Chapter 5, Problem 5.102QP

(a)

Interpretation Introduction

Interpretation: The questions based on the fuel value and the heat energy is to be stated.

Concept introduction: A fuel is a substance which releases energy on combustion. The released energy may in the form of heat, light or may in other form. Fuel value is one of the basic characteristic of the fuel. It is defined as the amount of energy released by the fuel when $1\text{g}$ of the substance/fuel undergoes complete combustion. The main source of fuels is the hydrocarbon. The unit of fuel value is $\text{kJ/g}$. Heat capacity is the calculated amount of energy of the specific mass of the compound to raise the temperature in a fixed range.

To determine: The amount of fuel value for the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$.

Answer to Problem 5.102QP

Solution

The amount of fuel value for the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is $\underset{\_}{-48.31kJ/g}$.

Explanation of Solution

Explanation

Given

The given amount of enthalpy of combustion $\Delta H_{\text{combustion}}^{\text{°}}$ for the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is $-\text{4163}\text{kJ/mol}$.

The given compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is a hydrocarbon. The major use of hydrocarbon in our daily life is as a fuel. A hydrocarbon generates energy on combustion with oxygen to generate electricity, to power vehicles, to cook our meals etc. The amount of enthalpy change during the combustion reaction is known as the $\Delta H_{\text{combustion}}^{\text{°}}$. As already defined above that the fuel value is the released amount of energy on complete combustion of per gram of the substance. For the above given hydrocarbons, fuel value is calculated by the formula,

$\text{Fuel}\text{value}=\frac{\Delta H_{\text{combustion}}^{\text{°}}}{M}$ (1)

Where,

• $\Delta H_{\text{combustion}}^{\text{°}}$ is the standard enthalpy of combustion.
• $M$ is the molar mass of the hydrocarbon.

The molar mass of the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is given as,

$\text{86}\text{.172}\text{g/mol}$

Substitute the values in the equation (1).

$\begin{array}{l}\text{Fuel}\text{value}=\frac{\Delta H_{\text{combustion}}^{\text{°}}}{M}\\ \text{Fuel}\text{value}=\frac{-\text{4163}\text{kJ/mol}}{\text{86}\text{.172}\text{g/mol}}\\ \text{Fuel}\text{value}=\underset{\_}{-48.31kJ/g}\end{array}$

(b)

Interpretation Introduction

To determine: The amount of heat released during the combustion of $1.00\text{kg}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$.

Answer to Problem 5.102QP

Solution

The amount of heat released during the combustion of $1.00\text{kg}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is $\underset{\_}{-48.31\times 10^{3}kJ}$

Explanation of Solution

Explanation

As already explained above that the amount of heat released by the fuel when $1\text{g}$ of the substance/fuel undergoes complete combustion is known as fuel value. It means to calculate the amount of heat released during the combustion of $1.00\text{kg}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ the fuel value in terms of per kilogram is to be calculated. Therefore, the amount of heat released during the combustion of $1.00\text{kg}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is the amount of fuel value for the combustion of $1.00\text{kg}$ compound and it is expressed as,

$\begin{array}{c}\text{Fuel}\text{value}=-\text{48}\text{.31}\text{kJ/g}\\ \text{For}\text{1}\text{.00}\text{g}\text{compound}\text{energy}\text{released}=-\text{48}\text{.31}\text{kJ}\\ \text{For}\text{1}\text{.00}\text{kg}\text{compound}\text{energy}\text{released}=\underset{\_}{-48.31\times 10^{3}kJ}\end{array}$

(c)

Interpretation Introduction

To determine: The amount of the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$.

Answer to Problem 5.102QP

Solution

The amount of the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is $\underset{\_}{5.189g}$.

Explanation of Solution

Explanation

The amount of heat $\left(q\right)$ required to warm $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is calculated by the formula,

$q=n{C}_{\text{p}}\Delta T$ (1)

Where,

• $n$ is the number of moles of the substance.
• $C_{\text{p}}$ is the molar heat capacity of the substance.
• $\Delta T$ is the change in temperature.

Molar mass of water is $18.02\text{g/mol}$.

Molar heat capacity of water is $75.3\text{J/mol}°\text{C}$.

For $1.00\text{kg}$ of water the number of moles $\left(n\right)$ of water is calculated by the formula,

$n=\frac{m}{M}$

Where,

• $m$ is the given mass of water.
• $M$ is the molar mass of water.

Substitute the values in the above equation.

$\begin{array}{l}n=\frac{m}{M}\\ n=\frac{1.00\times {10}^3\text{g}}{18.02\text{g/mol}}\\ n=55.49\text{moles}\end{array}$

Now, substitute the values in equation (1).

$\begin{array}{l}q=n{C}_{\text{p}}\Delta T\\ q=55.49\text{moles}\times 75.3\text{J/mol}°\text{C}\left(85-25\right)°\text{C}\\ q=250703.82\text{J}\\ q=250.70\text{kJ}\end{array}$

This calculated value of heat capacity is the required amount of heat to warm $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$.

As already calculated in part (a) that the amount of energy released by compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ on combustion of $1.00\text{g}$ is $-\text{48}\text{.31}\text{kJ}$. The negative sign shows the released amount of energy.

