Chapter 9, Problem 9.109QA

Interpretation Introduction

To find:

a) Calculate the fuel value of $C_5{H}_{12}$, given that ${∆H}_{comb}^o= -3535 kJ/mol$.

b) How much heat is released during the combustion of $1.00 kg$ of $C_5{H}_{12}$?

c) How many grams of $C_5{H}_{12}$ must be burned to heat $1.00 kg$ of water from $20.0^{o}C$ to $90.0^o C$? Assume that all the heat released during combustion is used to heat the water.

Answer to Problem 9.109QA

Solution:

a) Fuel value of $C_5{H}_{12}$, where ${∆H}_{comb}^o= -3535 kJ/mol$ is $48.99 kJ/g$

b) $4.90\times {10}^{4}kJ$ of heat is released during the combustion of $1.00 kg$ of $C_5{H}_{12}$.

c) $5.97 g C_5{H}_{12}$ must be burned to heat $1.00 kg$ of water from $20.0^{o}C$ to $90.0^o C$.

Explanation of Solution

1) Concept:

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

This can be calculated from the enthalpy change for combustion reaction $\left({∆H}_{comb}^o\right)$

The value for heat of combustion is in the units $kJ/mol$, so the amount of released heat is in $kJ$ when $1 mol$ of that substance is combusted.

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:

$q=m{c}_p∆T$

3) Given:

i) ${∆H}_{comb}^{o}of C_5{H}_{12}= -3535 kJ/mol$

ii) Amount of $C_5{H}_{12}=1.00 kg$

iii) Amount of water $=1.00 kg$

iv) Initial temperature of water = $20.0^{o}C$

v) Final temperature of water = $90.0^o C$

4) Calculations:

a) Calculating fuel value for $C_5{H}_{12}$

Combustion reaction of $C_5{H}_{12}$ is

$C_5{H}_{12}+8 O_2\to 5C{O}_2+6H_{2}O$

Heat of combustion ${∆H}_{comb}^o$ of $1$ mole of $C_5{H}_{12}$ is $-3535 kJ/mol$.

The molar mass of $C_5{H}_{12}=72.1498 g/mol$

Therefore, the fuel value of $C_5{H}_{12}$ $kJ/g$ is

$Fuel value = \frac{3535 kJ}{mol}\times \frac{1 mol C_5{H}_{12} }{72.1498 g}=48.99 kJ/g$

Thus, the fuel value for $C_5{H}_{12}$ is $48.99 kJ/g$.

b) Calculating heat released during the combustion of $1.00 kg$ of $C_5{H}_{12}$:

Convert the mass in $kg to g$.

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

Converting mass in grams into moles using molar mass as

$1.00\times {10}^3 g C_5{H}_{12}\times \frac{1 mol}{72.1498 g}=13.86 mol C_5{H}_{12}$

Heat released for $1$ mole of $C_5{H}_{12}$ is $3535 kJ/mol$.

Therefore, calculating heat released for $13.86 mol C_5{H}_{12}$ as

$\mathrm{ }13.86 mol C_5{H}_{12}\times \frac{3535 kJ}{mol}=48995.1 kJ=4.90\times {10}^{4}kJ$

Thus, $4.90 \times {10}^4 kJ$ of heat should be released when $1 kg$ of $C_5{H}_{12}$ is combusted.

c) Calculating the amount of $C_5{H}_{12}$ that must be burned to heat $1.00 kg$ of water from $20.0^{o}C$ to $90.0^o C$

We need to find here first the heat required to heat water, which can be calculated as

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

Therefore, energy required to increase the temperature of water from $20.0^{o}C$ to $90.0^o C$  is

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

$q_{water}=1.00\times {10}^3 g \times 4.18J/\left(g{.}^{o}C\right)\times ( 90.0^o C-20.0^{o}C)$

$q_{water}=292600 J =292.6 kJ$

Here we assume that all the heat, which is required to heat water is coming from the combustion of $C_5{H}_{12}$. From the given enthalpy of combustion for $C_5{H}_{12,}$ we can find moles of $C_5{H}_{12}$ that must be used to release $292.6 kJ$ of heat as

$292.6 kJ\times \frac{1 mol }{3535 kJ}=0.08277 mol C_5{H}_{12}$

Converting moles of $C_5{H}_{12}$ to mass using molar mass:

$0.08277 mol C_5{H}_{12}\times \frac{72.1498 g}{1 mol}=5.97 g C_5{H}_{12}$

Therefore, $5.97 g C_5{H}_{12}$ must be burned to heat $1.00 kg$ of water from $20.0^{o}C$ to $90.0^o C$.

Conclusion:

All the heat lost by a combustion reaction is gained by water, assuming the conservation of energy.