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10/27/2011 ENERGY Energy - the capacity to do work Reminder: Changes in matter Can Be Physical or Chemical Change When energy is involved when there is a change in matter. Every Change in matter involves a change in energy All physical and chemical changes involve a change in energy. Sometimes energy is released when a change in matter occurs. Sometimes absorbed. Process - Any change in matter in which energy is changed. THE SCIENTIFIC UNIVERSE • The system is the part of the universe we wish to focus our attention on. In the world of chemistry, the system is the chemical reaction. For example: • The system consists of those molecules which are reacting. • The surroundings are everything else; the rest of the universe. For example, say the above reaction is happening in gas phase; then the walls of the container are part of the surroundings. ENERGY TRANSFER • Endothermic (+) and Exothermic Processes (-) • Exo/Endo difference is direction of energy flow SURROUNDINGS EXO ENDO THE SCIENTIFIC UNIVERSE • You are the system (that is, the chemical reaction) and as you look outward, everything else are surroundings. • FOR THE SYSTEM: We will signify an increase in energy with a positive sign (+) and a loss of energy with a negative sign. (-) ENERGY TRANSFER • Energy is transferred back and forth between a system and its surroundings. • Endothermic Surroundings sends energy to system • The boiling of water & melting of ice physical endothermic. • Can also be chemical endothermic (Ice Packs). • Endothermic processes, (energy is absorbed, never destroyed.) SURROUNDINGS 1 10/27/2011 ENERGY TRANSFER • MOSH PIT ANALOGY • Exothermic - system releases heat to Surroundings – Burning paper (chemical exothermic changes) / freezing water of (exothermic physical changes). Can also have chemical exothermic. – Exothermic processes, energy is released, may make it seem as if energy is being created. THIS RELATIONSHIP BETWEEN ENERGY CHANGE AND CHANGE OF STATE IS VERY IMPORTANT TO OUR MILK AND SCIENCE http://sfgate.com/cgi-bin/object/article?f=/kron/archive/2001/05/03/frozen.DTL&o=0 Solid Particles • Fixed very close together orderly, fixed arrangement. • They are held in place by the attractive forces that are between all particles. • Because solid particles can vibrate only in place and do not break away from their fixed positions, solids have fixed volumes and shapes. States of Matter The physical properties of each state come from the arrangement of particles and bonds holding. Liquid Particles • Move easily past each other • Some liquids can flow very readily, some not (viscosity). • Held closely by attractive forces (fixed volume). Forces between particles, which gives liquids special properties: – Surface Wetting - Liquid particles have cohesion, attraction for each other. They can also have adhesion, attraction for particles of solid surfaces. The balance of these forces determines whether a liquid will wet a solid surface • Solid takes up the same amount of space. Solids usually exist in crystalline Gas 1. Particles are independent. 2. Gas particles are much farther apart than the solids and liquids. – Capillary Action - The adhesion of the water molecules pulls water molecules up the sides of the tube. Critical to Biology. – Surface Tension - Substances are liquids instead of gases because the cohesive forces. • Liquids tend to form spherical shapes, because a sphere has the smallest surface area for a given volume. (rain droplets are spherical) 2 10/27/2011 Changing States • Melting (endothermic) - Solid becomes a liQuid • Evaporation (endothermic) - liQuid becomes a gas takes a lot of energy. The liQuid particles gain energy when colliding. Sweating, low dew pt. steals energy. • Sublimation (endothermic)- solids to a gas, Quick change (mothballs) • Condensation (exothermic)- gas becomes a liQuid • Water on a pan lid – lid steals energy • Deposition (exothermic)- gas become solids. Energy is released (Night Frost/Iodine) • Freezing (exothermic) - liQuid becomes a solid as energy is stolen, particles slow down. Changing States • All state changes are physical changes. • During the change of phase, energy is used to realign. • Realignment is seen by presence of both states. • Chemical properties are the same in different states (but reaction rates will be affected) ENERGY VS TEMPERATURE 1) Temp increases as particle motion increases. 