Solution Found!
An adult eats food whose nutritional energy totals approximately 2.2 × 103 Cal per day
Chapter 3, Problem 59P(choose chapter or problem)
An adult eats food whose nutritional energy totals approximately \(2.2\times10^3\mathrm{\ Cal}\) per day. The adult burns \(2.0\times10^3\mathrm{\ Cal}\) per day. How much excess nutritional energy, in kilojoules, does the adult consume per day? If 1 lb of fat is stored by the body for each \(14.6\times10^3\mathrm{\ kJ}\) of excess nutritional energy consumed, how long will it take this person to gain 1 lb?
Questions & Answers
QUESTION:
An adult eats food whose nutritional energy totals approximately \(2.2\times10^3\mathrm{\ Cal}\) per day. The adult burns \(2.0\times10^3\mathrm{\ Cal}\) per day. How much excess nutritional energy, in kilojoules, does the adult consume per day? If 1 lb of fat is stored by the body for each \(14.6\times10^3\mathrm{\ kJ}\) of excess nutritional energy consumed, how long will it take this person to gain 1 lb?
ANSWER:Step 1 of 2
Total energy gained per day = \(2.2\times10^3\mathrm{\ cal}\)
Energy burned per day = \(2.0\times10^3\mathrm{\ cal}\)
Excess nutritional energy gained by the adult = Total energy gained per day - Energy burned per day = \(2.2\times10^3\mathrm{\ cal}-2.0\times10^3\mathrm{\ cal}=0.2\times10^3\mathrm{\ cal}\)
We know that 1 cal = 4.184 J
So the excess nutritional energy in KJ gained by the adult
\(\begin{array}{l}=4.184\times0.2\times10^{3\ }\mathrm{J}\\ =0.836\mathrm{\ J}\times10^3\mathrm{\ KJ}\end{array}\)
