Measuring the energy content of a fuel using standard laboratory setups
When scientists evaluate a new fuel, one important measure is its heat of combustion , which indicates the amount of energy released when a specific amount of the fuel undergoes complete combustion . Molar Enthalpy of Combustion, measured in kJ mol-1, can also be used for this evaluation. It represents the enthalpy change when one mole of fuel undergoes complete combustion. The molar enthalpy of combustion is always a negative value, reflecting that combustion is an exothermic process.
Use this page to revise the following concepts within measuring the energy content of a fuel using standard laboratory setups:
- Measuring molar enthalpy of combustion
- Measuring energy content of mixturess
- Heat loss during measurement
- Energy transformation efficiency
Measuring molar enthalpy of combustion
To calculate the amount of energy released from combusting fuel, the energy is used to heat up water using the following set-up:
The energy that is transferred to, or absorbed by, the water can be calculated using:
Heat energy = \(\text{mass of water}\times{\text{specific heat capacity}}\times{\text{temperature change}}\)
which is
\[q=mc\Delta{T}\]
All the measurements in this calculation are related to water:
- \(q\) is the amount of heat energy (in \(J\))
- \(m\) is the mass of water (in \(g\))
- Mass can be calculated from the volume of water given dentistry of water is \(1.0\,g\,mL^{-1}\)
- \(c\) is the specific heat capacity of the water (\(4.18\,J g^{-1} °C^{-1}\))
- \(\Delta T\) is the temperature change of water (in \(°C\) or \(K\)).
The heat derived from fuel (chemical energy) is calculated using \(q = n\Delta H\).
where all measurements are related to the fuel:
- \(n\) is the number of moles of the fuel
- \(\Delta\)H is the molar enthalpy of combustion of the fuel, in \(kJ\,mol^{-1}\).
Assuming there is no heat loss to the surrounding environment, the energy released from the fuel should be the same amount of energy absorbed by water. Using this relationship, we can calculate the \(\Delta{H}\) of the fuel.
Worked example
To determine the molar enthalpy of combustion of octane, 0.450g of octane (C8H18) undergoes complete combustion in a spirit burner. The thermal energy released is used to heat 50 mL of water. The temperature of the water rose from 22.45°C to 41.32°C. Calculate the heat of combustion of octane in kJ mol-1 and write the thermochemical equation for the reaction.
Solution
\[m(H_2O)=density\times{volume}=50\times1.0=50g\]
\[\Delta{T}=41.32-22.45=18.87^{\circ}C\]
\[q=m\times{c}\times\Delta{T}=50\times4.18\times18.87\]
\[=3.94\times10^3J=3.94kJ\]
Assuming no heat loss during energy transfer:
Energy released from combustion = energy absorbed by water
\[n(octane)=\frac{m(octane)}{M(octane)}=\frac{0.450}{114.0}=0.00395mol\]
Because combustion reactions are exothermic reactions, molar heat of combustion \(\Delta{H}<0\)
\[\Delta{H}=-\frac{q}{n}=-\frac{3.94}{0.00395}=-999KJ\,mol^{-1}\]
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Measuring energy content of mixtures
Foods and some fuels, such as diesel, are mixtures rather than pure substances. Since the molar mass of these mixtures cannot be determined, their quantities cannot be expressed in moles. To quantify the energy contained in a certain amount of fuel, we use energy content (\(kJ\,g^{-1}\)).
Energy content can be calculated using the formula:
Energy content = \(\frac{q}{\Delta{m}}\)
where
- \(q\): Energy released during the combustion of fuel, measured by determining the amount of heat absorbed by water
- \(\Delta{m}\): Mass of the fuel consumed during combustion, measured as the mass difference before and after combustion (in g).
Worked example
A 3.00g sample of biodiesel was burned under a steel can containing 200g of water. After combustion, the remaining biodiesel weighed 1.80g, and the water temperature rose by 35.0°C. Calculate the energy content of the biodiesel in kJ g-1.
Solution
Step 1: Calculate \(q\) absorbed by water.
\[q=mc\Delta{T}=200\times4.18\times35.0=2.93\times10^4J=29.3kJ\]
Assuming there is no heat loss, the energy released from combustion is \(2.93\times10^{4}J\), which which was absorbed by the water in the can
Step 2: Determine Δ\(m\) of fuel
\[\Delta{m}=m(\text{initial fuel})-m(\text{final fuel})=3.00-1.80=1.20g\]
Step 3: Calculate energy content
Energy content = \(\frac{q}{\Delta{m}}=\frac{29.3}{1.20}=24.4 kJ g^{-1}\)
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Heat loss during measurement
The calculations of molar enthalpy change and energy content assume that all the energy released during fuel combustion is absorbed by water. However, some energy is lost to the surroundings, making the energy absorbed by the water less than the total energy released. As a result, calculations of molar enthalpy changes and energy content based on this assumption are underestimated. To improve accuracy, we could minimise heat loss by using a lid and insulating the water container.
Energy transformation efficiency
There will always be energy loss during energy transformations and energy transfers. For example, only about 20% of the chemical energy in petrol is used to propel a car, with the rest lost as heat or other waste forms. Therefore, energy efficiency needs to be taken into account in calculations.
Energy efficiency can be calculated using:
Energy Efficiency (%) = \(\frac{\text{Useful Energy Output}}{\text{Total Energy} Input}\times100\)
Total energy input can be calculated from molar enthalpy change or energy content:
- Total Energy Input = \(n\Delta{H}\) or
- Total Energy Input = \(m\times(\text{Energy Content})\)
Useful energy output can be calculated by:
\(q=mc \Delta{T}\)
Worked example
The molar enthalpy of combustion of ethanol is -1370 kJ mol-1 . A student burns 0.050 mol of pure ethanol to heat 250 g of water. The % efficiency of the process is 15%, and the specific heat capacity of water is 4.18 J°C-1g-1. Calculate the expected temperature rise.
\(\text{Total energy input }= 1370 \times 0.050\,mol = 69 kJ\)
Useful energy output = Total energy input energy efficiency = \(69\times 15%\) \(= 10 kJ = 1.0\times 10^{4} J\)
The energy output = \(m\times{c}\times{\Delta{T}}=250\times4.18\times{\Delta{T}}\)
Hence, \(\Delta{T}=\frac{10275}{250\times4.18}=9.8^{\circ}C\)