1 MEASUREMENTS AND CALCULATIONS
1.1
Scientific Notation
Scientific work most often deals with very large numbers or very small numbers, obtained either by
measurement or through calculations. For example, it has been experimentally determined that in a
sample of 12.011 g of carbon atoms, there are
602200000000000000000000
that is, the number 6022 followed by 20 zeroes. This number, known as Avogadro’s Number, is very often
used in chemistry. Clearly it is unreasonable to write out 20 zeroes each time Avogadro’s Number is
used. Scientific notation has been devised to write out numbers such as Avogadro’s Number in a
compact form which conveys the same information more conveniently.
To see how to do this, note the following:
10 = 101
100 = 10 x 10 = 102
1000 = 10 x 10 x 10 = 103
…..
1000000 = 10 x 10 x 10 x 10 x 10 x 10 = 106
(one million)
and so on. Thus, for example, the number one million is a “1” followed by 6 zeroes. Scientific notation
makes use of the relationship between the number of zeroes and an associated power of 10. The
following example illustrates how numbers can be represented in scientific notation.
The number 1234 can be written as the following product:
1.234 x 1000 = 1.234 x 10 x 10 x 10 = 1.234 x 103
The number, 1.234 x 103 represents the number 1234 in standard scientific notation.
The simplest way to go about expressing a number in standard scientific notation may be observed in the
following example.
e.g.
The speed of light has been measured to be 299792458 metres per second (the unit metres per
–1
second can be abbreviated to m·s ). Write the speed of light in standard scientific notation.
299792458 m·s–1= 2.99792458 x 100000000 m·s–1 = 2.99792458 x 108 m·s–1
speed of light in
standard scientific
notation
A faster way to do this is simply by counting potential decimal places. We want to express
299792458 m·s–1 as:
2.99792458 times some power of 10 m·s–1
Identify the location of the decimal point in the
number, then move the decimal immediately to the
right of the first digit.
page 1
299792458 = 299792458.
Implicit location of the decimal
2 9 9 7 9 2 4 5 8.
The decimal will be moved to the left by
+8
8 places, therefore the factor is 10 .
Thus
299792458 = 2.99792458 x 100000000 = 2.99792458 x 10+8
8 zeroes
Applying the same principle to Avogadro’s Number:
602 200 000 000 000 000 000 000 = 6.022 10
+23
23 places to the left
For small numbers (i.e., numbers less than 1), note that
1
= 10 1
10
1 1
0.01 =
x
= 10 2
10 10
1 1 1
0.001 =
x x
= 10 3
10 10 10
.....
0.1 =
0.000001 = 10 6
For small numbers, movement of the decimal place is to the right.
e.g.
Write the number 0.1234 in standard scientific notation.
0.1234 = 1.234 x
e.g.
1
= 1.234 10
10
1
Write the number 0.000314159 in standard scientific notation.
0.000314159 = 3.14159 times some power of 10
We need to move the decimal place 4 positions to the right (this will result in a negative exponent)
Therefore
–4
0.000314159 = 3.14159 10
page 2
1.1.1
Operations Using Powers of 10
1.1.1.1 Summary of the Rules for Exponentiation
x0 = 1
1
=x a
xa
x a x b = x a +b
xa
1
= xa b = xa x b = xa b
b
x
x
[ xy ]a = x a y a
x
y
(x )
a
a
=
b
xa
ya
= x ab
1.1.1.2 Multiplying powers of ten: Add the exponents.
2
3
=
10
2+3
=
10
3
–1
=
10
3–1
=
10
4
5
=
2
3
2
–3
(10 × 10) × (10 × 10 × 10)
=
10 × 10
(10 × 10 × 10) × (1/10)
=
10 × 10
=
10 × 10
(10 × 10) × (1/10 × 1/10 × 1/10)
(1/10 × 1/10 × 1/10) × (1/10)
