Chapter 9, Problem 9.110QA

Interpretation Introduction

To find:

a) Calculate the fuel value of $C_6{H}_{14}$, where ${∆H}_{comb}^o=-4163 kJ/mol$.

b) How much heat is released during the combustion of $1.00 kg C_6{H}_{14}$?

c) How many grams of $C_6{H}_{14}$ are needed to heat $1.00 kg C_6{H}_{14}$ of water from $25.0^{o}C$ to $85.0^{o}C$? Assume that all the heat released during combustion is used to heat the water.

d) Assume white gas is 25% $C_5$ hydrocarbons (fuel value = 48.99 kJ/g) and 75% $C_6$ hydrocarbons; how many grams of white gas are needed to heat 1.00 kg of water from $25.0^{o}C$ to $85.0^{o}C$?

Answer to Problem 9.110QA

Solution:

a) Fuel value of $C_6{H}_{14}$, where ${∆H}_{comb}^o= -4163 kJ/mol$  is $48.3 kJ/g$.

b) $4.83\times {10}^{4}kJ$ of heat is released during the combustion of $1.00 kg$ of $C_6{H}_{14}$.

c) $5.19 g C_6{H}_{14}$ must be burned to heat $1.00 kg$ of water from $25.0^{o}C$ to $85.0^{o}C$.

d) $5.17 g$ of white gas is needed to heat $1.00 kg$ of water from $25.0^{o}C$ to $85.0^{o}C$.

Explanation of Solution

1) Concept:

Fuel value is the amount of energy generated by complete combustion of a particular mass of fuel (hydrocarbon, $1$ gram).

This can be calculated from the enthalpy change for combustion reaction.

From the heat of combustion of $1$ mole of substance, heat released during the combustion of a particular amount can be calculated.

The amount of energy required to raise the temperature can be calculated using mass, specific heat, and change in temperature,

i.e.$, q=m{c}_p∆T$

where,

$q$= amount of energy

$m$= mass

$c_p$= specific heat (the amount of energy required to raise the temperature per unit mass by ${ 1}^{o}C$.)

$∆T$= change in temperature (final temperature – initial temperature)

2) Formula:

i. $q=m{c}_p∆T$

 ii. $1 kg = 1000 g$

3) Given:

i) ${∆H}_{comb}^o=-4163 kJ/mol C_6{H}_{14}$

ii) Amount of $C_6{H}_{14}=1.00 kg$

iii) Amount of water $=1.00 kg$

iv) Initial temperature of water = $25.0^{o}C$

v) Final temperature of water = $85.0^o C$

vi) White gas = $25\% C_5{H}_{12} + 75\% C_6{H}_{14}$

vii) Fuel value of $C_5{H}_{12}= 48.99 kJ/g$ (From Q. 9.109)

4) Calculation:

a) Calculating fuel value of $C_6{H}_{14}$

Heat of combustion per mol is the amount of energy released in complete combustion of 1 mol of that substance.

Heat of combustion ${∆H}_{comb}^o$ of 1 mole of $C_6{H}_{14}$ is $-4163 kJ/mol$.

Mass of $1$ mol of $C_6{H}_{14}=86.1766 g$

Therefore, the fuel value of $C_6{H}_{14}$ in $kJ/g$ is

$\frac{4163 kJ}{mol}\times \frac{1 mol C_6{H}_{14} }{86.1766 g}=48.3 kJ/g$

Thus, the fuel value for $C_6{H}_{14}$ is $48.3\frac{kJ}{g}.$

b) Calculating the heat released during the combustion of $1.00 kg C_6{H}_{14}$:

Converting mass in $kg to g$.