2) The higher the temperature the more the particle movement. a) solids particles vibrate more rapidly. b) liquid particles move more Quickly. c) gas particles collide more often. Once realignment is complete, temperature changes. Heat As Energy Transfer We often speak of heat as though it were a material that flows from one object to another; it is not. Rather, it is a form of energy. 3) After a certain point, adding more energy will cause a substance to experience a change of state instead of a temperature increase. Heat As Energy Transfer Definition of heat: Heat is energy transferred from one object to another because of a difference in temperature. Unit of heat: calorie (cal) 1 cal is the amount of heat necessary to raise the temperature of 1 g of water by 1 Celsius degree. • Remember that the temperature of a gas is a measure of the kinetic energy of its molecules. Don’t be fooled – the calories on our food labels are really kilocalories (kcal or Calories), the heat necessary to raise 1 kg of water by 1 Celsius degree. 3 10/27/2011 Heat As Energy Transfer If heat is a form of energy, it ought to be possible to equate it to other forms. The experiment below found the mechanical equivalent of heat by using the falling weight to heat the water: ENERGY • Energy Types: Kinetic (motion) and Potential (stored) • Energy Forms: chemical, mechanical, light, thermal, electrical, and sound. http://www.youtube.com/watch?v=Ko5VpvE2btY&safety_mode=true&persist_safety_mode=1&safe=active ENERGY CONSERVATION OF ENERGY Energy cannot be destroyed or created. Potential Kinetic Chemical Mechanical Heater/Furnace Chemical Thermal Hydroelectric Gravitational Electrical Solar Optical Electrical Nuclear Nuclear Battery Chemical Electrical Food Chemical Heat, Kinetic Photosynthesis Optical Chemical Automobile Engine Heat, Kinetic, Optical During any physical or chemical change, the total Quantity of energy remains constant. Energy is never created or destroyed Energy Can Be Absorbed/Released Thermally As Heat HEAT VS. TEMPERATURE TEMPERATURE • Heat - Thermal energy transferred between objects that are at different temperatures. • 1. Measurement of the average kinetic energy of the random motion of particles in a substance. • 2. A measure of how hot (or cold) • 3. As kinetic energy increases so does temp. • Thermal energy is always transferred from a warmer object (higher energy) to a cooler (lower energy) object until thermal equilibrium is reached. • Heat Is Different from Temperature. Energy transfer can be measured by temperature. • The transfer of thermal energy (heat) does not always result in a change of temperature. 4 10/27/2011 TEMPERATURE Fahrenheito = (Co)(9/5)+32 NMC “Celsius has to work 9-5 for over a month to equal Fahrenheit.” TEMPERATURE • SI unit for temperature is the Kelvin, K. • Why do we have use Kelvin? The scale is a temperature scale where absolute zero—the coldest possible temperature where there is no heat energy—is defined as zero kelvin (0 K). • Usefull for scientific calculations, since it begins at absolute zero, meaning it has no negative numbers. • (The word "degrees" is NOT used with Kelvin – it is not in degree but is an absolute number from a fixed zero.) ABSOLUTE ZERO TEMPERATURE Absolute zero is the point on the thermodynamic (absolute) temperature scale where the heat energy is minimum, that is, no more heat can be removed from the system, corresponds to zero kinetic energy. 