=
=
10 × 10 × 10
=
10 × 10 × 10
=
1
=
2
=
=
1
5
=
10
2+3+1
=
=
=
=
=
1.1.1.3 Dividing powers of ten: Subtract the exponents.
2
3
=
10
2–3
=
10
–1
3
–1
=
10
3 – (–1)
=
10
4
4
5
=
(10 × 10) ÷ (10 × 10 × 10)
=
10 ÷ 10
(10 × 10 × 10) ÷ (1/10)
=
10 / 10
=
10 / 10
(10 × 10) / (1/10 × 1/10× 1/10)
=
=
=
(1/10 × 1/10 × 1/10) / (1/10)
=
=
=
2
1
=
–3
1
=
=
=
=
(10 / 10 )/ 10
2
=
3
10 × 10 / 10
10
(2 – 3) – 1
=
page 3
1.1.1.4 Raising a power of ten to another power: multiply the exponents.
2 3
=
10
2x3
=
10
6
–1 4
=
10
(–1) x 4
=
10
–4
3 –2
=
10
3 x (–2)
=
10
–6
=
10
(3 x 2) x (–1)
=
10
–6
(10 × 10) × (10 × 10) × (10 × 10)
=
(10 )
(1/10) × (1/10) × (1/10) × (1/10)
=
(10 )
=
(10 )
1/(10 × 10 )
=
[(10 ) ]
(10 × 10 × 10 × 10) × (10 × 10 × 10 × 10)
=
3
3
1/(10 ) × 1/(10 )
3
3
=
=
3 2 –1
=
=
–2 3
=
=
–2 –2
=
=
(10 )
(10 )
1.1.1.5 Fractional powers of ten: can be expressed as a root (
10
10
10
1/2
1/3
1/4
× 10
× 10
× 10
1/2
1/3
1/4
× 10
× 10
1/3
1/4
× 10
1/4
= 10
1
= 10
1
= 10
1
3
10
3/2
× 10
3/2
= 10
10
2/3
× 10
2/3
=
10
3/4
× 10
3/4
× 10
3/4
).
Therefore
10 =
1/2
10
Therefore
10 = 3
10
Therefore
1/4
10 =
Therefore
10 =
Therefore
10 =
1/3
10 3
3/2
2/3
=
1.1.1.6 Adding and subtracting powers of ten
Unless the two numbers are of the same order already, one has to:
1. factor out the smallest power of ten; then
2. add (or subtract) the two numbers in brackets.
2
2
= 1 × 10 + 1 × 10
2
4
= 1 × 10 + 10 × 10
4
3
= 10 × 10 – 1 × 10
3
6
10 + 10
10 + 10
10 – 10
10 + 10
10
–3
+ 10
2
10 + 10
10
–6
2
2
2
= (1 + 1) × 10
2
3
2
3
2
= 2 × 10
2
= (1 + 100) × 10
= (10 – 1) × 10
3
2
= 101 × 10
= 9 × 10
3
= 2 × 10
2
2
= 2 × 10
2
= (1.01 × 10 ) × 10
= 9 × 10
3
2
= 9 × 10
–6
–3
–2
1.3 × 102 + 2.793 × 104
4.55 × 10
–2
– 2.7935 × 10
2
= 1.01 × 10
–2
– 10
10 – 10
2
= 1.3 × 102 + 279.3 × 102
0
page 4
3
4
Exercise 1.1
n
1. Explain how you would write the numbers 10 and 10
–n
as normal numbers.
2. Write the following numbers in standard scientific notation.
0.0000000721
6319 x 10–2
0.005219
0.000000007251
5219
72.41 x 10–2 = (7.241 x 10+1) x 10–2 = 7.241 x 10+1–2 =
6.319 x 10–2
1.4325
4207
347.215
77000000
0.0043050
3. Write the following numbers as ordinary decimal numbers.
8.621 x 104
6.235 x 10–2
7.229 x 103
5.001 x 10–6
4. Show that
4
103 =
4
10
3
page 5
1.2 Units
Units (or dimensions) give the scale that is being used to express the result of a measurement. In
science, pure numbers rarely make an appearance. Numerical data in science always include the
numerical value together with a unit. Thus, for example, it is incorrect to say the length of an object is
15.02 – this is meaningless by itself because the 15.02 can refer to the length in cm, km, miles or some
other expression of length.
