Chapter 2, Problem 2.9PR

(a)

Interpretation Introduction

Interpretation:

The percentage of daily calorie requirement served by $\frac{3}{4}$ cup of pasta for a person on a $\text{2200}\text{calorie}$ diet has to be calculated.

Concept introduction:

Enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burned in oxygen with all the reactants and products in their standard state under standard conditions ($\text{298}\text{K}\text{and}\text{1}\text{bar}\text{pressure}$).

The food consumed by a person by the process of respiration reacts with oxygen to give energy.

Answer to Problem 2.9PR

The percentage of daily calorie requirement served by $\frac{3}{4}$ cup of pasta for a person on a $\text{2200}\text{calorie}$ diet is $7.64\%$.

Explanation of Solution

According to the question in most common diets $65\%$ of the food energy comes from carbohydrates.

It has been given that a $\frac{3}{4}$ cup of pasta contains $\text{40}\text{g}$ of carbohydrates.

Combustion of $1\text{g}$ pure carbohydrate gives $\text{4}\text{.20}\text{kCal}$ of energy.

According to the question the person has a daily diet of $\text{2200}\text{calorie}$.

Hence $\frac{3}{4}$ cup of pasta provides energy of,

  $\begin{array}{c}\text{E =}\text{4}\text{.20}\text{kCal}\text{×}\text{40}\text{g}\\ \\ \text{=}\text{168}\text{Cal}\text{(1}\text{kCal}\text{=}\text{1}\text{Cal)}\end{array}$

Hence the percentage is,

  $\begin{array}{l}\text{=}\frac{\text{168}\text{Cal}}{2200\text{Cal}}\times 100\%\\ \\ =7.64\%\end{array}$

The percentage of daily calorie requirement served by $\frac{3}{4}$ cup of pasta for a person on a $\text{2200}\text{calorie}$ diet is $7.64\%$.

(b)

Interpretation Introduction

Interpretation:

The energy released has to be calculated on the combustion of one glucose tablet of mass $\text{2}\text{.5}\text{g}$ in the air.

Concept introduction:

Enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burned in oxygen with all the reactants and products in their standard state under standard conditions ($\text{298}\text{K}\text{and}\text{1}\text{bar}\text{pressure}$).

Answer to Problem 2.9PR

The energy released on the combustion of one glucose tablet of mass $\text{2}\text{.5}\text{g}$ in the air is $\text{35}\text{.5}\text{kJ}$.

Explanation of Solution

Given that the mass of the glucose table is $\text{2}\text{.5}\text{g}$.

Combustion of $1\text{g}$ glucose gives $\text{14}\text{.2}\text{kJ}$ of energy.

Hence combustion of $\text{2}\text{.5}\text{g}$ glucose gives,

  $\begin{array}{l}\text{=}\left(\text{14}\text{.2}\text{kJ×2}\text{.5}\text{g}\right)\\ \\ \text{=}\text{35}\text{.5}\text{kJ}\end{array}$

Hence the energy released on the combustion of one glucose tablet of mass $\text{2}\text{.5}\text{g}$ in the air is $\text{35}\text{.5}\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

Assuming that $25\%$ of energy is available for work, how much height one can climb consuming one glucose tablet has to be calculated.

Concept introduction:

Enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burned in oxygen with all the reactants and products in their standard state under standard conditions ($\text{298}\text{K}\text{and}\text{1}\text{bar}\text{pressure}$).

The food consumed by a person by the process of respiration reacts with oxygen to give energy.

Potential energy:

Potential energy within an object is the amount of energy stored due to its position, arrangement or state.

It can be calculated as,

  $\text{P}\text{.E}\text{.}\text{=}\text{mgh}$

Where,

$\text{m}$ is the mass of the object’

$\text{g}$ is the gravitational constant

$\text{h}$ is the height of the object

Answer to Problem 2.9PR

The person can climb a height of $\text{13}\text{.93}\text{m}$.

Explanation of Solution

The energy released on the combustion of one glucose tablet of mass $\text{2}\text{.5}\text{g}$ in the air is $\text{35}\text{.5}\text{kJ}$.