0° Celsius scale is designated as the freezing point of water. 0 Kelvin scale is designated as absolute zero We will use both the Celsius and Kelvin scales. Temperature change is the same in Kelvin/Celsius. T(K) = T(°C) + 273 °C T(°C) = T(K) – 273 K NMC “Why can’t I see that Celsius will be smaller by 273.” TEMPERATURE CONVERSIONS 1. 126 °F = 2. 827 °F = 3. 323 °F = 4. 414 °F = 5. 443 °F = 6. 322 °F = 7. 322 °C = 8. 462 °C = 9. 486 °C = 10.52 °C = 11.32 °C = 12.-62 °C = _______ °C _______ °C _______ °C _______ °C _______ °C _______ °C _______ °F _______ °F _______ °F _______ °F _______ °F _______ °F 1. 126 °F = 2. 827 °F = 3. 323 °F = 4. 414 °F = 5. 443 °F = 6. 322 °F = 7. 322 °C = 8. 462 °C = 9. 486 °C = 10.52 °C = 11.32 °C = 12.-62 °C = __52__ °C __442__ °C __162__ °C __212__ °C __228__ °C __161__ °C __612__ °F __864__ °F __907__ °F _126__ °F __90__ °F _-80__ °F 5 10/27/2011 TEMPERATURE CONVERSIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 126 K = 827 K = 323 K = 414 K = 443 K = 322 K = 322 °C = 462 °C = 486 °C = 52 °C = 32 °C = -62 °C = _______ °C _______ °C _______ °C _______ °C _______ °C _______ °C _______ K _______ K _______ K _______ K _______ K _______ K Round Up 1. 126 K = __-147__ °C 2. 827 K = __554__ °C 3. 323 K = ___50__ °C 4. 414 K = ___141_°C 5. 443 K = __170__ °C 6. 322 K = ___49__ °C 7. 322 °C = __595__ K 8. 462 °C = __735__ K 9. 486 °C = __759__ K 10.52 °C = __325__ K 11.32 °C = __305__ K 12.-62 °C = __211__ K TEMPERATURE CONVERSIONS QUIZ 1) 32 °C = _______ °F 2) -62 °C = _______ °F 3) 32 °C = _______ K 4) -62 °C = _______ K 5) 283K = _______ °C 1) 32 °C = 2) -62 °C = 1. 32 °C = 2. -62 °C = 1) 283K = __90__ °F _-80__ °F __305__ K __211__ K _10__ °C A practical refrigerator cycle uses a special liquid which, when the pressure is reduced, evaporates to become a gas. Such a cycle is illustrated below. Uncertainty Accuracy is how close one measurement relates to the True Value; always considered in experiments Bulls Eye = high accuracy Precision -How closely several measurements agree in exactness to one measurement. How closely several measurements of the same quantity made in the same way agree with one another. Precision differs from accuracy. Dartboard The bull’s-eye of the dartboard represents the true value. 1) Darts in the bull’s-eye mean high accuracy and high precision. 2) Darts clustered within a small area but far from the bull’s-eye mean low accuracy and high precision. 3) Darts scattered around the target and far from the bull’s-eye mean low accuracy and low precision 6 10/27/2011 Dartboard Uncertainty Divide 9 by 55 Uncertainty Uncertainty Calculating Errors There are two common ways to calculate your error: Percentage Error is the most common way of measuring an error, and often the most easy to understand. 1. Absolute Error is when you subtract the accepted value from your measured value… Absolute Error = Accepted Value - Measured Value Percentage Error = Absolute Error/Accepted Value X 100 So, if a student measured a pencil to be 102mm long, and an independent lab with high tech equipment measured it as 104mm, the percentage error is… A negative answer means you are over the accepted value. A positive answer means you are under the accepted value Percentage Error = (102mm - 104mm) / (104mm) =-0.02 But take the absolute value of this number So, if a student measured a pencil to be 102mm long, and an independent lab with high tech equipment measured it as 104mm, the absolute error is… Which means you have a -2% error. The minus sign just means that you were under the accepted value. 104-102=2mm In high school labs, don’t be surprised if you obtain errors of 25%. The important part is, can you explain your errors! 