Chapter 6, Problem 6.98P

(a)

Interpretation Introduction

Interpretation:

The joules are in $1.00\text{ Btu}$ is to be calculated.

Concept introduction:

The calorie $\left(4.184\text{J}\right)$ is defined as the amount of energy required to increase the temperature of $1.00\text{ g}$ of liquid water by $1.00°\text{C}$. The British thermal unit is defined as the amount of energy required to increase the temperature of $1.00\text{ lb}$ of liquid water by $1.00°\text{F}$.

The conversion factor to convert $\text{lb}$ into $\text{g}$ is:

  $\text{1}\text{lb}=453.6\text{g}$

The conversion factor to convert $°\text{C}$ into $°\text{F}$ is:

  $\text{1}°\text{C}=1.8°\text{F}$

Answer to Problem 6.98P

$1.1\times {10}^3\text{J/Btu}$ are present in $1.00\text{ Btu}$.

Explanation of Solution

$1.00\text{ Btu}$ energy in joules is calculated as follows:

  $\begin{array}{c}\text{Energy}=\left(\frac{1.00\text{ lb }°\text{F}}{1.00\text{ Btu}}\right)\left(\frac{1.0°\text{C}}{1.8°\text{F}}\right)\left(\frac{453.6\text{ g}}{\text{1 lb}}\right)\left(\frac{4.184\text{ J}}{\text{1}\text{cal}}\right)\left(\frac{\text{1}\text{cal}}{\text{g}°\text{C}}\right)\\ =1054.368\text{J/Btu}\\ \approx 1.1\times {10}^3\text{J/Btu}\text{.}\end{array}$

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

(b)

Interpretation Introduction

Interpretation:

The joules in $1.00\text{ therm}$ is to be calculated.

Concept introduction:

The calorie $\left(4.184\text{J}\right)$ is defined as the amount of energy required to increase the temperature of $1.00\text{ g}$ of liquid water by $1.00°\text{C}$. The British thermal unit is defined as the amount of energy required to increase the temperature of $1.00\text{ lb}$ of liquid water by $1.00°\text{F}$.

The conversion factor to convert $\text{therm}$ into $\text{btu}$ is:

  $\text{1}\text{therm}=100,000\text{btu}$

The conversion factor to convert $\text{btu}$ into $\text{J}$ is:

  $\text{1}\text{btu}=1054.368\text{J}$

Answer to Problem 6.98P

$1.1\times {10}^8\text{J}$ are present in $1.00\text{ therm}$.

Explanation of Solution

$1.00\text{ therm}$ energy in joules is calculated as follows:

  $\begin{array}{c}\text{Energy}=1.00\text{ therm}\left(\frac{100,000\text{btu}}{\text{1}\text{therm}}\right)\left(\frac{1054.368\text{J}}{\text{1}\text{btu}}\right)\\ =1.054368\times {10}^8\text{J}\\ \approx 1.1\times {10}^8\text{J}\text{.}\end{array}$

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

(c)

Interpretation Introduction

Interpretation:

The mole of methane that must be burned to give $1.00 \text{therm}$ of energy is to be calculated.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactant at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum n\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

Answer to Problem 6.98P

$1.3\times {10}^2\text{mol}$ of methane must be burned to give $1.00 \text{therm}$ of energy.

Explanation of Solution

The balanced chemical equation for the reaction of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_2$ is as follows:

  ${\text{CH}}_{\text{4}}\left(g\right)+2{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The formula to calculate the standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{1\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\\ \left\{1\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{4}}\left(g\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]\right\}\end{array}\right]$        (1)

Substitute $-393.5\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CO}}_2\left(g\right)\right]$, $-285.840\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$, $-74.87\text{kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CH}}_{\text{4}}\left(g\right)\right]$ and 0 for $\Delta H_{\text{f}}^{°}\left[{\text{O}}_2\left(g\right)\right]$ in the equation (1).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(\text{1}\text{mol}\right)\left(-393.5\text{kJ/mol}\right)+\left(2\text{mol}\right)\left(-285.840\text{kJ/mol}\right)\right\}-\\ \left\{\left(\text{1}\text{mol}\right)\left(-74.87\text{kJ/mol}\right)+0\right\}\end{array}\right]\\ =–802.282\text{ kJ/mol}\end{array}$

The number of moles that must be burned to give $1.00 \text{therm}$ of energy is calculated as follows:

  $\begin{array}{c}\text{Number of moles}=\frac{1.00 \text{therm}}{–802.282\text{ kJ/mol}}\left(\frac{\text{1}\text{kJ}}{1000\text{J}}\right)\left(\frac{1.054368\times {10}^8\text{J}}{\text{1}\text{therm}}\right)\\ =131.4211\text{ mol}\\ \approx 1.3\times {10}^2\text{mol}\text{.}\end{array}$

Conclusion

A combustion reaction is a reaction in which reactant is reacted with molecular oxygen to form the product.

