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

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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 =

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