7 10/27/2011 Uncertainty Anytime an experiment is conducted, a certain degree of uncertainty must be expected. There are three reasons you might have an error in a measurement. 1. Physical errors in the measuring device Example 1: Your thermometer was dropped and has small air bubbles in it. 2. Improper or sloppy use of measuring device Example 2: When you used your thermometer, you measured the values in Fahrenheit instead of Celsius or don’t place thermometer in center of beaker/same spot. 3. Ambient conditions (temperature, pressure, etc.) Example 3: Measuring the length of a piece of wood outdoors in the winter using a metal ruler, you forget that metal contracts in the cold making the ruler shorter. • ACCURACY AND PRECISION LAB Scientific Method Scientific Method A series of steps followed to: Form a Hypothesis Construct a Theroy No No Ask Questions Make Observations Test the Hypothesis Analyze the Results Do they support your Hypothesis? Draw Conclusions Yes Can others confirm yourYes results? Publish Results 1. 2. 3. 4. 5. solve problems collecting data formulating a hypothesis testing the hypothesis stating conclusions WHY???? Experiments May Not Turn Out As Expected LAW ? Roger Bacon (1220-1292) For hundreds of years alchemists tried to change metals such as lead into gold. Roger Bacon, though didn't "make" gold, made many worthwhile scientific discoveries like gunpowder and optic devices and had many interesting ideas about flying machines. He is regarded by many as the forerunner of experimental science. 8 10/27/2011 Scientific Method Scientific Method Hypothesis - Theory - A testable explanation for observations. short and specific statement guide scientists in making their observations about the natural world Takes cause effect, if-then statement Explanation for observations Result of repeated experiments (and refined hypotheses) Can be discarded for better theories Scientific Method Scientific Method Variable: anything in the natural world that has levels which are measurable. It could affect the results of an experiment. Law – behavior of natural world how things work example: 1. law of conservation of matter - matter cannot be created or destroyed (mostly) 2. law of conservation of energy - energy cannot be created or destroyed True until an observation/experiment proves false. SCIENTIFIC NOTATION Chemistry deals with very large and very small numbers. Consider this calculation: (0.000000000000000000000000000000663 x 30,000,000,000) ÷ 0.00000009116 The mess above is easily rewritten to: (6.63 x 10¯ 31 x 3.0 x 1010) ÷ 9.116 x 10¯ 8 SN makes it much more compact, better represents significant figures, and easier to manipulate mathematically. The trade-off, of course, is that you have to be able to read scientific notation. What you should remember how to write numbers in scientific notation how to convert to-from scientific notation. As you work, keep in mind that a number like 9.116 x 10¯ 8 is ONE number (0.00000009116) represented as a number 9.116 and an exponent (10¯ 8). A scientist changes variables one at a time to see which affects experimental outcome Independent (manipulated) variable Dependent (responding) variable Hypothesis – If I heat a 2 Liter bottle in an oven then it will increase in pressure. SCIENTIFIC NOTATION Format for Scientific Notation 1. Used to represent positive numbers only. 2. Every positive number X can be written as: (1 < N < 10) x 10 some positive or negative integer Where N represents the numerals of X with the decimal point after the first nonzero digit. 9 10/27/2011 SCIENTIFIC NOTATION 1. Write, 1,202.5 in scientific notation. 1,202.5 = 1.2025 X 103 SCIENTIFIC NOTATION 2. Write, 0.0005203 in scientific notation. 