This problem aims to familiarize us with the **daily consumption** of **calories** using the **fuel value** of daily **macronutrients.** The concept required to solve this problem is mostly related to the **fuel value** or **calorific calculation,** and the **energy** values of **macronutrients.**

The **calorie range,** or the **calorific value,** of an edible substance is the part of the **energy** that is released when the compound is fully **“burned,”** or **experiences** complete **combustion. Combustion** usually happens in the existence of **oxygen** gas, with a reacting agent such as **fuel.**

The **calorific** value of a compound is mostly given based on per **amount,** that is, per mole or gram of **compound burned.**

## Expert Answer

The **caloric value** of a substance can be resolved by **calculating** the **heat** produced when a given **amount** is fully **fumed** in **oxygen.** The energy received as an effect of full **combustion** is the **potential energy** but the energy released in the body is not** identical.**

According to the given** information,** a **pound** of **candies** contains:

**Fat** $=96g$,

**Carbohydrates** $=320g$,

**Protein** $=21g$,

The **mass** of serving is $m = 42g$

We are asked to find the value of **one serving.**

For this, we will first find the **fuel value** of **candies** per pound:

\[Fat \space= 98g \space \times \space 38 \dfrac{kJ}{g} \]

\[= 3648 kJ \]

\[Carbohydrates \space= 320g \space \times \space 17 \dfrac{kJ}{g} \]

\[= 5440 kJ \]

\[Protein \space= 21g \space \times \space 17 \dfrac{kJ}{g} \]

\[= 357 kJ \]

Total fuel **value** per **pound** is the sum of the **fuel** values of **Fat, carbohydrates,** and **protein.**

\[= 3648 kJ + 5440 kJ + 357 kJ \]

Total **fuel value** comes out to be: $9445 Kj$ per pound.

After calculating the total **fuel value** per pound, calculate the total **fuel value** per gram.

The **total fuel value** per **gram** is given as the total fuel value per **pound** into $\dfrac{1lb} {453.6g}$

That is:

\[=9445 \times \dfrac{1lb}{453.6}\]

The total **fuel value** per **gram** comes out to be $20.82 \space \dfrac{kJ}{g}$.

Now calculating the **fuel value** of one **serving:**

That is given as the **total fuel** value per gram times the mass. That is:

\[= 20.82 \dfrac{kJ}{g} \times 42g \]

\[= 874.44 kJ \]

## Numerical Answer

The **total** fuel value per **pound** is $9445 Kj$.

The **total** fuel value per **gram** is $20.82 \space \dfrac{kJ}{g}$.

The **fuel** value of one **serving** is $874.44 kJ$

## Example:

a) A $45-g$ golf ball is **moving** at $61 m/s$. Calculate its **kinetic** energy in **joules.**

b) Convert this **energy** to calories.

c) When the ball **lands** in a sand trap, what happens to this **energy?**

**Part a:**

\[ 1 J = (1 kJ) (m^2/s^2) \]

\[ E_k = \dfrac{1}{2} mv^2 \]

\[ E_k = \dfrac{1}{2} (45) (\dfrac{1kg}{1000g}) (61 \dfrac{m}{s})^2 \]

\[ = 84 kg (\dfrac{m^2}{s^2}) \]

\[ = 84 J \]

**Part b:**

\[ Calories = \dfrac{84}{4.184} \]

\[ Calories = 20 Cal \]

**Part c:**

The speed and the **kinetic** energy of the ball **drop** to zero as it hits the **sand.**