(d)

Interpretation Introduction

Interpretation:

The cost per mole of methane is to be calculated.

Concept introduction:

The calorie $\left(4.184\text{J}\right)$ is defined as the amount of energy required to increase the temperature of $1.00\text{ g}$ of liquid water by $1.00°\text{C}$. The British thermal unit is defined as the amount of energy required to increase the temperature of $1.00\text{ lb}$ of liquid water by $1.00°\text{F}$.

Answer to Problem 6.98P

$\$0.0050\text{/mol}$ is the cost per mole of methane.

Explanation of Solution

The number of moles in $1.00 \text{therm}$ is $131.4211\text{ mol}$.

The cost per mole of methane is calculated as follows:

  $\begin{array}{c}\text{Cost}=\left(\frac{\text{\$}0.66}{\text{therm}}\right)\left(\frac{\text{1}\text{therm}}{131.4211\text{ mol}}\right)\\ =\$0.005022\text{/mol}\\ \approx \$0.0050\text{/mol}\text{.}\end{array}$

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

(e)

Interpretation Introduction

Interpretation:

The cost to warm $\text{318 gal}$ of water in a hot tub from $15.0°\text{C}$ to $42.0°\text{C}$ is to be calculated.

Concept introduction:

Specific heat capacity $\left(c\right)$ of a substance is the amount of heat needed to raise the temperature of $1\text{g}$ of a substance by $1\text{K}$. The formula to calculate heat required is as follows:

  $q=\left(\text{mass}\right)\left(c\right)\left(T_2-T_1\right)$        (2)

Here,

$T_2$ is the final temperature.

$T_1$ is the initial temperature.

$q$ is the heat released or absorbed.

$c$ is the specific heat capacity of the substance.

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is ${\text{kg/m}}^{\text{3}}$. The formula to calculate density is,

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$        (3)

The conversion factor to convert $\text{gal}$ to $\text{qt}$ is as follows:

  $1\text{gal}=4\text{qt}$

The conversion factor to convert $\text{qt}$ to $\text{L}$ is as follows:

  $1\text{L}=1.057\text{qt}$

Answer to Problem 6.98P

The cost to warm $\text{318 gal}$ of water is $\$0.85$.

Explanation of Solution

Rearrange the equation (3) to calculate mass of water.

  $\text{Mass}\text{of water}=\left(\text{Density}\right)\left(\text{Volume}\right)$        (4)

Substitute $1\text{g/mL}$ for density and $318\text{gal}$ for volume in the equation (4).

  $\begin{array}{c}\text{Mass}\text{of water}=\left(1\text{g/mL}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(318\text{gal}\right)\left(\frac{4\text{qt}}{1\text{gal}}\right)\left(\frac{1\text{ L}}{1.057\text{qt}}\right)\\ =1.203406\times {10}^6\text{g}\end{array}$

Substitute $1.203406\times {10}^6\text{g}$ for mass, $4.184\text{J/g}\cdot °\text{C}$ for $c_{\text{water}}$, $42°\text{C}$ for $T_2$ and $15°\text{C}$ for $T_1$ in the equation (3).

  $\begin{array}{c}q=\left(1.203406\times {10}^6\text{g}\right)\left(4.184\text{J/g}\cdot °\text{C}\right)\left(42°\text{C}-15°\text{C}\right)\\ =1.359464\times {10}^8\text{J}\end{array}$

The cost to warm $\text{318 gal}$ of water is calculated as follows:

  $\begin{array}{c}\text{Cost}=1.359464\times {10}^8\text{J}\left(\frac{\text{1}\text{therm}}{1.054368\times {10}^8\text{J}}\right)\left(\frac{\text{\$}0.66}{\text{therm}}\right)\\ =\$0.850999\\ \approx \$\mathrm{0.85.}\end{array}$

Conclusion

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.