0.0005203 = 5.203X10-4 OPTION #1: Move the decimal left so that # is between 1 and 10. OPTION #2: Move the decimal to the right so that that # is between 1 and 10. The number of places the decimal moves left is the positive exponent of 10. The number of places the decimal moves right is the negative exponent of 10. WRITE IN SCI NOTATION Write in Scientific Notation 0.00087 8.7 x 10-4 9.8 9.8 x 100 (100 is seldom written) 23,000,000 2.3 x 107 0.000000809 8.09 x 10-7 250,000,000,000 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 2.10 x 10-2 1.0254 x 108 3.00 x 10-8 2.0 x 100 2.309 x 103 1.03 x 102 1.3 x 102 1.0001 x 103 3.03 x 100 1.0 x 10-1 2.0 x 101 7.3 x 100 0.0210 102540000 0.0000000300 2.0 23.09 x 102 103 130 1000.1 2 + 1.03 4.00 – 3.9 12 x 1.65 12 / 1.65 2.50 x 1011 Convert to Scientific Notation: 1) 2.8 x 10 1) 28,000,000 2) 305,000 3) 0.000000463 (6) 4) 0.000201 5) 3,010,000 6) 0.000000000000057 (13) 7) 20,100 8) 0.00025 9)65,000,000,000,000,000 (15) 10) 8.54 x 1012 11) 2101 x 10-16 12) 305.1 x 107 13) 0.0000594 x 10-16 (4) 14) 0.00000827 x 1019 (5) 15) 386 x 10-22 16) 2511 x 1012 17) 0.000482 x 10-12 (3) 18) 0.0000321 x 1012 (4) 19) 288 x 105 20) 4.05 x 1011 7 2) 3.05 x 105 3) 4.63 x 10-7 4) 2.01 x 10-4 5) 3.01 x 106 6) 5.7 x 10-14 7) 2.01 x 104 8) 2.5 x 10-4 9) 6.5 x 1016 10) In SN 11) 2.101 x 10-13 12) 3.051 x 109 13) 5.94 x 10-21 14) 8.27 x 1013 15) 3.86 x 10-20 16) 2.511 x 1015 17) 4.82 x 10-16 18) 3.21 x 107 19) 2.88 x 107 20) In SN MOLE DAY • SMORE LAB • ABSOLUTE ZERO VIDEO START http://www.youtube.com/watch?v=y2jSv8PDDwA&oref=http%3A%2F%2Fwww.youtube.com%2Fresults%3Fsearch_query%3Dabsolute%2Bz ero%2Bnova%26aq%3Df&ytsession=jUiZHm YrNwZw8kvsUrZtCU8gqGz ZfizRmTTIsNVCF2ofZCvHCJPihkrd4Shswa4Cq8rFhbbEhxwdoGWz OdXGjJmsDIuEwSm5_W1WIYwMX3Yamk0fjfvE3nuXWCUuYQucAhkKggD27Xw9FFpHpk1z 2B_b599pbafTHO198w53j43JZ7Wf 2Fcgcw1oTT25AIxgjOz WxWtxV3H10XiCRaNjAs_ZYmQwXZ915YPz R5uz XV08wUqlz Zt-egJEa0wCODftSH4LgPpUjuhoy76f6z nr0D4wJmYZ4_PfYy3aI_dF64&has_verified=1 10 10/27/2011 Significant Figures • DIVIDE 7 by 22 • DIVIDE 65 by 88 Calculators Do Not Identify Significant Figures If I have something that weights .005g and with a volume of .001mL, what is the density? All non-zero digits are significant. Zeros in between non-zero digits are significant. Example 1: zeros in 4506002 are all significant Leading zeros (before the first non-zero digit) are not sig. Example 2: in 0.0012, the zeros are not significant (only 2 significant figures) Trailing zeros are significant only if a decimal point is present. Example 3: in 13900 the trailing zeros are not significant, but in 13900.0 all zeros are significant 6.295 g has four significant figures. 46.3 m has three significant figures. 40.7 L has three significant figures. 87009 km has five significant figures. 0.009587 m has four significant figures. 0.0009 kg has one significant figure. Rule for Additions and Subtraction Determine the Number of SF: 1) 8 1) 28,000,000. 2) 305,000 3) 0.000000463 (6) 4) 0.000201 (3) 5) 3,010,000 6) 0.000000000000057 (13) 7) 20,100 8) 0.00025 9)65,000,000,000,000,000 (15) 10) 8.54 x 1012 11) 2101 x 10-16 12) 305.1 x 107 13) 0.0000594 x 10-16 (4) 14) 0.00000827 x 1019 (5) 15) 386 x 10-22 16) 2511 x 1012 17) 0.000482 x 10-12 (3) 18) 0.0000321 x 1012 (4) 19) 288 x 105 20) 4.05 x 1011 2) 3 3) 3 4) 3 5) 3 6) 2 7) 3 8)2 9) 2 10)3 11) 4 12) 4 13) 3 14) 3 15) 3 16) 4 17) 3 18) 3 19) 3 20) 3 SIG FIGS The resulting number has the same number of decimal places than the number with the least decimal places. Example 4: 5.36 + 99.124 104.48 (2 decimal places) Rule for multiplications and divisions The resulting number has the same number of significant figures than the number with the least significant figures. Example 5: 5.5 - 1.12 4.4 (2 decimal places) Example 6: 15.322 x 3.12 47.8 (3 significant figures) 11 10/27/2011 SIG FIG ROUNDING SIG FIG ROUNDING Example 7: • Rules for Rounding • Keep the extra digits until the end of the calculations then round the final digit to the correct number of significant digit. • Round the extra digit after the last significant one. • If the extra digit is smaller than 5 the last significant digit remains. • If the extra digit is greater or equal than 5, add round up last significant digit. SIG FIG ROUNDING Example 8: Round 12.576 to 3 significant digits 12.576 the extra digit is 7 therefore the last significant digit is increased by 1. The result is 12.6 COMPLEX SIG FIGS Example 9: 34.53 + 2.0342 x 16.2 According to the usual rules of arithmetic, we do multiplication before the addition. PEMDAS (PLEASE EXSCUSE MY DEAR AUNT SALLY) How many significant figures should be in the result of this multiplication? 