Chapter 13, Problem 13.135AP

Interpretation Introduction

Interpretation: The grams of liquid methanol required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ and grams of carbon dioxide produced in the combustion reaction is to be calculated.

Concept introduction: A reaction in which combustion of a hydrocarbon takes place to form carbon dioxide and water is known as combustion reaction. It is an exothermic reaction. Heat is always liberated in combustion reaction.

To determine: The grams of liquid methanol required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ and grams of carbon dioxide produced in the combustion reaction.

Answer to Problem 13.135AP

The grams of liquid methanol required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ is $\underset{\_}{757.02g}$ and grams of carbon dioxide produced in the combustion reaction is $\underset{\_}{1039.51g}$.

Explanation of Solution

The heat required to raise the temperature of water ( $Q$) is calculated by the formula,

$Q=cm\Delta T$ (1)

Where,

• $c$ is specific heat of water.
• $m$ is mass of water.
• $\Delta T$ is change in temperature.

The specific heat of water is $\text{4}\text{.186}\text{J/g}{}^{\text{ο}}\text{C}$.

Mass of water is $\text{454}\text{g}$.

The change in temperature is calculated by the formula,

$\Delta T=T_2-T_1$

Where,

• $T_2$ is final temperature.
• $T_1$ is initial temperature.

The value of $T_2$ is $\text{50}\text{.0}{}^{\text{ο}}\text{C}$.

The value of $T_1$ is $\text{20}\text{.0}{}^{\text{ο}}\text{C}$.

Substitute the value of $T_2$ and $T_1$ in the above equation.

$\begin{array}{c}\Delta T=\text{50}\text{.0}{}^{\text{ο}}\text{C}-\text{20}\text{.0}{}^{\text{ο}}\text{C}\\ =\text{30}\text{.0}{}^{\text{ο}}\text{C}\end{array}$

Substitute the value of $\Delta T$, $m$ and $c$ in equation (1).

$\begin{array}{c}Q=\text{4}\text{.186}{\text{J/g}}^{\text{ο}}\text{C}\times \text{454}\text{g}\times \text{30}\text{.0}{}^{\text{ο}}\text{C}\\ =57013.32\text{J}\end{array}$

The heat transfer to water is $100\%$ efficient. Therefore, the heat required for the combustion is $57013.32\text{J}$.

The mass of liquid methanol is calculated by the formula,

$Q=cm\Delta T$

Where,

• $c$ is specific heat of methanol.
• $m$ is mass of methanol.
• $\Delta T$ is change in temperature.
• $Q$ is heat required for combustion.

The specific heat of methanol is $2.5104{\text{J/g}}^{\text{ο}}\text{C}$.

Substitute the value of $c$, $Q$, $\Delta T$ in the above equation.

$\begin{array}{c}57013.32\text{J}=2.5104{\text{J/g}}^{\text{ο}}\text{C}\times m\times \text{30}\text{.0}{}^{\text{ο}}\text{C}\\ m=\frac{57013.32\text{J}}{2.5104{\text{J/g}}^{\text{ο}}\text{C}\times \text{30}\text{.0}{}^{\text{ο}}\text{C}}\\ =\underset{\_}{757.02g}\end{array}$

Number of moles of liquid methanol is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

The molar mass of methanol is $\text{32}\text{.04}\text{g/mol}$.

Substitute the value of mass and molar mass of methanol in the above equation.

$\begin{array}{c}\text{Moles}=\frac{\text{757}\text{.02}\text{g}}{\text{32}\text{.04}\text{g/mol}}\\ =23.62\text{mol}\end{array}$

Methanol ( ${\text{CH}}_3\text{OH}$) burns according to the thermochemical equation:

${\text{2CH}}_3\text{OH}\left(l\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since, $\text{2}\text{mol}$ of methanol produces $\text{2}\text{mol}$ of carbon dioxide.

Therefore, $23.62\text{mol}$ of methanol produces $23.62\text{mol}$ of carbon dioxide.

The amount of carbon dioxide produced is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

The molar mass of carbon dioxide is $\text{44}\text{.01}\text{g/mol}$.

Substitute the value of number of moles and molar mass of carbon dioxide in the above equation.

$\begin{array}{c}23.62\text{mol}=\frac{\text{Mass}}{\text{44}\text{.01}\text{g/mol}}\\ \text{Mass}=23.62\text{mol}\times \text{44}\text{.01}\text{g/mol}\\ =\underset{\_}{1039.51g}\end{array}$

Conclusion

The grams of liquid methanol required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ is $\underset{\_}{757.02g}$ and grams of carbon dioxide produced in the combustion reaction is $\underset{\_}{1039.51g}$