There are two systems of measurements used in science and engineering. These are the British
System and the metric system. The British System of units, the older of the two, is obsolete and in the
developed world, used only in the USA. The metric system was originally developed in post-revolutionary
France at the end of the 18th century; it has been updated by international agreement starting in 1960 into
the Système International, SI system of units. A comparison of the fundamental set of units in the SI
and British systems are presented in Table 1 below.
Table 1:
Fundamental Set of Units in the SI and British Systems
SI System
Physical Quantity
Name of Unit
British System
Abbreviation
Name of Unit
Abbreviation
Length
metre
m
foot
ft
Mass
kilogram
kg
Pound mass
Time
second
s
Second
s
Temperature
Kelvin
K
Rankine
°R
Current
Ampere
A
Ampere
A
lbm
All SI units may be expressed as combinations of the basic units of length, mass, time, temperature and
electrical current.
Often the standard SI unit does not provide a convenient scale for representing a measurement. In such
cases, a prefix is attached to the basic SI unit to indicate the order of magnitude (i.e., power of 10) of the
particular measurement. The standard prefixes are provided below.
1
Table 2:
Prefixes for the Powers of Ten
Power Prefix Abbreviation Power Prefix Abbreviation
24
10
21
10
18
10
15
10
12
10
10
9
10
6
10
3
10
1
Yotta
Zetta
exa
peta
tera
giga
mega
kilo
deka
Y
Z
E
P
T
G
M
k
da
10
–1
deci
d
10
–2
centi
c
10
–3
milli
m
10
–6
micro
µ
10
–9
nano
n
10
–12
pico
p
10
–15
femto
f
10
–18
atto
a
10
–21
zepto
z
10
–24
yocto
y
1
This list is provided for completeness. It is necessary to memorize only those prefixes (and their
12
–15
abbreviations) that are highlighted (i.e., those between tera, 10 , and femto, 10 ).
page 6
These prefixes are used as a fast way of expressing the power of 10 to which a number is expressed.
Observe the use of prefixes, abbreviations, symbols and powers of ten:
1
–2
m = 10 m
100
One centimetre is one hundredth of a metre
1 cm =
One megawatt is a million watts
1 MW = 106 W,
One kilohertz is a thousand hertz
1 kHz = 10 Hz.
3
For each unit below, write the abbreviation, the power of ten and the basic unit if it is not already written:
1. one milligram
2. one microsecond
=
g
=
s
3. one kilocal
=
=
cal
4. one nanocoulomb
=
5. one gigawatt
=
=
=
=
C
W
CONVERTING FROM ONE UNIT SCALE TO ANOTHER
Units are treated exactly the same as if they are number. This means that you can multiply units, divide
them, cancel them etc.
Example 1:
How many kilograms are there in one gram?
Step 1: Write the given datum on the left hand side of a proposed inequality. On the right hand
side, write the required unit.
1g
? kg
Step 2: A conversion factor (CF) is needed to relate the two sets of units. First examine the
units of the CF – first examine the given unit. This unit must be cancelled and replaced
with the required unit.
1g
kg
g
? kg
The units of the appropriate CF, relating kg to g, appear within the parentheses.
Step 3: The numerical portion of the CF must now be determined. In this example, the
definition of “kilo”, k, serves as the numerical portion of the CF.
1g
Example 2:
kg
103 g
= 10 3 kg
How many nanometres are there in one micrometre?
Step 1: Write the given datum on the left hand side of a proposed inequality. On the right hand
side, write the required unit.
1µ m
? nm
page 7
Step 2: A conversion factor (CF) is needed to relate the two sets of units. First examine the
units of the CF – first examine the given unit. This unit must be cancelled and replaced
with the required unit.
1µ m
nm
µm
? nm
The units of the appropriate CF, relating µm to nm, appear within the parentheses.
Step 3: The numerical portion of the CF must now be determined. In this case, the definitions
of “micro” and “nano” serve as the numerical portion of the CF. The solution is actually
obtained by breaking the CF into two CF’s. The first related µm to m and the second
relates m to nm.