32.95404 2.0342 x 16.2 =32.954 Next, we do the addition. How many significant figures should be in the result of this addition? 34.53 + 33.0=67.53 The correct answer should have: 67.5 (3 SF) Round 12.536 to 3 significant digits In this examples: Digits in green are significant, the digit in blue is the "extra" digit and the digit in red is ignored. 12.536 the extra digit is 3 therefore the last significant digit remains. COMPLEX SIG FIGS While operating within the same rules (i.e. X and ÷ ) use all the numbers in your calculator. At the end of these type of operations determine your significance. You must do this before your move into a different type of operations ( + and - ). COMPLEX SIG FIGS Example 10: (34.53 + 2.0342) x 16.2 The parentheses around the first two terms indicate that the addition should be done first. How many significant figures should be in the result of this addition? 34.53 + 2.0342 =36.564 Next, we do the multiplication. How many significant figures should be in the result of this multiplication? 36.564 x 16.2 = 592.337 The correct answer should have: 3 significant figures. 592 12 10/27/2011 COMPLEX SIG FIGS 1) 3.461728 + 14.91 + 0.980001 + 5.2631 24.61 2) 23.1 + 4.77 + 125.39 + 3.581 156.8 3) 22.101 - 0.9307 21.170 4) 0.04216 - 0.0004134 0.04175 5) 564,321 - 264,321 300,000. = 3.00000 x 105 COMPLEX SIG FIGS 10) 5.26 x 5.25 x 6.34 +0.11 175 (3 SF) 11) 5.25 x1.1 + 6.342 12.1 (3SF) COMPLEX SIG FIGS 6) (3.4617 x 107) ÷ (5.61 x 10-4) (3 SF) 6.17 x 1010 7) [(9.714 x 105) (2.1482 x 10 -9)] ÷ [(4.1212) (3.7792 x 10-5)] 1 1.3398 x 10 = 13.40 (4SF) 8) (4.7620 x 10-15) ÷ [(3.8529 x 1012)(2.813 x 10-7) (9.50)] 4.62 x 10-22 (3 SF) ÷[(3.8529 x 1012) (2.813 x 10-7) 9) (4.7620 x 10-15)÷ (9.50)] 2.30 x 103 (3 SF) COMPLEX SIG FIGS 1) 3.461728 + 14.91 + 5.2631x 3.46 ÷ 6.81 18.37 + 18.2 ÷ 6.81 = 21.04 2) 23.1 + 4.77 + 125.39 + 3.581 x 9.7 x 2.142 ÷ 4.12 153.26 + 18.1= 171.4 3) 9.23 x 2.6 ÷ 4.999 + 22.101 - 0.9307 4.8 + 21.170 = 25.9 4) 0.04216 ÷ 8.945 + 22.101 - 0.0004134 26.274 5) 564,321 - 264,321 x 0.544 ÷ 8.654 + 82.4533 0.0004134 18817.5 = 18818 COMPLEX SIG FIGS LAB PARTNER EXERCISE 6) (3.4617 x 107) ÷ (5.61 x 10-4) 6.17 x 1010 (3 SF) 7) [(9.714 x 105) (2.1482 x 10 -9)] ÷ [(4.1212) (3.7792 x 10-5)] 1.3398 x 101 = 13.40 (4SF) 8) (4.7620 x 10-15) ÷ [(3.8529 x 1012)(2.813 x 10-7) (9.50)] -22 4.62 x 10 (3 SF) 11)Why are significant figures important when taking data in the laboratory? 12)Why are significant figures NOT important when solving problems in your math class? 13) Using two different instruments, I measured the length of my foot to be 27 centimeters and 27.00 centimeters. Explain the difference between these two measurements. 14) I can lift a 20 kilogram weight over my head ten times before I get tired. Write this measurement to the correct number of significant figures. 13 10/27/2011 LAB PARTNER EXERCISE 11) Why are significant figures important when taking data in the laboratory? Significant figures indicate the precision of the measured value to anybody who looks at the data. For example, if a weight is measured as being “1100 grams”, this means that the mass has been rounded to the nearest hundred grams. If a weight is measured as being “1100.0 grams”, this means that the mass has been rounded to the nearest tenth of a gram. Though the numbers plug into the calculator in exactly the same way, they mean very different things. 