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Propagation of Uncertainty in Calculations -Uses uncertainty (or precision) of each measurement, arising from limitations of measuring devices. - The importance.

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Presentation on theme: "Propagation of Uncertainty in Calculations -Uses uncertainty (or precision) of each measurement, arising from limitations of measuring devices. - The importance."— Presentation transcript:

1 Propagation of Uncertainty in Calculations -Uses uncertainty (or precision) of each measurement, arising from limitations of measuring devices. - The importance of estimating errors is due to the fact that errors in data propagate through calculations to produce errors in results. Uncertainty propagation is required in IB labs and should help you direct the evaluation part of your conclusion. (DCP, CE) *this simplified version should be all that is needed for IB

2 4 rules for Propagating Uncertainties 1) Addition or subtraction of numbers with uncertainty 2) Multiplication and division of numbers with uncertainty 3) Add or subtract by a pure number 4) Multiplying or dividing by a pure number

3 1) Addition or subtraction of numbers with uncertainty When values are added or subtracted, the absolute uncertainty (AU) of each value is added. –For analog measurements (things with markings and a physical scale) the AU is typically half of the smallest division on the apparatus used ie. 25.0 mL (±0.5 mL) if markings are every 1 ml; thus measurement can be estimated to the nearest half of a mL –For digital measurements, the AU is typically ± smallest division ie. ±0.001 g for a milligram balance reading 3.426 g

4 Example: The change in temp of a mixture can be found by subtracting the initial from the final temperature (ΔT = Tf – Ti). What is the change in temperature for a liquid that started at 18.0°C (±0.5°C) and ended up at 25.0°C (±0.5°C)?  T = 25.0°C (±0.5°C) - 18.0°C (±0.5°C) = 7.0°C (±1°C) To understand this, consider that the real original temp of the liquid must lie between 17.5 °C and 18.5 °C and thus the change in temp can be as high as 8 or as low as 6 …thus difference is 7.0°C (±1°C) Sig. figs. of uncertainty values carry through the calculations independent from the sig. figs. of the measured values; thus, there will always be only one sig. fig. listed for the absolute uncertainty.

5 Practice Problems 1) 10.0 cm 3 of acid is delivered from a 10cm 3 pipette (  0.1 cm 3 ), repeated 3 times. What is the total volume delivered? 2) When using a burette (  0.02 cm 3 ), you subtract the initial volume from the final volume. Final volume = 38.46 cm 3  0.02 cm 3 Initial volume = 12.15 cm 3  0.02 cm 3 What is the total volume delivered?

6 Answers 1)10.0  0.1 cm 3 10.0  0.1 cm 3 Total volume delivered = 30.0 cm 3  0.3 cm 3 2) (38.46 cm 3  0.02 cm 3 ) – (12.15 cm 3  0.02 cm 3 ) = 26.31 cm 3  0.04 cm 3

7 2) Multiplication and division of numbers with uncertainty Percentage (relative) uncertainties are added. –Percentage uncertainty is the ratio of the absolute uncertainty of a measurement to the best estimate. It expresses the relative size of the uncertainty of a measurement (its precision). It is important to know about relative uncertainties so that you can determine if the apparatus used to generate the data is up to the task. % uncertainty = (A.U. / recorded value) x 100

8 EXAMPLE: What is the %U of 2.30g (±0.05g)? (0.05/2.30) x 100 = 2.2%  2% Percent uncertainties should be to 1 sig fig also

9 How does %U help w/ your CE? So now when we look at a calculation and see percentage uncertainties of 10%, 0.1%, 0.05% and 2% in it, hopefully you will realize that you need to lower the 10% uncertainty if you want to reduce the uncertainty of your answer. This should lead you to know exactly how to improve the method significantly (and this should be part of your conclusion & evaluation, CE)

10 Temperature MUST BE in KELVIN when converting to % uncertainty K =  C + 273 So the % uncertainty of a temperature recorded as 2.0°C (±0.5°C) is not 25%, but rather it is (0.5/275) x 100 = 0.2% so final answer is 2.0°C (±0.2%)

11 Example: An object has a mass of 9.01 g (±0.01 g) and when it is placed in a graduated cylinder it causes the level of water in the cylinder to rise from 23.0 cm 3 (±0.5 cm 3 ) to 28.0 cm 3 (±0.5 cm 3 ). Calculate the density of the object. (Recall that cm 3 = mL). Hint: First calculate volume Second calculate density

12 Density = mass / volume First calculate the volume of the object: volume object = final vol. water – initial vol. water volume = [ 28.0 cm 3 (±0.5 cm 3 )] – [23.0 cm 3 (±0.5 cm 3 )] volume = ( 28.0 cm 3 - 23.0 cm 3 ) ± (0.5 cm 3 + 0.5 cm 3 ) Volume = 5.0 cm 3 (± 1 cm 3 ) Rule #1 (add AUs)

13 Density = mass / volume Rule #2 (add %U’s) we need %U Notice that the uncertainty of the balance (mass) did not contribute significantly to the overall uncertainty of the calculated value; the graduated cylinder is therefore responsible for most of the random error. Something to state in CE. Not done yet… need AU Work that MUST be shown.

14 Density = mass / volume Finally, convert back to AU Density = 1.8 g/cm 3 (± 20.1%) → write as 1.8 g/cm 3 (± 20%) Density = 1.8 g/cm 3 (±.201x1.802 g/cm 3 ) Density = 1.8 g/cm 3 (± 0.362 g/cm 3 ) → write as 1.8 g/cm 3 (± 0.4 g/cm 3 ) always leave final answer in terms of absolute uncertainty

15 3) Adding or Subtracting by a pure number A “pure number” is a number without an estimated uncertainty The absolute uncertainty stays the same.

16 4) Multiplying or dividing by a pure number When doing this, multiply or divide the AU by the pure number.

17 4) Multiplying or dividing by a pure number Rule #3: divide AU by pure #


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