1µ m
nm
m
= 1µ m
µm
µm
nm
10 6 m
= 1µ m
µm
m
nm
10 6
=
nm = 10
9
10 m
10 9
6
10+9 nm = 10 6+9 nm = 10 +3 nm
Exercises 1.2
1. How many mL are there in one kiloliter?
2. How many centimetres in 1 µm?
3. How many nanoseconds in 1 decisecond?
4. How many GeV in 1TeV?
5. How many Mcal in 3.8 picocalories?
6. Convert 4.5 kg to grams.
7. Convert 6379 milliwatts to kW.
8. Convert 24 000 volts to kilovolts.
page 8
1.3 Significant Figures
In science there are two methods of acquiring numbers. One method is by counting; the other is by
measurement. Counting is by its very nature exact. Measurement on the other hand is done by
comparison with a calibrated instrument. Since there are limitations on the scales of all calibrated
instruments, there is a limit to the accuracy to which any measurement may be made. The use of
significant figures expresses the extent of this accuracy. The use of significant figures is really only an
approximate method for handling uncertainty in measurement and its propagation. For a more complete
and formal treatment of error, see the section on Percentage Error and Percentage Difference in the
following section on Error Analysis.
Generally the number of significant figures to which a measurement should be reported includes all of
those digits which are certain and only one digit, the last, which is uncertain. Thus, for example, if one
were to measure the width of a page of paper using a 30-cm ruler graduated to the nearest 0.1 cm, one
might find that it measures somewhere between 21.6 cm and 21.7 cm. Furthermore, one may estimate
that this width lies approximately halfway between the two and therefore record the width as 21.65 cm. In
this instance, the last 0.05 cm is uncertain.
The following set of rules is helpful in determining which of the digits in a number are significant.
Justification for Significant Figures
1.65 ± 0.02 cm
A=LxW
1.65 ± 0.02 cm
A = (1.67 cm) x (1.67 cm) = 2.7889 cm2
A = (1.65 cm) x (1.64 cm) = 2.7060 cm2
A = (1.67 cm) x (1.66 cm) = 2.7722 cm2
A = (1.65 cm) x (1.63 cm) = 2.6895 cm2
2
A = (1.67 cm) x (1.65 cm) = 2.7555 cm
A = (1.64 cm) x (1.67 cm) = 2.7388 cm2
A = (1.67 cm) x (1.64 cm) = 2.7388 cm2
A = (1.64 cm) x (1.66 cm) = 2.7224 cm2
A = (1.67 cm) x (1.63 cm) = 2.7221 cm2
A = (1.64 cm) x (1.65 cm) = 2.7060 cm2
2
A = (1.66 cm) x (1.67 cm) = 2.7722 cm
A = (1.64 cm) x (1.64 cm) = 2.6896 cm2
A = (1.66 cm) x (1.66 cm) = 2.7556 cm2
A = (1.64 cm) x (1.63 cm) = 2.6732cm2
A = (1.66 cm) x (1.65 cm) = 2.739 cm2
A = (1.63 cm) x (1.67 cm) = 2.7221cm2
A = (1.66 cm) x (1.64 cm) = 2.7224 cm2
2
A = (1.63 cm) x (1.66 cm) = 2.7058 cm2
A = (1.66 cm) x (1.63 cm) = 2.7058 cm
A = (1.63cm) x (1.65 cm) = 2.6895 cm2
A = (1.65 cm) x (1.67 cm) = 2.7555 cm2
A = (1.63 cm) x (1.64 cm) = 2.6732 cm2
A = (1.65 cm) x (1.66 cm) = 2.739 cm2
A = (1.63 cm) x (1.63 cm) = 2.6569 cm2
A = (1.65 cm) x (1.65 cm) = 2.7225 cm2
Range: 2.6569 cm2 – 2.7889
Average Area = 2.7225 ± 0.0664 cm2
page 9
1.3.1
DETERMINING THE NUMBER OF SIGNIFICANT FIGURES
1.3.1.1
Non-zero Digits
All non-zero digits in a number are significant.
1.3.1.2
Zeroes
There are three classes of zeroes to be considered
a)
Leading Zeroes – these are never significant
e.g.
b)
– 4 significant figures, one for each non-zero digit.