12) Why are significant figures NOT important when solving problems in your math class? Math classes don’t deal with measured values. As a result, all of the numbers are considered to be infinitely precise. LAB PARTNER EXERCISE . 13) Using two different instruments, I measured the length of my foot to be 27 centimeters and 27.00 centimeters. Explain the difference between these two measurements. As in problem 11, the first measurement implies that my foot is somewhere between 26.5 and 27.4 cm long. The second measurement implies that my foot is between 26.995 and 27.004 cm long. Again, though the numbers plug into the calculator in the same way, they imply different precisions. 14) I can lift a 20 kilogram weight over my head ten times before I get tired. Write this measurement to the correct number of significant figures. The answer is written as 20.0, with a line drawn above the zero in the tenths place. This is one of the few cases where you can measure data with infinite significant figures. After all, I can either lift it or I can’t – there’s no “half-lift” that would result in a decimal. LAB PARTNER EXERCISE How many significant figures are in each of the following numbers? 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 5.40 210 801.5 1,000 101.0100 1.2 x 103 0.00120 0.0102 9.010 x 10-6 2,370.0 3 2 4 1 7 2 3 3 4 5 Specific Heat (c) The specific heat - energy transferred as heat to a substance vs. that substance’s temperature change 1. The Quantity of energy as heat that must be transferred to raise the temperature of 1 g of a substance (1°C). 2. Specific heat is expressed in joules (energy) per gram Kelvin (J/gK). 3. Metals tend to have low specific heat (less energy must be transferred as heat to raise their temperatures. ) Water’s specific heat it is the highest of most common substances. Specific Heat (s) • Transfer of heat affects dissimilar substances in different ways. • Why does a pan of water heat slower than an empty pan? • Why does a tongue stick on a cold metal flagpole vs. a cold telephone pole? • Why does a 400 mL heat slower than 40 mL? Specific Heat (S) Q=cm∆ ∆ T NMC: Q-SMAT Q s m ∆T The relationship between heat and temperature change is usually expressed in the form shown above where c is the specific heat. The relationship does not apply if a phase change is encountered, because the heat added or removed during a phase change does not change the temperature. 14 10/27/2011 Specific Heat (c) Specific Heat (c) If the total energy required to increase the temperature of a substance is much larger, its specific heat is much larger. J/kg/o C cal/g/o C Water (0 oC to 100 oC) 4186 1.000 Methyl Alcohol 2549 0.609 2093 0.500 Steam (100 C) 2009 0.480 Wood (typical) 1674 0.400 Soil (typical) 1046 0.250 Air (50 oC) 1046 0.250 Aluminum 900 0.215 Marble 858 0.205 Glass (typical) 837 0.200 Iron/Steel 452 0.108 Copper 387 0.0924 Silver 236 0.0564 Mercury 138 0.0330 Gold 130 0.0310 Lead 128 0.0305 Substance o o Ice (-10 C to 0 C) o Specific Heat (s) Q unknown Step 4. Substitute your values into the formula Q= sm(∆ ∆ T) Q = 23.984 g x 393.0 oC x 0.902 J/g x oC Step 5. Cross out units where possible, and solve for unknown. Q= 23.984 g x 393.0 oC x 0.902 J/g x oC Q= 8501.992224 J Step 6. Round to the correct number of significant digits and check answer makes sense. Q = 8.50 x 103 J Specific Heat (s) m unknown Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which. Q = 24 500 J m = ? ∆ T = 69.5 oC s = 4.18 J/goC Step 4. Substitute your values into the formula. m = Q/(∆ ∆ T)s m = 24 500 J/69.5 oC x 4.18 J/goC Step 5. Cross out units where possible, and solve for unknown. m = 24500 J/(69.5 oC x 4.18 J/goC) m = 84.3344463184 g Example #1 If Heat Transferred (Q) is the unknown: Ex. Aluminum has a specific heat of 0.902 J/g x