Many problems in chemistry, math, physics, and engineering involve by nak14542


									Mr. C's Dimensional Analysis                                                                 Page 1 of 5

Many problems in chemistry, math, physics, and engineering involve dimensional analysis. Dimensional
analysis involves calculations where the UNITS CANCEL ONE ANOTHER OUT, leaving only the
desired unit. It is best to approach each problem the same way, starting with a given, and proceeding in
such a way that the unit in the first step cancels with the unit in the second step.

You are already familiar with dimensional analysis although you might not have realized it. For example,
if your teacher asks you to bring in 2-dozen marbles, what would you do? Chances are that you would find
some marbles and count out 24 of them. So what did you do?

What you did was dimensional analysis in your head. You knew that a dozen marbles is really just 12
marbles, and since you need 2-dozen, you multiplied 12 by 2. Easy, right!?

Let’s look at a more systematic way to do this that can be applied to a number of science and engineering
problems. The approach is to always start with the GIVEN. In our case, it is 2-dozen marbles. Then we
will convert (switch from one thing to another). To do this, multiply by a fraction (really just both
multiplication and division at the same time). In every step after the given, take the units from the previous
step (in our case dozen) and PUT IT ON THE BOTTOM.

Given: 2 doz. Next step, multiply by a fraction and make sure that the doz is on the bottom.

2 doz marbles     x __________
                    doz marbles

Now we need to find a CONVERSION FACTOR . This is just something that converts dozen to
something else. The conversion factor we will use is 12 marbles / doz marbles.

2 doz marbles     x 12 marbles
                    doz marbles

Notice what happened. Since we have the same thing on top and bottom, the units canceled. Note that the
number 2 did not cancel, just the doz marbles. Now all we have to do is multiply 2 x 12 marbles to get 24

There is one more “trick” that we can use, and that is to remember that we cannot add apples to oranges,
Example: You are 4 miles from school heading due north at a constant velocity of 15km/hr due south.
How many meters from the school will you be (and which direction) after traveling for 45 minutes?

Solution: We know that we must add meters to meters because the question specified meters for the
 4mi  5280ft  12in  2.54cm     1m          –15km  1000m  1hr  45min
          1mi    1ft      1in    100cm        +   1hr      1km 60min                         =–4812.6m

  mi       ft      in       cm     6437.4m           km/hr      m/hr    m/min -11250m               4812m S

Lets look at some ridiculous examples. Lets say that there are 22 boleighs in every 3 siefs. There are also
2 runts in every 13 siefs and 6 runts in every yaunts. So our conversion factors are:
        22 boleighs        2 runts   6 runts Or, just the same 3 siefs              13 siefs     1 yaunt
        3 siefs.           13 siefs 1 yaunt                      22 boleighs        2 runts      6 runts
Say we want to know how many yaunts are in 3300 boleighs. Lets start with this as the given and use our
conversion factors making sure that the unit from the first step is in the bottom of the next.

3300 boleighs x 3 siefs          x 2 runts x       1 yaunt
                22 boleighs        13 siefs        6 runts

Now lets make sure that all of the units cancel and leave us with what we are looking for.
Mr. C's Dimensional Analysis                                                                Page 2 of 5

3300 boleighs x 3 siefs         x 2 runts x       1 yaunt    = 11.54 yaunts
                22 boleighs       13 siefs        6 runts

If there is more than one given: make sure the unit you are looking for ends up in the numerator. If not,
just invert (“Flip”) your answer.

Practice (show work and cancel units)                                   Name:______________
For all of the following, use the format shown showing that units cancel when one is on top and another on

               2.345km     1000m       1000mm        1cm           1in        1ft          1mi
Conversions Given           1km          1m         10mm         2.54cm      12in         5280ft
  Value and
 unit for this
     step      2.345km     2345m 2345000mm 234500cm 92322.8in 7693.6                     1.457mi
Look up the conversion factors needed and work each of the following. Round to the 3 rd decimal.