0.001234
– 4 significant figures, one for each non-zero digit.
Trapped Zeroes – these are always significant
e.g.
c)
0001234
– 4 significant figures; the zeroes are trapped and are therefore significant
1001
Trailing Zeroes – these are may or may not be significant
If a decimal point exists somewhere in the number, the trailing zeroes are significant.
e.g.
10.0
– 3 significant figures; a decimal points exists.
0.010200
– 5 significant figures; the leading zeroes are not significant, the trapped
zero is and since there is a decimal point in the number, the trailing
zeroes are significant.
If there is no decimal point anywhere in the number, the situation is ambiguous.
e.g.
100
1 10
– Ambiguous. The only way to properly express the correct number of
significant figures in this situation is to express the number in scientific
notation
2
1.0 10
– 1 significant figure
2
– 2 significant figures
2
1.3.2
1.00 10
– 3 significant figures
100.
– 3 significant figures
SIGNIFICANT FIGURES IN CALCULATIONS
When numbers are multiplied (or divided), the error in the accuracy of these numbers is also multiplied.
The product of a multiplication typically has more numbers in it than are really significant. The following
rules are used to determine the number of significant digits following various mathematical operations.
When “truncating” large numbers to their proper number of significant figures, the rules for rounding off
(next section) should be observed.
1.3.2.1
Multiplication and Division:
Under multiplication and division, the final answer must have the fewest number of significant figures
among the numbers involved in the calculation.
e.g.
24.015
5 sig. figs.
x
6.50
3 sig. figs.
=
156
3 sig. figs.
page 10
1.3.2.2
Addition and Subtraction
Under addition and subtraction, the final answer must have the fewest number of decimal places among
the numbers involved in the calculation.
2.4015 10-6
e.g.
+
6.50 10-4
=
+ 6.50 ) 10-4
( 0.024015
6 decimal places
=
6.524015 10
=
6.52 10-4
2 decimal places
-4
Final answer has only 2 decimal places
Note: Conversion factors are regarded as infinitely precise, therefore should never be considered when
the number of significant figures in the final answer is determined.
1.3.2.3
ROUNDING OFF
In calculations, the number obtained usually has many more digits than would correctly express the
precision to which the original measurements were made. Thus a calculated number must be reduced to
the proper number of significant figures; this process is known as rounding off. The following set of rules
represent one convention for rounding off numbers to the proper number of significant figures. In the
examples that follow, the last appropriate significant digit is underlined.
First determine the correct number of significant figures to which the calculated number is to be rounded
off. Inspect the digit immediately to the right of the last significant digit.
If this digit is less than 5
Ignore the digit.
e.g.
0.0257539
0.02575
If this digit is greater than, or equal to, 5
Increase the last significant digit by one.
e.g.
0.02999793
0.03000
page 11
Exercises 1.3
1.
How many significant digits are present in the following numbers?
102
000 000 034 89
2500
0.000 3004
0.00 520700
2.
Round off the following numbers to 3 significant figures.
435
76145
22752
9997
2500
46459
3.
Perform the following calculations
149.2+ 0.034 + 2000.34
1.0322 x 103 + 4.34 x 103
4.03 x 10
–2
–3
– 2.44 x 10
5
2.094 x 10 – 1.073 x 10
6
8
(0.0432)(2.909)(4.43 x 10 )
4
2
1
(2.9932 x 10 )(2.4443 x 10 + 1.0032 x 10 )
2
–1
(2.34 x 10 + 2.443 x 10 ) / 0.0323
–3 2
(4.38 x 10 )
–6 ½
(5.9938 x 10 )
page 12
1.4 Error Analysis
All measurements involve an element of uncertainty. In experiments, therefore, it will be important to
determine quantitatively how these uncertainties affect the values that are computed from the data.
There are two aspects of error analysis. The first involves the comparison of a measured value with an
accepted or literature value. The second involves an analysis of the uncertainty in the actual
measurement. Consequently there will be an uncertainty in any final result based on this measurement.
This uncertainty exists whether or not an accepted value is known.
