 Download Presentation Unit 2: Energy Icons are used to prioritize notes in this section.

# Unit 2: Energy Icons are used to prioritize notes in this section.

Download Presentation ## Unit 2: Energy Icons are used to prioritize notes in this section.

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2. Chapter 3 Energy Transfer •••••••••••••••••••••••••••••••• •••••••••••••••• ••••••••••••••••

3. Overview: Heat, work and temperature are frequently confused by students. The terms are closely related to each other, but they are not the same. Heat and work are a forms of energy transfer, temperature is a measure of molecular agitation. Heat and temperature both involve random movement of particles, while work is an orderly movement. Heat and work measurements depend on the mass of a material, while temperature does not. 3.1 Heat, Work & Temperature

4. Energy Work and • Energy is sometimes defined as the ability to do work or to produce heat. • This means that energy, work and heat can all be measured with the same unit: the joule (J) • Work is a transfer of kinetic energy that results in an orderly movement of particles, such as the motion of an object, or the expansion of a balloon.

5. Heat • Heat is a transfer of thermal energy that occurs when two systems of different temperatures come into contact. • Heat is transferred as a result of molecular agitation, that is random motion of particles. • Click on the boxes below to link to more on heat

6. Heat Transfer PREVIEW Equal Hot Object Cold Object Temperature Thermal Energy When two systems of different temperatures are placed in contact… Thermal energy (or heat) flows from the hotter one into the colder one… Until their temperature is the same. This concept will be developed fully in a later lesson.

7. Temperature • Temperature is a measurement of the agitation of the particles (atoms or molecules) in a system. • While temperature is related to the kinetic energy of the individual particles, it does not reflect the energy of the system as a whole. It cannot be measured in joules. It is measured in Kelvins (K) or in degrees (°C) • If particles were sufficiently cooled, they would (theoretically) reach a temperature where they had no agitation at all. This temperature is called absolute zero, and is –273.15°C. You remember this from the Gas Laws. We often round this to 3 significant digits: -273°C

8. Assignments • Exercises page 130 # 1 to 4

9. Overview: The law of conservation of energy states that energy can be transformed or transferred, but that it cannot be created or destroyed in any chemical or physical change*. Conservation of energy is the first law of thermodynamics, the study of energy changes. In this section we will study thermodynamic systems. 3.2 Law of Conservation of Energy *It’s a bit more complicated with nuclear changes, since matter can change into energy and vice-versa at the nuclear level, as Einstein calculated E=mc2

10. Thermodynamic Systems Nothing is lost, Nothing is created, Everything is transformed. • A system is the location where we are observing an energy transfer or change. • A system can be as simple as a beaker with a thermometer, or as complex as a Dewar flask or a calorimeter with an array of thermocouples. • There are three different types of system. Antoine Lavoisier Promoter of the Laws of Conservation

11. Three Types of System • An Open System • Both matter and energy can easily enter or leave the system, for example, a beaker. • A Closed System • Energy can enter or leave the system, but matter cannot, for example, a sealed balloon. • An Isolated System • Neither energy nor matter can easily enter or leave the system. For example, an insulated container or calorimeter.

12. Imperfection of Isolated Systems • In the real world it is impossible to create a perfectly isolated system. • Even the best Dewar flask or Thermos™ bottle will eventually let some heat in or out. • However, if experiments are conducted quickly, an insulated container is close enough to an isolated system to give us acceptable results. A Dewar Flask

13. Calorimeters • A calorimeter is a device used to measure heat transfer. • An inner chamber (AKA “bomb”) where the system (material or reaction that produces the heat) is placed. • A thermometer . • An outer chamber that holds water(or another substance of known specific heat capacity) • Insulation to prevent loss of heat to the surroundings. Simplified calorimeter Typical calorimeter

14. Homemade Calorimeters • For quick experiments a simple, but effective calorimeter can be made out of Styrofoam cups and a good thermometer. • The chemicals are mixed in the cup, and the Styrofoam insulates long enough to measure temperature differences. • Measurements must be made quickly with this type of calorimeter, before heat loss can occur. When using a homemade calorimeter promptness is important. Every minute you waste will reduce the accuracy of your measurements. Make sure you gather your data quickly but carefully!

15. Overview: When a substance changes temperature, thermal energy is absorbed or released. The thermal energy is related to the mass and specific heat capacity of the substance by the familiar formula: Q = mc As a general rule, the amount of thermal energy lost by one system is equal to the energy gained by another. –Q1 = Q2 3.3 Energy Relationships

16. Specific Heat Capacity (c) • Specific heat capacity is the amount of energy required to raise temperature of one gram of a substance by 1°C. • Specific heat is a characteristic property, it is different for each substance, if measured at a standard temperature. Remember the specific heat capacity of water. Textbooks sometimes use 4.19 or 4.18 if the problem requires no more than 3 significant figures.

17. Heat Gained or Lost by a Substance • The heat gained or lost by a substance can be measured by the calorimeter formulas: Where: Q = thermal energy (heat), in joules m=mass of a substance, in grams c = specific heat capacity of a substance, in J/(g∙°C) ΔT = temperature change (Tf-Ti), in °C Where: Tf = the final temperature of the substance, in °C Ti = the initial temperature of the substance, in °C

18. Sample Question • Calculate the energy that is absorbed by a 3.00 kg block of aluminum when its temperature changes from 17.1 °C to 35.5 °C. (From the internet, the specific heat capacity of aluminum is found to be 0.900 J/g°C) Data: m= 3.00kg = 3000g c =0.900 J/(g∙°C) Ti =17.1 °C Tf=35.5 °C To find: ΔT= Q= Step 1: Calculate ΔT: ΔT = Tf– Ti = 35.5 °C – 17.1 °C = 18.4 °C Always show equation Note: large amounts of energy should be converted to kilojoules and answers rounded to a sensible number of significant digits. Step 2: Calculate Q: Q = mcΔT = 3000 g • 0.900 J/(g∙°C) • 18.4 °C = 18.4 °C Always show equation 49 680 J = 49.68 kJ 49 680 J Answer: The aluminum block must absorb 49.7 kilojoules of thermal energy. (3 significant figures were used to match m and c precision)

19. Read pages 134-136 • Examine sample questions on page 136. • Do questions 1 to 16 on page 137

20. Overview: Energy transfer occurs when energy moves from one body to another. When two systems at different temperatures come in contact, the thermal energy from the hotter system is transferred into the cooler system until both systems reach the same temperature. The amount of heat lost by the hotter system should exactly equal the amount of heat gained by the cooler system, unless heat is lost to the surroundings. 3.4 Calculating Energy Transfer

21. Heat Content of an Object • The heat content of an object or system that is available for transfer is calculated by the same formula used in calorimetry, which was described in the previous lesson: • Now we are going to combine two systems, so we will be calculating

22. Heat Transfer When two systems of different temperatures are placed in contact… Thermal energy (or heat) flows from the hotter one into the colder one… Until their temperature is the same. The heat lost by the warmer object will equal the heat gained by the cooler one. