2. 2
3 Parts of a Measurement
70 kilograms =
154
pounds
numerical value
unit
Significant figure
1 SF
3 SF
Unit is also called dimension
Degree of precision
3. USE OF UNITS: COMMUNICATION TECHNIQUE
2 X 3 = 6
This calculation communicates
less information
(2 candy mints bought) X 3 cents paid
----------------------------- = 6 cents paid
1 candy mint bought
(2 yd deep) X 3 feet
----------= 6 feet deep
1 yard
Note the use of
adjectives and
verbs to
communicate more
clearly
4. parameter Metric English
Length meter Inches, foot, mile, yard
mass gram Pounds, ton, slug
volume M3 Pint, gallon, fluid ounce
temperature oC oF
6. 6
• The metric or International System (SI, Systeme
International) is a decimal system of units.
• It is built around standard units.
• It uses prefixes representing powers of 10 to
express quantities that are larger( multiples)
or smaller (submultiples) than the standard
units.
7. 7
International System’s
Standard Units of Measurement
Quantity Name of Unit Abbreviation
Length meter m
Mass kilogram kg
Temperature Kelvin K
Time second s
Amount of substance mole m
Electric Current ampere A
Luminous Intensity candela cd
8. 8
Prefixes and Numerical Values for SI Units
Power of 10
Prefix Symbol Numerical Value Equivalent
exa E 1,000,000,000,000,000,000 1018
peta P 1,000,000,000,000,000 1015
tera T 1,000,000,000,000 1012
giga G 1,000,000,000 109
mega M 1,000,000 106
kilo k 1,000 103
hecto h 100 102
deca da 10 101
— — 1 100
9. 9
Prefixes and Numerical Values for SI Units
deci d 0.1 10-1
centi c 0.01 10-2
milli m 0.001 10-3
micro 0.000001 10-6
nano n 0.000000001 10-9
pico p 0.000000000001 10-12
femto f 0.00000000000001 10-15
atto a 0.000000000000000001 10-18
Power of 10
Prefix Symbol Numerical Value Equivalent
11. 11
Dimensional Analysis
Dimensional analysis converts one
unit to another by using conversion
factors.
unit1 x conversion factor = unit2
Technique of treating units as numbers and manipulated
mathematically to get rid of unwanted units and introduce units
that are wanted
15. CONVERSION OF UNITS
• Dimensional Analysis
1. Express the conversion problem as a
mathematical equation
2. Multiply the right side of the equation with
one or more conversion factors of
appropriate forms until its unit is the same as
the unit that is being sought.
3. Perform the needed arithmetic operations
16. 16
How many millimeters are there in 2.5
meters?
• It must cancel
meters, (previous
unit.)
• It must introduce
millimeters, ( unit
being sought)
unit1 x conversion factor = unit2
m x conversion factor = mm
The conversion factor must
accomplish two things:
17. 17
Convert 16.0 inches to centimeters.
16.0 in
2.54 cm
x
1 in
= 40.6 cm
2.54 cm
1 in
Use this
conversion
factor
18. 18
Convert 3.7 x 103 cm to micrometers.
3
3.7 x 10 cm
1 m
x
100 cm
6
10 μm
x
1 m
7
= 3.7 x 10 μm
Centimeters can be converted to micrometers
by writing down conversion factors in
succession.
cm m meters
19. 19
Convert 3.7 x 102 cm to micrometers.
Centimeters can be converted to micrometers
by two stepwise conversions.
cm m meters
3
3.7 x 10 cm
1 m
x
100 cm
1
= 3.7 x 10 m
6
10 μm
x
1 m
7
= 3.7 x 10 μm
1
3.7 x 10 m
20. A frequent comment is , “It looks easy when my
professor does it, but I don’t know where to start
when I try it on my own. “
FOUR STEP RECIPE
1. Write down the given number with its unit
2. Write a ratio with the given unit in the denominator (at
the bottom) and the unit sought in the numerator (0n
top)
3. Insert numbers into the ratio so that the numerator
and the denominator are equal
4. Multiply Steps 1 and 3 together
COOKBOOK DIMENSIONAL ANALYSIS
21. Formula approach : Pros and Cons
Speed and convenience are the main advantages of using memorized formulas
Example : If you need 10 gallons of gasoline that sells for $2.00 per gallon, how much is
the cost ?
Formula : (number of gallons purchased ) X (cost in dollars per gallon) = cost of gasoline in dollars
10 gallons X $ 2.00 per gallon = $10
One major disadvantage with formulas, Careless use of units, formulas don’t always work
Example : You need exactly 3 qt gasoline in order to start your
vacation with a full tank and the gasoline sells by the pint for 25
cents per pint, How much money will it cost you to top off your gas
tank ?
22. REWARDS THAT COME TO MASTERY OF DIMENSIONAL ANALYSIS
1. Powerful tool that will be very valuable to work with most problems .
2. The method is a valuable communication device when your
calcul ation show both numbers and units
3. You will find yourself solving problems in areas in which you
are relatively ignorant . Your one method of attack (converting
given units into desired units ) can sometimes be enough to
solve difficult and unfamiliar problems.
4. Taken together, your ability to solve unfamiliar problems and
your ability to solve problems without memorized formulas
increase your effective intelligence.
5. You can derive “ very quicly” the formulas that Newton etc. took decades to
derive. (adding hindsight to your mastery of the dimensional analysis. )
24. PRECISION AND
SIGNIFICANT FIGURES
• Rules in counting number of significant
figures
1. All non zero digits are significant
2. Zeros may or may not be significant
depending upon the kind of the
zero.
a. Leading zeros are those before
non zero digits , are never
significant.
b. Confined zeros are those in
between non zero digits, are
significant.
c. Trailing zeros are those after
non zero digits , are significant
if there is an explicit decimal
point. Not significant if there is
no explicit decimal point.
Significant Figure=
number of digits in a
measurement
determined with
certainty plus one
digit that is uncertain.
25. 25
Significant Figures
• The number of digits that are known plus
one estimated digit are considered
significant in a measured quantity
estimated5.16143
known
36. 36
The results of a calculation cannot be more
precise than the least precise measurement.
37. ARITHMETIC OPERATIONS AND SIGNIFICANT FIGURES.
When arithmetic operations are to be done on the measurement, they should
not change the degree of precision of the measurement. So that there are
rules to be applied.
• Division and Multiplication
Number of S.F (final answer) =
number of SF of the
measurement with the least
number of SF
• Addition and Substraction
• Number of decimal places right of
decimal point (final answer) =
number of decimal places right of
decimal point of the
measurement with the least
number of decimal places to the
right of the decimal point.
39. 39
In multiplication or division, the answer must
contain the same number of significant
figures as in the measurement that has the
least number of significant figures.
40. 40
(190.6)(2.3) = 438.38
438.3
8
Answer given by
calculator.
2.3 has two significant figures.
190.6 has four significant
figures.
The answer should have two significant figures
because 2.3 is the number with the fewest significant
figures.
Drop these three
digits.
Round off this digit to
four.
The correct answer is 440 or 4.4 x 102
42. 42
The results of an addition or a
subtraction must be expressed to the
same precision as the least precise
measurement.
43. 43
The result must be rounded to the same
number of decimal places as the value
with the fewest decimal places.
44. 44
Add 125.17, 129 and 52.2
125.17
129.
52.2
306.37
Answer given by
calculator.
Least precise number.
Round off to the nearest
unit.
306.37
Correct answer.
45. 45
1.039 - 1.020
Calculate
1.039
1.039 - 1.020
= 0.018286814
1.039
Answer given by
calculator.
1.039 - 1.020 = 0.019
0.019
= 0.018286814
1.039
The answer should have two significant figures
because 0.019 is the number with the fewest
significant figures.
0.018286814
Drop these 6 digits.
Two significant
figures.
46. 46
Rounding Off Numbers
• Often when calculations are performed
extra digits are present in the results.
• It is necessary to drop these extra digits so
as to express the answer to the correct
number of significant figures.
• When digits are dropped the value of the
last digit retained is determined by a
process known as rounding off numbers.
47. 47
80.873
Rule 1. When the first digit after those you
want to retain is 4 or less, that digit and all
others to its right are dropped. The last digit
retained is not changed.
4 or less
Rounding Off Numbers
48. 48
5 or greater
5.459672
Rule 2. When the first digit after those you
want to retain is 5 or greater, that digit and all
others to its right are dropped. The last digit
retained is increased by 1.
drop these
figuresincrease by 1
6
Rounding Off Numbers
49. 49
• Very large and very small numbers are often
encountered in science.
602200000000000000000000
0.00000000000000000000625
• Very large and very small numbers like
these are awkward and difficult to work
with.
50. 50
602200000000000000000000
A method for representing these numbers in a
simpler form is scientific notation.
