Topic 6 · page 7 of 15
Errors, uncertainty and assumptions
To evaluate a calorimetry result, name the source of error, say which way it moves the answer, and suggest an improvement. Uncertainty is the limit of your apparatus, written as a percentage; assumptions are what the calculation takes for granted.
Read first: From joules to kJ mol⁻¹
Sources of error 6.8
The burner experiment on the last page gave −601 kJ mol⁻¹ against a data-book value of −1367 kJ mol⁻¹. A result so far from the accepted value is a systematic problem, not bad luck. In these experiments the main problem is the same one every time: energy does not all go where you assume it goes.
| Source of error | Effect on the result | Improvement |
|---|---|---|
| Heat lost to the surroundings (the air, the cup, the can) | ΔT too small, so ΔH less exothermic than the true value | lid, polystyrene or foam insulation, a draught shield, lagging; apply a cooling-curve correction |
| Incomplete combustion (soot on the can, a smoky flame) | less energy released per mole of fuel burned | adjust the air supply so the flame is clean |
| Fuel evaporates or the burner cap is left off | the mass loss overstates the moles burned, so ΔH is too small | cap the burner each time it is weighed |
| Heat absorbed by the cup, the can or the thermometer is ignored | ΔT too small | allow for the heat capacity of the apparatus |
| Water leaves as vapour, not liquid water | less energy released than for the standard combustion (which forms liquid water) | a limit of the method; mention it, then compare with the data book |
| Not standard conditions (temperature, concentration) | the result differs slightly from the standard value | quote the conditions used |
A heat-loss or incomplete-combustion error acts the same way every time. It is a systematic error, and repeating the experiment does not remove it. Scatter between repeats, for example from reading a thermometer, is random error. Repeats and a mean reduce random error only.
How far a result is from the accepted value can be given as a percentage:
For the burner result this is (1367 − 601) ÷ 1367 × 100 = 56.0% too small in size.
Uncertainty 6.8
Every measurement has a limit set by the apparatus.
Uncertainty on its own does not show how much it matters. A ±1 °C uncertainty is huge on a 7 °C rise and tiny on a 70 °C rise. Compare them as a percentage:
A temperature change uses two readings, so the two uncertainties add. A mass found by weighing before and after also uses two readings.
Worked exampleUncertainty in ΔT and in a mass
The temperature rise in the neutralisation experiment was 6.8 °C. Find the percentage uncertainty with (a) a thermometer with 1 °C divisions; (b) a thermometer with 0.1 °C divisions. (c) A balance reading to ±0.01 g is used before and after burning 0.80 g of fuel. Find the percentage uncertainty in the mass burned.
(a) Each reading ±0.5 °C, so ΔT is ±(0.5 + 0.5) = ±1.0 °C. Percentage = 1.0 ÷ 6.8 × 100 = 15%.
(b) Each reading ±0.05 °C, so ΔT is ±0.1 °C. Percentage = 0.1 ÷ 6.8 × 100 = 1.5%.
(c) Two readings, so ±0.02 g. Percentage = 0.02 ÷ 0.80 × 100 = 2.5%.
Two conclusions follow. A thermometer with finer divisions is a cheap, large improvement: the uncertainty in ΔT falls from 15% to 1.5%. And a bigger temperature change makes the same ±0.1 °C smaller as a percentage, which is why you choose masses and volumes that give a rise of several degrees.
To see which measurement limits the final answer, work out the percentage uncertainty of every measurement used. The largest one matters most. When several measurements are multiplied or divided, you add their percentage uncertainties to get the uncertainty in the result.
Assumptions 6.8
The calculation q = mcΔT takes several things for granted. Each is an assumption, and each is slightly untrue.
| Assumption | Why it is not exactly true |
|---|---|
| all the energy goes into the water or solution | some is lost to the air, the cup, the can and the thermometer |
| the solution has the specific heat capacity of water (4.18 J g⁻¹ °C⁻¹) and the density of water (1.00 g cm⁻³) | a dissolved solute changes both slightly |
| the cup, can and thermometer absorb no heat | they do absorb some |
| the reaction goes to completion (or the fuel burns completely) | some reactant may be left, or soot may form |
| no heat is lost while the reaction is happening | the mixture starts to cool as soon as it warms |
Check yourself
- A thermometer reads to ±0.05 °C. In an experiment the temperature rises by 11.0 °C. Calculate the percentage uncertainty in the temperature change.
- A balance reads to ±0.001 g. Magnesium is weighed before and after transferring it, and the mass used is 0.120 g. Calculate the percentage uncertainty in this mass.
- State two sources of error that make the value of the enthalpy change of combustion of ethanol from a burner experiment too small, and explain how each does this.
- A student finds ΔcH of ethanol to be −752 kJ mol⁻¹. The data-book value is −1367 kJ mol⁻¹. Calculate the percentage difference.
- Explain why replacing a thermometer with 1 °C divisions by one with 0.1 °C divisions gives a better result when the temperature rise is about 7 °C.
- State two assumptions made in calculating the enthalpy change of neutralisation, and explain the effect on the result if one of them is not true.
Answers cover these until you have tried the questions
- Two readings: ±(0.05 + 0.05) = ±0.1 °C. Percentage = 0.1 ÷ 11.0 × 100 = 0.91%.
- Two readings: ±(0.001 + 0.001) = ±0.002 g. Percentage = 0.002 ÷ 0.120 × 100 = 1.7%.
- Any two of: heat lost to the surroundings (the can and the air), so the water warms less than it should; incomplete combustion, so less energy is released per mole of fuel; evaporation of fuel, so the mass loss is greater than the mass burned; water vapour formed, so less heat is given out than for liquid water. Each makes the energy transferred per mole smaller than it should be, so ΔH is less negative.
- (1367 − 752) ÷ 1367 × 100 = 615 ÷ 1367 × 100 = 45.0%.
- The uncertainty in ΔT falls from ±1.0 °C to ±0.1 °C. On a rise of about 7 °C the percentage uncertainty falls from about 15% to about 1.5%, so the calculated ΔH is far more reliable.
- Any two of: all the energy goes into the solution (some is lost, so ΔT is too small and ΔH is less exothermic); the solution has the specific heat capacity and density of water (the true values differ slightly, so q is slightly wrong, in a direction that depends on the solution); the cup absorbs no heat (some is absorbed, so ΔT is too small and ΔH is less exothermic).