Now in the question it is given that $25\%$ of energy is available for work.

Hence the amount of energy available for work is,

  $\begin{array}{l}=\frac{25}{100}\times \text{35}\text{.5}\text{kJ}\\ \\ =8.875\text{kJ}\end{array}$

Now with the help of this amount of energy a person can climb up and that can be stored in the person as the potential energy. So according to the formula of potential energy,

  $\text{mgh}\text{=}\text{8}\text{.875}\text{kJ}$

Now the person to be considered is myself and hence,

  $\begin{array}{l}\text{m}\text{=}\text{65}\text{kg}\\ \\ \text{g}\text{=}\text{9}\text{.8}{\text{ms}}^{\text{-2}}\end{array}$

Now the calculation is,

  $\begin{array}{l}\text{mgh}\text{=}\text{8}\text{.875}\text{kJ}\\ \\ \text{65}\text{kg×9}\text{.8}{\text{ms}}^{\text{-2}}\text{×h}\text{=}\text{8}\text{.875}\text{kJ}\\ \\ \text{h}\text{=}\frac{\text{8}{\text{.875×10}}^{\text{3}}\text{J}}{\text{65}\text{kg×9}\text{.8}{\text{ms}}^{\text{-2}}}\\ \\ \text{h}\text{=}\text{13}\text{.93}\text{m}\end{array}$

Hence the person can climb a height of $\text{13}\text{.93}\text{m}$.

(d)

Interpretation Introduction

Interpretation:

Standard enthalpy of combustion of glucose is whether higher or lower at the blood temperature than in normal temperature of ${\text{25}}^{\text{ο}}\text{C}$.

Concept introduction:

Enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burned in oxygen with all the reactants and products in their standard state under standard conditions ($\text{298}\text{K}\text{and}\text{1}\text{bar}\text{pressure}$).

Enthalpy:

Enthalpy is a property of a thermodynamic system that is equal to the sum of the internal energy of the system and the product of pressure and volume. For a closed system where transfer of matter between system and surroundings is prohibited, for the processes that occur at constant pressure, the heat absorbed or released equals to the change in enthalpy.

It is denoted as $\text{H}$. The molar enthalpy is defined as the enthalpy per mole. It is denoted as ${\text{H}}_{\text{m}}$.

From thermodynamics,

  $\text{H}\text{=}\text{U}\text{+}\text{PV}$

Where,

$\text{H}$ is the enthalpy of the system

$\text{U}$ is the internal energy of the system

$\text{P}$ is the pressure and $\text{V}$ is the volume of the system

Specific heat capacity at constant pressure:

Specific heat capacity at constant pressure can be defined as the amount of energy required to increase the temperature of a substance by ${\text{1}}^{\text{ο}}\text{C}$.

It is denoted as ${\text{C}}_{\text{p}}$. The molar specific heat capacity is defined as the specific heat capacity per mole and it is denoted by ${\text{C}}_{\text{p,m}}$.

  ${\text{C}}_{\text{p}}={\left(\frac{\partial \text{H}}{\partial \text{T}}\right)}_{\text{p}}$

Explanation of Solution

Temperature of blood is $\text{37}{\text{.2}}^{\text{ο}}\text{C}$.

So on combustion of glucose in blood temperature the value will change and it will be dependent upon the ${\text{C}}_{\text{P}}$ of the glucose.

${\text{C}}_{\text{P}}\text{=}\text{115}{\text{JK}}^{\text{-1}}$

Hence enthalpy of combustion at $\text{37}{\text{.2}}^{\text{ο}}\text{C}$ is,

  $\begin{array}{l}{\displaystyle \int \text{dH}}\text{=}{\text{C}}_P{\displaystyle \int \text{dT}}\\ \\ \text{H}\text{=}{\text{C}}_{\text{P}}\text{T}\end{array}$

Putting the values,

  $\begin{array}{l}\text{H}\text{=}\text{115}{\text{JK}}^{\text{-1}}\text{×(37}\text{.2+273)K}\\ \\ \text{H}\text{=}\text{35673}\text{J}\\ \\ \text{H}\text{=}\text{35}\text{.673kJ}\end{array}$

Combustion of $1\text{g}$ glucose gives $\text{14}\text{.2}\text{kJ}$ of energy at ${\text{25}}^{\text{ο}}\text{C}$.