From these two calculations the amount of the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is calculated as,

$\begin{array}{c}\text{48}\text{.31}\text{kJ}\text{heat}\text{obtained}=1.00\text{g}{\text{C}}_6{\text{H}}_{\text{14}}\\ 1\text{kJ}\text{heat}\text{obtained}=\frac{1.00\text{g}{\text{C}}_6{\text{H}}_{\text{14}}}{\text{48}\text{.31}\text{kJ}}\\ 250.70\text{kJ}\text{heat}\text{obtained}=\frac{1.00\text{g}{\text{C}}_6{\text{H}}_{\text{14}}}{\text{48}\text{.31}\text{kJ}}\times 250.70\text{kJ}\\ 250.70\text{kJ}\text{heat}\text{obtained}=\underset{\_}{5.189g}\end{array}$

(d)

Interpretation Introduction

To determine: The amount of the white gas $\left(25\%{\text{C}}_5+75\%{\text{C}}_{\text{6}}\right)$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$.

Answer to Problem 5.102QP

Solution

The amount of the white gas $\left(25\%{\text{C}}_5+75\%{\text{C}}_{\text{6}}\right)$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is $\underset{\_}{5.17g}$.

Explanation of Solution

Explanation

According to the assumption the white gas has $25\%$ ${\text{C}}_{\text{5}}$ hydrocarbons and $75\%$ ${\text{C}}_6$ hydrocarbons. Therefore, the amount of hydrocarbons in calculated as,

For $25\%$ ${\text{C}}_{\text{5}}$ hydrocarbons,

$\begin{array}{c}1\text{g}\text{White}\text{gas}\text{contains}=25\%\text{hydrocarbon}\\ \text{=}\frac{25}{100}\text{g}\text{hydrocarbon}\\ \text{=0}\text{.25}\text{g}\text{hydrocarbon}\end{array}$

For $75\%$ ${\text{C}}_6$ hydrocarbons,

$\begin{array}{c}1\text{g}\text{White}\text{gas}\text{contains}=75\%\text{hydrocarbon}\\ \text{=}\frac{75}{100}\text{g}\text{hydrocarbon}\\ \text{=0}\text{.75}\text{g}\text{hydrocarbon}\end{array}$

Molar mass of ${\text{C}}_{\text{5}}$ hydrocarbons $\left({\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\right)$ is $72.148\text{g/mol}$.

The enthalpy change of combustion $\left(\Delta H_{\text{combustion}}^{\text{°}}\right)$ is $-\text{803}\text{.3}\text{kJ/mol}$.

The amount of fuel value of ${\text{C}}_{\text{5}}$ hydrocarbons is calculated as,

$\begin{array}{l}\text{Fuel}\text{value}=\frac{\Delta H_{\text{combustion}}^{\text{°}}}{M}\\ \text{Fuel}\text{value}=\frac{-\text{803}\text{.3}\text{kJ/mol}}{72.148\text{g/mol}}\\ \text{Fuel}\text{value}=-\text{48}\text{.997}\text{kJ/g}\end{array}$

The energy obtained by the $0.25\text{g}$ ${\text{C}}_5$ hydrocarbon is calculated as,

$\begin{array}{c}\text{Enrgy}\text{obtained}\text{by}\text{1}\text{g}=-48.997\text{kJ}\\ \text{Enrgy}\text{obtained}\text{by}\text{0}\text{.25}\text{g}=-48.997\text{kJ}\times \text{0}\text{.25}\\ \text{Enrgy}\text{obtained}\text{by}\text{0}\text{.25}\text{g}=-\text{12}\text{.249}\text{kJ}\end{array}$

The calculated amount of fuel value of ${\text{C}}_{\text{6}}$ hydrocarbon is $48.31\text{kJ/g}$.

The energy obtained by the $0.75\text{g}$ of ${\text{C}}_{\text{6}}$ hydrocarbon is calculated as,

$\begin{array}{c}\text{Enrgy}\text{obtained}\text{by}\text{1}\text{g}=-48.31\text{kJ}\\ \text{Enrgy}\text{obtained}\text{by}\text{0}\text{.75}\text{g}=-48.31\text{kJ}\times \text{0}\text{.75}\\ \text{Enrgy}\text{obtained}\text{by}\text{0}\text{.75}\text{g}=-\text{36}\text{.23}\text{kJ}\end{array}$

Therefore, the total amount of heat released when $1.00\text{g}$ white gas undergoes combustion is $\left(-12.249\text{kJ}\right)+\left(-36.23\text{kJ}\right)=-48.479\text{kJ}$.

From all these calculations the amount of the compound ${\text{C}}_6{\text{H}}_{\text{14}}$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is calculated as,

$\begin{array}{c}\text{48}\text{.479}\text{kJ}\text{heat}\text{obtained}=1.00\text{g}\text{mixture}\text{of}\text{hydrocarbon}\\ 1\text{kJ}\text{heat}\text{obtained}=\frac{1.00\text{g}\text{mixture}\text{of}\text{hydrocarbon}}{\text{48}\text{.479}\text{kJ}}\\ 250.70\text{kJ}\text{heat}\text{obtained}=\frac{1.00\text{g}{\text{C}}_6{\text{H}}_{\text{14}}}{\text{48}\text{.479}\text{kJ}}\times 250.70\text{kJ}\\ 250.70\text{kJ}\text{heat}\text{obtained}=\underset{\_}{5.17g}\end{array}$

Conclusion

1. a) The amount of fuel value for the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is $\underset{\_}{-48.31kJ/g}$.
2. b) The amount of heat released during the combustion of $1.00\text{kg}$ of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ is $\underset{\_}{-48.31\times 10^{3}kJ}$
3. c) The amount of the compound ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is $\underset{\_}{5.189g}$.
4. d) The amount of the white gas $\left(25\%{\text{C}}_5+75\%{\text{C}}_{\text{6}}\right)$ to heat $1.00\text{kg}$ of water from $25.0°\text{C}$ to $85.0°\text{C}$ is $\underset{\_}{5.17g}$