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- 1. Scientific Measurement • numerical description of properties and changes in matter , using numbers
- 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
- 5. 5 The Metric System
- 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
- 10. 10 Problem Solving
- 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
- 12. 12 conversion factor conversion factor The conversion factor is derived from the equality. 1 m = 1000 mm Divide both sides by 1000 mm Divide both sides by 1 m 1 m 1000 mm = 1 1 m 1 m 1 m 1000 mm = 1 1000m 1000 mm
- 13. 1 m = 1000 mm 1 m 1000 mm = 1 1000m 1000 mm 1 m 1000 mm = 1 1 m 1 m Conversion factors come in pairs Pair are reciprocals
- 14. 14 The conversion factor takes a fractional form. mm m x = mm m
- 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. )
- 23. Metric English Meter inchescentimeter millimeterLenght Time X C F X CF X equivalent factor Conversion which involves quantities of same parameters Conversion which involves quantities of different parameters parameter
- 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
- 26. 26 146 3 Significant Figures All nonzero numbers are significant. Significant Figures
- 27. 27 140 3 Significant Figures A zero is significant when it is between nonzero digits. Significant Figures
- 28. 28 A zero is not significant when it is before the first nonzero digit. 1 Significant Figure 600.0 Significant Figures
- 29. 29 A zero is significant at the end of a number that includes a decimal point. 5 Significant Figures 000.55 Significant Figures
- 30. 30 A zero is not significant when it is at the end of a number without a decimal point. 1 Significant Figure 00005 Significant Figures
- 31. 31 Significant Figures on Reading a Thermometer
- 32. 32 Temperature is estimated to be 21.2oC. The last 2 is uncertain. The temperature 21.2oC is expressed to 3 significant figures.
- 33. 33 Temperature is estimated to be 22.0oC. The last 0 is uncertain. The temperature 22.0oC is expressed to 3 significant figures.
- 34. 34 Temperature is estimated to be 22.11oC. The last 1 is uncertain. The temperature 22.11oC is expressed to 4 significant figures.
- 35. 35 Significant Figures in Calculations
- 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.
- 38. 38 Multiplication or Division
- 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
- 41. 41 Addition or Subtraction
- 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.
- 59. 59 Degree Symbols degrees Celsius = oC Kelvin (absolute) = K degrees Fahrenheit = oF
- 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
- 61. 61 180 Farenheit Degrees = 100 Celcius degrees 180 =1.8 100
- 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
- 64. 64 Density
- 65. 65 Density is the ratio of the mass of a substance to the volume occupied by that substance. mass d = volume
- 66. 66 Mass is usually expressed in grams and volume in ml or cm3. g d = mL3 g d = cm The density of gases is expressed in grams per liter. g d = L
- 67. 67 Density varies with temperature o 2 4 C H O 1.0000 g g d = = 1.0000 1.0000 mL mL o 2 80 C H O 1.0000 g g d = = 0.97182 1.0290 mL mL
- 68. 68
- 69. 69 Examples
- 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
- 76. 76

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