Chapter 4, Problem 122QRT

(a)

Interpretation Introduction

Interpretation:

Volume of tank car has to be estimated. The mass of propane in the tank car has to be calculated. Also the energy transferred at burning of propane has to be identified.

Answer to Problem 122QRT

Volume of tank car is ${\text{35 m}}^{\text{3}}$

The mass of propane in the 5 tank cars are $\text{1}\text{.0}\times {\text{10}}^{\text{5}}{\text{ kg C}}_{\text{3}}{\text{H}}_{\text{8}}$.

The energy transferred at burning of propane is $\text{4}\text{.6}\times {\text{10}}^{\text{9}}\text{ kJ}$.

Explanation of Solution

The dimensions of tank car are 20 feet $\left(\text{6}\text{.1m}\right)$ long, 8 feet $\left(\text{2}\text{.4 m}\right)$ high and 8 feet $\left(\text{2}\text{.4 m}\right)$ wide. Now the volume of tank car is determined as follows,

    $\begin{array}{c}\text{Volume = 6}\text{.1m × 2}\text{.4 m × 2}\text{.4 m}\\ \\ {\text{= 35 m}}^{\text{3}}\end{array}$

The liquid propane density at its boiling point is found to be ${\text{582 kg/m}}^{\text{3}}$.

    $\begin{array}{l}\Rightarrow {\text{5 Cars × 35m}}^{\text{3}}\times \frac{582{\text{ kg C}}_{\text{3}}{\text{H}}_{\text{8}}}{1{\text{m}}^{\text{3}}{\text{ C}}_{\text{3}}{\text{H}}_{\text{8}}}\\ \\ \Rightarrow \text{ 1}\text{.0}\times {\text{10}}^{\text{5}}{\text{ kg C}}_{\text{3}}{\text{H}}_{\text{8}}\end{array}$

The balanced chemical equation for combustion of propane is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(g\right){\text{ + 5O}}_{\text{2}}\left(g\right)\to 3{\text{CO}}_{\text{2}}\left(g\right){\text{ + 4H}}_{\text{2}}\text{O}\left(g\right)$.

Now, the reaction enthalpy for above reaction is shown as below,

    $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{ = 3}\left(\text{-393}\text{.509 kJ/mol}\right)\text{ + 4}\left(\text{-241}\text{.818 kJ/mol}\right)\text{ - }\left(\text{-103}\text{.8 kJ/mol}\right)\text{-5}\left(\text{0 kJ/mol}\right)\\ \\ =\text{ -2044}\text{.0 kJ}\end{array}$

Finally the energy evolved by the mass of propane present in 5 tank cars is calculated as follows,

    $\begin{array}{l}\Rightarrow \text{1}\text{.0 }\times {\text{10}}^{\text{5}}{\text{ kg C}}_{\text{3}}{\text{H}}_{\text{8}}\text{ × }\frac{1000\text{ g}}{1\text{ kg}}\times \frac{1{\text{ mol C}}_{\text{3}}{\text{H}}_{\text{8}}}{44.0953{\text{ g C}}_{\text{3}}{\text{H}}_{\text{8}}}\times \frac{2044.0\text{ kJ}}{1{\text{ mol C}}_{\text{3}}{\text{H}}_{\text{8}}}\\ \\ \Rightarrow \text{ 4}\text{.6}\times {\text{10}}^{\text{9}}\text{ kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The metric kilotons of $\text{TNT}$ that provide energy equal to the energy produced by combustion of propane in five car tanks has to be determined.

Answer to Problem 122QRT

The amount of $\text{TNT}$ is $1.0$ metric kilotons of $\text{TNT}$.

Explanation of Solution

First, the molar mass of $\text{TNT}$(${\text{C}}_{\text{7}}{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{6}}$) is calculated.

    $\begin{array}{c}\text{Molar mass = 7}\left(12.0107\text{ g/mol C}\right)+5\left(1.0079\text{ g/mol H}\right)+3\left(14.0067\text{ g/mol N}\right)\\ +6\left(15.9994\text{ g/molO}\right)\\ \\ =\text{ 227}\text{.1309 g/mol TNT}\end{array}$

Now the energy evolved by the propane mass in 5 tank cars is converted into moles which then using metric mass relationships the metric kilotons of $\text{TNT}$ is determined as shown below,

    $\begin{array}{l}\Rightarrow \text{4}{\text{.7 ×10}}^{\text{9}}\text{ kJ ×}\frac{\text{1 mol TNT}}{\text{1066}\text{.1 kJ}}\text{× }\frac{\text{227}\text{.1309 g TNT}}{\text{1 mol TNT}}\text{× }\frac{\text{1 Mg}}{{\text{1 ×10}}^{\text{6}}\text{ g}}\text{×}\frac{\text{1 metric ton}}{\text{1 Mg}}\times \frac{1\text{ m}\text{.k}\text{.t}}{\text{100 metric ton}}\\ \\ \Rightarrow \text{ 1}\text{.0 metric kilotons}\text{.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The energy transferred from Hiroshima and Nagasaki nuclear fission bombs has to be determined. Also up to date the largest nuclear weapon detonated has to be identified and whether evacuating the town serves as good idea or not has to be indicated.

Answer to Problem 122QRT

The energy of Hiroshima bomb transferred is $6.7\text{ \%}$.

The energy of Nagasaki bomb transferred is $5\text{ \%}$.

The conclusion obtained from this is to evacuate the town serves as the good idea.

Explanation of Solution

The bomb dropped at Hiroshima is $15\text{ kilotons}$ and the one dropped at Nagasaki is $20\text{ kilotons}$.

So far the biggest hydrogen bomb ever tested was the Tsar Bomb. It was tested over the Mityushikha Bay nuclear testing range north of Arctic Circle over the Nocaya Zemlya archipelago in Arctic Sea. About $57$ million tons of $\text{TNT}$ was released from this.

Using percentages the energy released from propane in tank cars are compared with the above bomb as follows,

    $\begin{array}{l}\Rightarrow \frac{1.0\text{ m}\text{.k}\text{.t}}{\text{15 m}\text{.k}\text{.t}}\times 100\%\\ \\ \text{ = 6}\text{.7 \% the energy of Hiroshima bomb}\\ \\ \Rightarrow \frac{1.0\text{ m}\text{.k}\text{.t}}{\text{20 m}\text{.k}\text{.t}}\times 100\%\\ \\ \text{ = 5}\text{.0 \% the energy of Nagasaki bomb}\end{array}$

Therefore, evacuating the town was the good idea.

(d)

Interpretation Introduction

Interpretation:

The energy released by propane combustion has to be compared with the hurricane energy.

Explanation of Solution

The energy released from the hurricane is found to be around $\text{1}{\text{.3×10}}^{\text{17}}\text{ Joules/day}$ which was determined having average wind speed in the inner core of hurricane of radius $60\text{ km}$ as $40\text{ m/s}$.

The energy evolved by the mass of propane present in 5 tank cars is calculated as $\text{4}{\text{.6×10}}^{\text{9}}\text{ kJ}$.

    $\begin{array}{l}\Rightarrow \text{4}{\text{.6×10}}^{\text{9}}\text{ kJ×}\frac{\text{1000J}}{\text{1kJ}}\times \frac{\text{ 1 day}}{1.3\times {10}^{17}\text{ J}}\times \frac{24\text{ h}}{\text{ 1 day}}\times \frac{3600\text{ s}}{1\text{ h}}\\ \\ \Rightarrow \text{ 3}\text{.1 s}\end{array}$

Therefore, $3.1\text{ s}$ of kinetic energy is produced by hurricane.