Comparison with a Literature Value: If a generally accepted value of a quantity, Y, is known to exist
(referred to as the literature value of Y) and one or more experiments are performed to measure Y, then it
is usual to compare the experimentally measured value of Y with the literature value according to:
% error =
F Experimental Value - Literature Value I 100%
H
K
Literature Value
This %error represents the accuracy to which the experimental value has been measured. Accuracy is
defined to be how close the experimental value is to the true value.
Comparison of Experimental Values: If a literature value is unavailable but the results of two different
measurements are, then it is useful to compute the percentage between the results. Three methods for
computing the percentage difference follows:
1.
The % difference between Y1 and Y2 relative to Y1
%difference =
2.
Y1
Y1
100%
The % difference between Y1 and Y2 relative to Y2
%difference =
3.
Y2
Y1 Y2
100%
Y2
The % difference between Y1 and Y2 relative to their arithmetic mean
%difference =
F
H
Y1 Y2
Y1 + Y2
2
I
K
100%
The %difference represents the precision to which the experimental values have been measured.
Precision can mean two things in science:
a) It is a measure of the number of significant digits to which a experimental value is known;
b) It is a measure of how close several experimental values on the same object are to each other
using the same measurement techniques.
page 13
1.5 Dimensional Analysis
Dimensional analysis is a problem solving technique based on an analysis of units. It is used to solve two
types of problems:
1. Converting from one system of units to another; and
2. Relating one physical quantity to another using a definition.
Dimensional analysis has already been used to convert between different powers, e.g., from µm to nm.
The technique is always the same: find a conversion factor that relates one unit to another or one
physical quantity to another. Finding the appropriate conversion factor may not be a simple process – at
times, several related conversion factors must be linked to obtain the desired conversion factor.
Example 1: How many ounces are in 24.3 pounds?
Note that this is a conversion of one unit of mass (ounces) to another (pounds).
Step 1: Write the given datum on the left hand side of a proposed inequality. On the right hand
side, write the required unit.
24.3 lb
? oz
Step 2: A conversion factor (CF) is needed to relate the two sets of units. First examine the
units of the CF – first examine the given unit. This unit must be cancelled and replaced
with the required unit.
24.3 lb
oz
lb
? oz
The units of the appropriate CF, relating lb to oz, appear within the parentheses.
Step 3: The numerical portion of the CF must now be determined. In this case, you must know,
or look up in tables, the relationship between pounds and ounces (N.B., you know that
you have to look up this relation because you are converting from one unit scale to
another). The appropriate conversion is 1 lb = 16 oz.
27.3 lb
16 oz
= 388.8 oz = 389 oz
1 lb
The last equality is because we must observe the proper use of significant figures.
Example 2: How many kilograms are in 71.24 pounds?
Note that this is a conversion of one unit of mass (pounds) to another (kilograms). Thus we will need a
conversion factor that relates mass in the British system of units (lb) to mass in the SI system (kilograms).
Tables of conversion factors show that 1lb =453.593 g. The unit of grams can easily be related to
kilograms using the definition of “kilo”.
71.24 lb
? kg
kg
71.24 lb
lb
? kg
71.24 lb
g
lb
kg
g
71.24 lb
453.593 g
lb
? kg
kg
1000 g
= 32.31396532 kg = 32.31 kg
page 14
1.5.1
CONVERTING UNITS : AREA
Shape
Name
h
Formula
Rectangle
bh
Square
s2
Parallelogram
bh
h
b
s
h
b
Triangle
½bh
h
b
Circle
r2
r rr
Example 1:
Your bathroom floor is 1.5 m by 2.0 m. You want to cover it with ceramic tiles that are 2 cm by 2 cm each.
How many tiles should you order? (Ignore the spaces left between the tiles).