oC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a temperature of 415.0 oC to a temperature of 22.0 oC? Step 1: First read the Question and try to understand what they are asking you. Step 2: Write the original formula. Q= sm(∆ ∆ T) Step 3: List the known and unknown factors. Q= ? m=23.984g ∆ T=(415.0oC-22.0oC)=393.0oC (remember, asked for the change in temperature) s = 0.902 J/goC Specific Heat (s) m unknown Example #2 If mass (m) is the unknown: The temperature of a sample of water increases by 69.5 o C when 24,500 J are applied. The specific heat of liquid water is 4.18 J/g x oC. What is the mass of the sample of water? Step 2: Write the original formula, and then modify it isolate the unknown. Q= sm(∆ ∆ T) Q = m(∆T)c (∆T)s (∆T)s m = Q/(∆ ∆ T)s Specific Heat (s) m unknown Step 6. Round to the correct number of significant figs and check to see that you answer makes sense. m = 84.3 g Our answer makes sense because grams are the correct units for mass, and the value should be positive. 15 10/27/2011 Specific Heat (s) ∆ T unknown 850 joules of heat are applied to a 250 g sample of water with an initial temperature of 13.0 oC. Find a) the change in temperature and b) the final temperature. (remember, the specific heat of liquid water, in joules, is 4.186 j/g x oC.) Q = m(∆ T)s Q = m(∆T)s ms ms ∆ T = Q/ms Specific Heat (s) ∆ T unknown Step 5. Cross out units where possible, and solve for unknown. ∆ T = 850 cal/250 g x 4.186/g x oC Answer to step a) ∆ T = .81 o C Answer to step b) Tf= 13.0 oC + .81 oC = 13.8 oC Step 6. Round to the correct number of significant digits and check to see that you answer makes sense. 13.8 oC Specific Heat (s) s is unknown Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which Q = 34700 J m = 350 g s = ? ∆ T = (173.0oC - 22.0oC) = 151.0 oC Step 4. Substitute your values into the formula s = Q/m(∆ T) s = 34700 J/(350 g x 151.0 oC) Specific Heat (s) ∆ T unknown Step 2: Write the original formula, and then modify it isolate the unknown. Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which. Q = 850 J m = 250 g s = 4.86/goC ∆ T = ? Step 4. Substitute your values into the ∆ T = Q/mc formula ∆ T = 850 cal/(250 g x 4.86/g oC) Specific Heat (s) c is unknown When 34, 700 J of heat are applied to a 350 g sample of an unknown material the temperature rises from 22.0 oC to 173.0 oC. What must be the specific heat of this material? Step 2: Write the original formula, and then modify it to isolate the unknown. s = Q/m(∆ ∆ T) Specific Heat (s) s is unknown Step 5. Cross out units where possible, and solve for unknown. s = 34 700 J/350 g x 151.0 oC s = 0.65657521286 J/g x oC Step 6. Round to the correct number of significant digits and check to see that you answer makes sense. s= 0.66 J/g x oC 16 10/27/2011 Specific Heat Problem 1 Specific Heat Problem 2 Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25oc to 60oc m = 204.75g m = 250g s = 4.18 J/goC Tf = 56oC Ti = 20oC Q = s x m x (Tf - Ti) Q = 250g x 4.18 J/goC x (56 - 20)oC Q = 250g x 4.18 J/goC x 36oC Q = 37620 J = 38000J Specific Heat Problem 3 216 J of energy is required to raise the temperature of aluminum from 15oC to 35oC. Calculate the mass of aluminum. (Specific Heat Capacity of aluminum is 0.90 J/goC. Q = 216 J s = 0.90 J/goC Ti = 15oC Tf = 35oC Q = m x x (Tf - Ti) 216J = m x 0.90 J/goC x (35 - 15) oC 216J = m x 0.90 J/goC x 20 oC 216 216J = m x 18g m = 216J ÷ 18J/g= 12g m = 15g Tf = 60oC Ti = 25oC Q = s x m x (Tf - Ti)oC 204.75J = 15g x s x (60 - 25)oC 204.75J = 15g x s x 35oC 204.75J = 525goC x s s = 204.75J/goC ÷ 204.75 = 0.39 J/goC Specific Heat Problem 4 The initial temperature of 150g of ethanol was 22oC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol? Specific heat capacity of ethanol is 2.44 J/goC). Q = 3240 J m = 150g s = 2.44 J/goC Ti = 22oC Q = m x s x (Tf - Ti) 3240J = 150 x 2.44 x (Tf - 22) 3240J = 366 (Tf - 22) 8.85 = Tf - 22 Tf = 30.9oC 17