1.   How many oz are in 2.5 cups?

2.   How many oz are in 2.5 gallons?

3.   How many cm are in 3 feet?

4.   How many feet are in 2.5 miles?

5.   How many inches are in 2.5 miles?

6.   How many cm are in 2.5 miles?

7.   How many km are in 2.5 miles?

8.   How many seconds are in 0.5 hours?

9.   If you travel 2.5 miles in 0.5 hours, what is your speed in mi/hr? In km/hr? In m/s?

10. How many yaunts are in 4500 runts?

11. How many seifs?

12. How many boleighs?

13. How many boleighs are in 99 runts?

14. How many runts are in 99 boleighs?

15. How many yaunts are in 2110 siefs?
Mr. C's Dimensional Analysis                                                              Page 3 of 5
16. How many siefs are in 2110 yaunts?

17. How many siefs are in 90 runts?

18. How many cups are in 12 quarts?

19. How many pints?

20. How many gallons are in 112 cups?

21. pints?

22. quarts?

23. oz.?

The specific heat of liquid water is 4.184 J/g OC.

24. How many Joules (how much energy) are required to heat 4184g of water 20 OC?

25. If you can supply 3000 J to water at 40 OC, how many grams could you heat to 70OC?

26. The density of gold is 19.3 g/mL. If you have a ring that is 45g of pure gold, what’s the ring’s volume
    of gold?

27. If you have 0.055 L of pure gold, how many grams of gold do you have? How many kg?

28. We know from the periodic table that there are 197 grams of gold per mole of gold. If you have 0.80
    moles of gold, what is its mass?

29. If you have 45 grams of pure gold, how many moles of gold do you have?

Review Questions
For any given dimensional analysis problem:

1 What do we always start with? 2      What do we do in each subsequent step? 3 When do we know when

to stop? 4 What should happen to the units? How? 5 What units will the answer have?
Mr. C's Dimensional Analysis                                                               Page 4 of 5
AP and Honors Physics ONLY

The book cover provides many fundamental constants and derived units. The appendix may provide other
useful information. In this exercise, you will use these quantities to develop relationships (formulas).

Example: The units for acceleration are m/s2. Put in terms of a) length and time and b) velocity and time.
                                                                                    2                         2
Solution: length is in meters, so it must go on top. Time is in seconds, so we need t on the bottom since s
                         2                                                2
is on the bottom. a = /t . In terms of velocity: v = /t, but we need /t so we just need another t on the
bottom, so we end up with v/t. Use substitution to show the dimensional analysis:
Start with v/t. Now substitute v = /t giving v/t = (/t)/t which equals /t

Fill in all boxes

Derived Unit         Expressed as Fundamental Unit              Complete the equation          In Terms of
                                     2 2
   Farad                          C s                                                     potential difference,
                                                                       C = Q/V
(Capacitance)                    kg  m 2                                                  V, and charge,Q.

                                 kg  m2                                                      current, I and
      Volt                                                                V=
                                  Cs 2                                                        resistance, R
                                                                                              current, I and
      Watt                                                                P=
                                                                                              resistance, R
                                                                                              work, w and
      Watt                                                                P=
                                                                                                 time, t
                                                                                              current, I and
      Watt                                                                P=
                                                                                               potential, V
                                 kg  m                                                        force, F and
                                   s2                                                            length, 
                                                                                               mass, m and
      Joule                                                               E=
                                                                                                velocity, v
                                                                                                 mass, m,
      Joule                                                               E=                  acceleration, a
                                                                                                 length, 
                                                                                             Plank’s const, h
   Plank’s                       kg  m                                                         velocity, v
  Constant, h                       s                                                            length, 
Mr. C's Dimensional Analysis                                                              Page 5 of 5

1. 750 runts
2. 29250 seifs
3. 214,500 boleighs
4. 4719 boleighs
5. 2.08 runts
6. 54.1 yaunts
7. 82290 yaunts
8. 585 seifs
9. 48 cups
10. 24 pints
11. 7 gal
12. 56 pints
13. 28 qt
14. 896 oz
15. 4184g x 20OC x 4.184J/g OC = 350,117J
16. You need to first realize that it is heated 30OC.
    To end up with g on top, you need to invert the heat capacity
    1gOC/4.184J it can now be seen that to cancel the J and the OC that you will need a J on top and a OC
    on the bottom: 1gOC/4.184J x 3000J/30 OC = 23.9g
17. 2.33 g
18. 1062g = 1.062 kg
19. 157.6 g
20. .228 mol

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