The total area to cover is
(1.5 m ) x ( 2.0 m ) = 3.0 m
2
= 3.0 m 2
cm
10 2 m
2
= 3.0 x10 4 cm 2
Since each tile of ceramic is (2 cm × 2 cm) = 4 cm2
Therefore, you will need 3.0 x104 cm 2
1 tile
= 7.5 x103 tiles
4 cm 2
Example 2:
An artist is patiently working on a tapestry, progressing at the average rate of 15.2 cm2 a day. If the tapestry
is to be 134 mm wide and 2.20 m long, how long will it take for the work to be completed? We have three
different units here. Let us transform everything into cm. The total surface is:
134 mm × 2.20 m = (134 × 10–1 cm) × (2.20 × 102 cm) = (134 × 2.20) × 10–1×102 cm2 = 295×101 cm2
= 2.95×103cm2
or
134 mm × 2.20 m = (13.4 cm) × (220. cm) = 13.4 × 220. cm2 = 2.95 × 103 cm2.
How many times does 15.2 cm2 enter in 2.95×103 cm2 ?
(2.95 × 103 cm2 )/( 15.2 cm2/day) = 0.194 × 103 day = 194 day
In other words, it will take 194 days to complete the work.
Note how we took care of the significant figures here. The surface and the rate at which the artist works are
known to 3 significant figures. The result should also be given with 3 significant figures.
page 15
Exercises:
2
1. A circle has a radius of 25.0 cm. What is its area in mm ?
2
2
2. A parallelogram has an area of 1547 mm . What is its area in m ?
2
3. A sidewalk is 2.53 km long and 75.0 cm wide. What is its surface in m ?
2
4. The base of a triangle measures 1.345 m and its height 256 mm. Give its surface in cm .
1.5.2
CONVERTING UNITS: VOLUME
Shape
Name
Formula
h
lwh
Box
w
l
s3
s
r
Cube
l
Cylinder
r2l
Sphere
4 3
r
3
The conversion of units goes as in the surface case, but there is one more dimension here. Volumes will be
3
3
3
expressed in m , cm , mm etc…
In chemistry, it is usual to express volumes in litres. The formula for the conversion is:
3
1 cm = 1 mL
Exercises:
3
1. What is the volume in cm of a box having the following dimensions: base = 13 mm,
width = 2.5 cm, height = 1.3 m?
3
2
3
3
422.5 cm = 4.2 x 10 cm =420 cm
4
3
3
2. If the volume of a sphere is 2.7 × 10 cm , what is it in m ?
2.7 × 10
–2
cm
3
3
3. How many litres are there in 1 m ?
3
10 L
4. What is the volume in mL of a cylinder of 25.7 mm radius and 1.78 dm length?
369.3483147 cm3 = 369 cm3 = 369 mL
page 16
1.5.3
CONVERTING UNITS: DENSITY
Density is an intrinsic property of a pure substance. It relates the mass of the substance to its volume at a
given temperature and pressure.
density =
mass
volume
In symbols
=
m
V
The formula for density can be manipulated according to the requirements of the problem.
m
V
m= V
=
V=
m
Example 1a)
You want to built a raft from a new material which, according to the manufacturer, has a density of 4.56
g/cm3. Will this raft float? Note that for the raft to float, its density should be less than that of water: 1.00 × 103
3
kg/m .
3
3
3
To compare 4.56 g/cm with 1.00 × 10 kg/m we have to convert some of the units. Among the various
3
3
possibilities, let us choose the conversion: g kg and cm
m.
4.56
g
cm
3
=
4.56 × 10 3 kg
(10
2
3
m)
=
4.56 × 10 3 kg
10 6 m 3
= 4.56 × 10 3
kg
m3
Therefore the density of the revolutionary material is significantly higher than that of water and, unless we
trap in some air pockets or use other tricks, the raft will sink.
Example 1b)
What will be the total mass of the raft if it is 1.25 m wide, 1.95 m long and 6.50 cm high?
With the data given, we can easily calculate the volume of the raft:
Volume = 1.25 m × 1.95 m × .0650 m = 0.158 m3
3
3
3
Now that we know the volume (0.158 m ) and the density (4.56 × 10 kg/m ) how can we calculate the mass?
mass
Since:
then:
mass = density × volume
density =
volume
3
3
3
3
Therefore, the raft has a mass of 0.158 m × 4.56 × 10 kg/m = 0.722 × 10 kg = 722 kg.
Exercises:
1.
