Answer :
Let's address the question step-by-step.
### Part A: Calculation of Excess Nutritional Energy in Kilojoules
1. Determine the excess energy consumed in Calories per day:
The adult consumes \(2.4 \times 10^3\) Calories per day but burns \(2.1 \times 10^3\) Calories per day.
[tex]\[ \text{Excess Energy (in Calories)} = 2.4 \times 10^3 \text{ Calories} - 2.1 \times 10^3 \text{ Calories} = 0.3 \times 10^3 \text{ Calories} \][/tex]
2. Convert the excess energy from Calories to kilojoules:
Knowing that 1 Calorie = 4.184 kilojoules, we convert 0.3 \(\times 10^3\) Calories to kilojoules:
[tex]\[ \text{Excess Energy (in kilojoules)} = 0.3 \times 10^3 \text{ Calories} \times 4.184 \frac{\text{kJ}}{\text{Cal}} \][/tex]
[tex]\[ = 1255.2 \text{ kJ} \][/tex]
So, the excess nutritional energy consumed by the adult per day is 1255.2 kJ.
### Part B: Number of Days to Gain 1 lb
1. Determine the kilojoules needed to gain 1 pound:
It is given that \(1 \text{ lb}\) of fat is stored by the body for each \(1.46 \times 10^4\) kilojoules of excess energy consumed.
2. Calculate the number of days to gain 1 lb:
Using the previously calculated excess energy of 1255.2 kJ per day:
[tex]\[ \text{Days to gain 1 lb} = \frac{1.46 \times 10^4 \text{ kJ}}{\text{Excess Energy per Day in kJ}} \][/tex]
[tex]\[ = \frac{1.46 \times 10^4 \text{ kJ}}{1255.2 \text{ kJ/day}} \][/tex]
[tex]\[ \approx 11.63 \text{ days} \][/tex]
So, it will take approximately 11.6 days for the person to gain 1 pound.
Thus:
- The excess nutritional energy consumed per day is 1255.2 kJ.
- It will take the person 11.6 days to gain 1 pound of fat.
### Part A: Calculation of Excess Nutritional Energy in Kilojoules
1. Determine the excess energy consumed in Calories per day:
The adult consumes \(2.4 \times 10^3\) Calories per day but burns \(2.1 \times 10^3\) Calories per day.
[tex]\[ \text{Excess Energy (in Calories)} = 2.4 \times 10^3 \text{ Calories} - 2.1 \times 10^3 \text{ Calories} = 0.3 \times 10^3 \text{ Calories} \][/tex]
2. Convert the excess energy from Calories to kilojoules:
Knowing that 1 Calorie = 4.184 kilojoules, we convert 0.3 \(\times 10^3\) Calories to kilojoules:
[tex]\[ \text{Excess Energy (in kilojoules)} = 0.3 \times 10^3 \text{ Calories} \times 4.184 \frac{\text{kJ}}{\text{Cal}} \][/tex]
[tex]\[ = 1255.2 \text{ kJ} \][/tex]
So, the excess nutritional energy consumed by the adult per day is 1255.2 kJ.
### Part B: Number of Days to Gain 1 lb
1. Determine the kilojoules needed to gain 1 pound:
It is given that \(1 \text{ lb}\) of fat is stored by the body for each \(1.46 \times 10^4\) kilojoules of excess energy consumed.
2. Calculate the number of days to gain 1 lb:
Using the previously calculated excess energy of 1255.2 kJ per day:
[tex]\[ \text{Days to gain 1 lb} = \frac{1.46 \times 10^4 \text{ kJ}}{\text{Excess Energy per Day in kJ}} \][/tex]
[tex]\[ = \frac{1.46 \times 10^4 \text{ kJ}}{1255.2 \text{ kJ/day}} \][/tex]
[tex]\[ \approx 11.63 \text{ days} \][/tex]
So, it will take approximately 11.6 days for the person to gain 1 pound.
Thus:
- The excess nutritional energy consumed per day is 1255.2 kJ.
- It will take the person 11.6 days to gain 1 pound of fat.
