8368 kiloJoules
Amount of fat gained by an adult per day =
= [tex]0.8368 kiloJoules\times \frac{1}{14.6\times 10^3}=5.7315\times 10^{-5} lb[/tex] of fat
In 1 day an adult gains = [tex]5.7315\times 10^{-5} lb[/tex] of fat
Time taken to gain 1 lb fat:
[tex]\frac{1}{5.7315\times 10^{-5}} day=17,447.41 days[/tex]
It will take 17,447.41 days long to gain 1 lb of fat by this person.