0.00000000000000000000625
6.022 x 1023
6.25 x 10-21
51. 51
Scientific Notation
• Write a number as a power of 10
• Move the decimal point in the original
number so that it is located after the first
nonzero digit.
• Follow the new number by a multiplication
sign and 10 with an exponent (power).
• The exponent is equal to the number of
places that the decimal point was shifted.
52. 52
Write 6419 in scientific notation.
64196419.641.9x10164.19x1026.419 x 103
decimal after
first nonzero
digit
power of 10
53. 53
Write 0.000654 in scientific notation.
0.0006540.00654 x 10-10.0654 x 10-20.654 x 10-36.54 x 10-4
decimal after
first nonzero
digit
power of 10
54. 54
Heat
• A form of energy associated with
small particles of matter.
Temperature• A measure of the intensity of heat,
or of how hot or cold a system is.
55. 55
• The SI unit for heat
energy is the joule
(pronounced
“jool”).
• Another unit is
the calorie.
4.184 J = 1 cal
(exactly)
4.184 Joules = 1 calorie
This amount of heat energy will raise the
temperature of 1 gram of water 1oC.
56. 56
A form of energy
associated with
small particles
of matter.
A measure of
the intensity of
heat, or of how
hot or cold a
system is.
An Example of the Difference
Between Heat and Temperature
57. 57
Twice as much heat
energy is required to
raise the
temperature of 200 g
of water 10oC as
compared to 100 g of
water.
200 g water
20oC
A
100 g water
20oC
B
100 g water
30oC
200 g water
30oC
heat beakers 4184 J 8368 J
temperature
rises 10oC
58. 58
Temperature Measurement
• The SI unit of temperature is the Kelvin.
• There are three temperature scales: Kelvin,
Celsius and Fahrenheit.
• In the laboratory temperature is commonly
measured with a thermometer.
60. 60
o o o
F - 32 = 1.8 x C
To convert between the scales use the following
relationships.
o o o
F = 1.8 x C + 32
o
K = C + 273.15
o
o F - 32
C =
1.8
62. 62
It is not uncommon for temperatures in the Canadian
planes to reach –60oF and below during the winter.
What is this temperature in oC and K?
o
o F - 32
C =
1.8
o o60. - 32
C = = -51 C
1.8
63. 63
It is not uncommon for temperatures in the Canadian
planes to reach –60oF and below during the winter.
What is this temperature in oC and K?
o
K = C + 273.15
o
K = -51 C + 273.15 = 222 K
70. 70
A 13.5 mL sample of an unknown liquid has a mass of
12.4 g. What is the density of the liquid?
M
D
V
0.919 g/mL
12.4g
13.5mL
71. 71
46.0 mL
98.1 g
A graduated cylinder is filled to the 35.0 mL mark with water. A
copper nugget weighing 98.1 grams is immersed into the cylinder and
the water level rises to the 46.0 mL. What is the volume of the
copper nugget? What is the density of copper?
35.0 mL
copper nugget final initialV = V -V = 46.0mL - 35.0mL = 11.0mL
g/mL8.92
mL11.0
g98.1
V
M
D
72. 72
The density of ether is 0.714 g/mL. What is the
mass of 25.0 milliliters of ether?
Method 1
(a) Solve the density equation for mass.
mass
d =
volume
(b) Substitute the data and calculate.
mass = density x volume
0.714 g
25.0 mL x = 17.9 g
mL
73. 73
The density of ether is 0.714 g/mL. What is the
mass of 25.0 milliliters of ether?
Method 2 Dimensional Analysis. Use density as a
conversion factor. Convert:
0.714 g
25.0 ml x = 17.9 g
mL
mL → g
g
mL x = g
mL
The conversion of units is
74. 74
The density of oxygen at 0oC is 1.429 g/L. What is the
volume of 32.00 grams of oxygen at this temperature?
Method 1
(a) Solve the density equation for volume.
mass
d =
volume
(b) Substitute the data and calculate.
mass
volume =
density
2
2
32.00 g O
volume = = 22.40 L
1.429 g O /L
75. 75
The density of oxygen at 0oC is 1.429 g/L. What is the
volume of 32.00 grams of oxygen at this temperature?
Method 2 Dimensional Analysis. Use density as a
conversion factor. Convert:
2 2
2
1 L
32.00 g O x = 22.40 L O
1.429 g O
g → L
L
g x = L
g
The conversion of units is