2.
What is the volume in mL of 45.7 g of a liquid that has a density of 2.14 g/mL?
3
How many litres will be occupied by 1.50 kg of water? 1.50 L
3.
The density of iron is 7.86 × 10 kg/m . What is the mass of a 1.00 m long iron bar that has a
–3
3
m = 13.6 kg
circular section of 2.35 cm radius?
(V = 1.73 x 10 m )
4.
What is the total mass of the air contained in a room 4.00 m × 3.75 m × 3.50 m? The density of
air is 1.29 × 10-3 g/cm3. (V = 52.5 m3) 6.7725 x 104 g = 67.7 kg
3
21.355 mL
3
page 17
1.5.4
CONVERTING UNITS: TEMPERATURE
Temperature is a physical quantity that indicates whether heat will flow from one body to another (and in
what direction) when they are brought together. Temperature scales are a way of assigning numbers to
bodies to indicate whether there will be heat flow, and if so, in what direction. Experience tells us that heat
will flow in direction from high temperature to low temperature.
There are three common temperature scales in use. They are the:
1. Fahrenheit scale
2. Celsius scale; and the
3. Absolute (Kelvin) temperature scale.
All first two scales are based on fixing the freezing and boiling points of water.
1.5.4.1
The Fahrenheit Scale
At a pressure of exactly one atmosphere, the fixed points of water are established at:
Freezing point = 32°F
Boiling point = 212°F
Range: 212°F – 32°F = 180°F
1.5.4.2
The Celsius Scale
At a pressure of exactly one atmosphere, the fixed points of water are established at:
Freezing point = 0°C
Boiling point = 100°C
Range: 100°C – 0°C = 100°C
1.5.4.3
The Kelvin Temperature Scale
The absolute temperature scale is based on the Celsius scale in that the range between the fixed points
of water is 100 K. At a pressure of exactly one atmosphere, the fixed points of water are established at:
Freezing point = 273.15 K
Boiling point = 373.15 K
The importance of this temperature scale is that the Kelvin temperature is directly proportional to the
molecular kinetic energy.
1.5.4.4
Relationship Between the Temperature Scales
212°F
100°C
373.15 K
Boiling Point
32°F
0°C
273.15 K
Freezing Point
Fahrenheit
Celsius
Kelvin
page 18
It is observed from the data, that a change of 180°F corresponds to a change of 100°C but also that the
zero point is 0°C whereas it is 32°F. Thus:
T° F =
180° F
T°C + 32° F
100°C
For the conversion between Celsius and Kelvin, observe that the 100°C range between the fixed points of
water also corresponds to a 100 K range. Thus the magnitude of 1°C equals 1 K. However, the zero point
in the Kelvin scale begins at 273.15 K. Thus
TK = T° C + 273.15 K
Exercise 1: What is the Fahrenheit temperature that corresponds to 80.0°C?
T° C = 80.0° F
T° F = 1.8T° C + 32° F
T° F = 1.8(80.0° F ) + 32° F = 176° F
Exercise 2: What is the Celsius temperature that corresponds to 75.0°F?
T° F = 75.0° F
T° F = 1.8T°C + 32° F
T° C =
T° F
32° F 75.0 32.0
=
= 23.9°C
1.8
1.8
Exercise 3: At what value are the Celsius temperature and Fahrenheit temperatures equal?
T° F = T°C
T° F = 1.8T° C + 32° F
T° F = 1.8T° F + 32° F
(1 1.8)T° F = 32° F
T° F =
32° F
= 40° F
0.8
Exercise 4: What is the absolute (Kelvin) temperature that corresponds to 92.0°C?
T° C = 92.0°C
TK = 92.0 + 273.15
TK = 365.2 K
Exercise 5: What is the Celsius temperature that corresponds to 345 K?
TK = 345 K
T° C = 345 273.15
T° C = 71.8°C
Exercise 5: What is the absolute (Kelvin) temperature that corresponds to 100°F?
T° F = 100° F
T° F 32 100 32
=
= 37.8°C
1.8
1.8
TK = 37.8 + 273.15 = 310.9 K
T° C =
page 19