$1.00 kg C_6{H}_{14}\times \frac{1000 g}{1 kg}=1.00\times {10}^3 g C_6{H}_{14}$

Calculating moles from given mass using molar mass as

Molar mass of $C_6{H}_{14}=86.1766 g/mol$

$1.00\times {10}^3 g C_6{H}_{14}\times \frac{1 mol}{86.1766 g}=11.6040 mol C_6{H}_{14}$

Heat released for $1$ mole of $C_6{H}_{14}$ is $4163 kJ$

$\mathrm{H}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{ }11.6040 mol C_6{H}_{14}\times \frac{4163 kJ}{mol}=48370.77 kJ=4.83\times {10}^{4}kJ$

 

c) Calculating amount of $C_6{H}_{14}$ that must be burned to heat $1.00 kg$ of water from $25.0^{o}C$ to $85.0^o C$:

Assume that all the heat released during combustion is used to heat the water. Hence, the amount of energy gained by water will be equal to amount of energy released by the combustion of $C_6{H}_{14}$.

i.e., ${- q}_{C_6{H}_{14}}=q_{water}$

$c_p$ for liquid water is $4.18 J/\left(g{.}^{o}C\right)$, initial temperature of water =$25.0^{o}C$, final temperature of water =$85.0^{o}C$.

Converting the mass of water from $kg to g$,

$1.00 kg \times \frac{1000 g}{1 kg}=1.00\times {10}^3 g$

Therefore, energy required to increase the temperature of $1.00 kg$ water from $25.0^{o}C$ to $85.0^o C$  is

$q_{water}=m_{water}\times c_p\times ∆T$

Plug in the given values:

$q_{water}=1.00\times {10}^3 g \times 4.18J/(g \times ℃)\times ( 85.0^o C-25.0^{o}C)$

$q_{water}=250800 J =250.8 kJ$

Therefore, ${- q}_{C_6{H}_{14}}=250.8 kJ,$ i.e., $C_6{H}_{14}$ must release $250.8 kJ$ of heat.

Here, we assume that all heat, which is required to heat water, is coming from the combustion of $C_6{H}_{12}$. From the given enthalpy of combustion for $C_6{H}_{12},$ we can find moles of $C_6{H}_{12}$ that must be used to release $250.8 kJ$ of heat as

$250.8 kJ\times \frac{1 mol }{4163 kJ}=0.060245 mol C_6{H}_{14}$

Converting moles to mass using molar mass as

$0.060245 mol C_6{H}_{14}\times \frac{86.1766 g}{1 mol}=5.19 g C_6{H}_{14}$

Therefore, $5.19 g C_6{H}_{14}$ must be burned to heat $1.00 kg$ of water from $25.0^{o}C$ to $85.0^{o}C$.

d) Calculating the amount of white gas needed to heat $1.00 kg$ of water from $25.0^{o}C$ to $85.0^{o}C$:

White gas = $25\% C_5{H}_{12} + 75\% C_6{H}_{14}$

Fuel value of $C_5{H}_{12}= 48.99 kJ/g$ (from Q. 9.109)

Fuel value of $C_6{H}_{14}=48.31 kJ/g$ (calculated above)

$Fuel value for 25\% C_5{H}_{12}=\frac{48.99 kJ}{g}\times \frac{25}{100}=12.2475 kJ/g$

$Fuel value for 75\% C_6{H}_{14}=\frac{48.31 kJ}{g}\times \frac{75}{100}=36.2325kJ/g$

$Total fuel value = 12.2475 kJ/g+36.2325 kJ/g=48.48 kJ/g$

Thus, the total fuel value of white gas is $48.48 kJ/g$.

The energy required to increase the temperature of $1.00 kg$ water from $25.0^{o}C$ to $85.0^o C$  is

$250.8 kJ$.

So, the mass of white gas required to released $250.8 kJ$ energy is

$250.8 kJ\times \frac{1 g}{48.48 kJ}=5.17 g of white gas$

Therefore, $5.17 g$ of white gas is needed to heat $1.00 kg$ of water from $25.0^{o}C$ to $85.0^{o}C$.

Conclusion:

Fuel value for the mixture of hydrocarbons is calculated using the proportion in which they are mixed.