Using calorimeters for accurate heat measurement
Calorimetry is an experiment method by which the heat energy released or absorbed is measured. Conducting the experiment in a calorimeter provides more precise heat measurements. With a thick layer of insulation, a calorimeter is a device that can provide more accurate energy change values.
A bomb calorimeter, as shown below, is a device that has an insulated container in which a sealed, oxygen-filled reaction vessel is surrounded by a known volume of water. This allows a complete combustion reaction to happen within the calorimeter and then the heat released from the combustion vessel can be transferred directly to the surrounding water, increasing the accuracy of measurements.
To determine the enthalpy change of a chemical reaction occuring in a solution, a solution calorimeter is used. In this device, the reactants are sealed in a glass bulb immersed in water. The reaction is initiated by breaking the glass bulb containing reactant, allowing the reaction to proceed without opening the lid and minimising the heat loss.
Both bomb calorimeters and solution calorimeters give a more accurate measurement of energy content than a measurement performed in a beaker. The thick insulation minimises the heat loss, ensuring more reliable results.
Both bomb calorimeters and solution calorimeters can be calibrated as some heat loss will still occur. Calibration establishes the relationship between the observed temperature change and the energy change in a reaction, and is represented as a calibration factor when calculating enthalpy change.
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Calculating energy content using calorimetry
The calibration factor (CF) represents the amount of energy required to change the temperature of the calorimeter’s contents by 1 oC. It is expressed in J/°C or kJ/°C.
Calibration involves providing a known amount of heat and observing the temperature change. Calibration should be carried out before each use of the calorimeter. For consistency, the same amount of water must be used during both the calibration and the actual experiment to minimise errors and ensure accuracy.
There are two different calibration approaches:
The calibration factor is determined by passing a known current through a heater for a set time and measuring the temperature rise.
\[E=Vlt\]
where:
- E is the energy released from the heater (joules, \(J\))
- V is the voltage provided to the heater (volts, \(V\))
- I is the current flow through the heater (amperes, \(A\))
- t is the time that electrical energy is applied (seconds, \(s\)).
The calibration factor is calculated by combusting or dissolving a substance with a known enthalpy change (ΔH) and measuring the temperature change.
\[E=n\Delta{H}\]
where:
- \(E\) is the energy released from the given amount of chemical (kilojoules, kJ)
- \(n\) is the number of moles of the known chemical involved (moles, mol)
- \(\Delta{H}\) is the molar enthalpy change of the known compound (molar enthalpy, kJ mol-1).
Calibration Factor
After energy change during the calibration step is determined, the calibration factor (CF) can be calculated:
\[CF=\frac{E}{\Delta{T}}\]
where
- \(E\) is the energy change involved in the calibration process (J or kJ)
- \(\Delta{T}\) is the temperature change during the calibration process (oC or K)
Testing an unknown sample
After calibration, the unknown sample will be tested and the resulting temperature change (T) will be recorded. The energy due to the tested sample can be calculated by :
\[q=CF\times\Delta{T}\]
Depending on what the desired measurement is, the following formulas could be used:
- Energy content of sample \((\text{ J g}^{-1}\text{ or kJ g}^{-1})=\frac{q}{m}\)
- Molar enthalpy change of tested sample \((\text{J mol}^{-1}\text{ or kJ mol}^{-1}) =\frac{q}{n}\)
Worked example
A solution calorimeter that contains 80 g H2O was calibrated using electrical calibration. A current of 1.25 A was passed for 2.0 mins at a potential difference of 5.50 V. The temperature of the water in the calorimeter rose by 2.9 oC.
- Calculate the calibration factor (CF) of the calorimeter in J oC-1.
- If the calibrated calorimeter recorded a temperature increase of 2.5 oC when 2.0 g de-icing powder (with a key active component CaCl2) is added. Determine the energy released during this dissolving process.
Solution
The following table can be used to organise the given information and plan your calculation:
| Calibration Process | Measuring unknown sample |
|---|---|
\(V=5.50V\) \(l=1.25A\) \(t=1.2\times10^2s\) \(\Delta{T}=2.9^{\circ}C\) | \(\Delta{T}=2.5^{\circ}C\) |
Step 1: Calculate the energy released during calibration
\[q=V\times{l}\times{t}=5.50\times1.25\times1.2\times10^2=825J\]
Step 2: Determine CF
\[CF=\frac{825}{2.9}=2.8\times10^2J^{\circ}C^{-1}\]
Step 3: Calculation involving the measurement of the unknown sample
\[q=CF\times\Delta{T}=2.8\times10^2\times2.5=7.1\times10^2J\]
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Worked example
The molar enthalpy of combustion of ethanol (C2H5OH) is -1370 kJ mol-1. A bomb calorimeter is calibrated by combusting 1.00 g of ethanol, releasing 29.7 kJ of energy, causing a temperature rise of 5.5°C in the calorimeter. After calibration, the calorimeter is then immediately used to determine the energy content of a biodiesel sample.
When 1.31 g of this biodiesel is combusted, the temperature of the calorimeter increases by 7.0°C. What is the energy content of the biodiesel sample in kJ g-1?
Solution
Step 1: Calibration process
\[n=\frac{m}{M}=\frac{1.00}{46.0}=0.0217mol\]
\[q=n\times{\Delta{H}}=0.0217\times1370=29.8kJ\]
Step 2: Calculating calibration factor
\[CF=\frac{q}{\Delta{T}}=\frac{29.8}{5.5}=5.4kJ^{\circ}C^{-1}\]
Step 3: Calculation involving the test of unknown sample
\[q=CF\times{\Delta{T}}=5.4\times7.0=38kJ\]
\[\text{Energy Content}=\frac{q}{m}=\frac{38}{1.31}=29\text{kJ g}^{-1}\]