Hence at blood temperature the heat of combustion of glucose is more than heat of combustion at ${\text{25}}^{\text{ο}}\text{C}$.

(e)

Interpretation Introduction

Interpretation:

The energy released has to be calculated on the combustion of one typical sugar cube of mass $\text{1}\text{.5}\text{g}$ in the air.

Concept introduction:

Enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burned in oxygen with all the reactants and products in their standard state under standard conditions ($\text{298}\text{K}\text{and}\text{1}\text{bar}\text{pressure}$).

Answer to Problem 2.9PR

The energy released on the combustion of one typical sugar cube of mass $\text{1}\text{.5}\text{g}$ in the air is $12.315\text{kJ}$.

Explanation of Solution

Given that the mass of one typical sugar cube is $\text{1}\text{.5}\text{g}$.

Combustion of $1\text{g}$ sugar cube gives $8.21\text{kJ}$ of energy.

Hence combustion of $\text{1}\text{.5}\text{g}$ glucose gives ,

$\begin{array}{l}\text{=}\left(8.21\text{kJ×1}\text{.5}\text{g}\right)\\ \\ \text{=}12.315\text{kJ}\end{array}$

Hence the energy released on the combustion of one typical sugar cube of mass $\text{1}\text{.5}\text{g}$ in the air is $12.315\text{kJ}$.

(f)

Interpretation Introduction

Interpretation:

Assuming that $25\%$ of energy is available for work, how much height one can climb consuming one table sugar cube has to be calculated.

Concept introduction:

Enthalpy of combustion:

Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burned in oxygen with all the reactants and products in their standard state under standard conditions ($\text{298}\text{K}\text{and}\text{1}\text{bar}\text{pressure}$).

The food consumed by a person by the process of respiration reacts with oxygen to give energy.

Potential energy:

Potential energy within an object is the amount of energy stored due to its position, arrangement or state.

It can be calculated as,

  $\text{P}\text{.E}\text{.}\text{=}\text{mgh}$

Where,

$\text{m}$ is the mass of the object’

$\text{g}$ is the gravitational constant

$\text{h}$ is the height of the object

Answer to Problem 2.9PR

The person can climb a height of $4.83\text{m}$.

Explanation of Solution

From the part (e) the energy released on the combustion of one typical sugar cube of mass $\text{1}\text{.5}\text{g}$ in the air is $12.315\text{kJ}$.

Now in the question it is given that $25\%$ of energy is available for work.

Hence the amount of energy available for work is,

  $\begin{array}{l}=\frac{25}{100}\times 12.315\text{kJ}\\ \\ =3.08\text{kJ}\end{array}$

Now with the help of this amount of energy a person can climb up and that can be stored in the person as the potential energy. So according to the formula of potential energy,

$\text{mgh}\text{=}3.08\text{kJ}$

Now the person to be considered is myself and hence,

  $\begin{array}{l}\text{m}\text{=}\text{65}\text{kg}\\ \\ \text{g}\text{=}\text{9}\text{.8}{\text{ms}}^{\text{-2}}\end{array}$

Now the calculation is,

  $\begin{array}{l}\text{mgh}\text{=}3.08\text{kJ}\\ \\ \text{65}\text{kg×9}\text{.8}{\text{ms}}^{\text{-2}}\text{×h}\text{=}3.08\text{kJ}\\ \\ \text{h}\text{=}\frac{3.08{\text{×10}}^{\text{3}}\text{J}}{\text{65}\text{kg×9}\text{.8}{\text{ms}}^{\text{-2}}}\\ \\ \text{h}\text{=}4.83\text{m}\end{array}$

Hence the person can climb a height of $4.83\text{m}$.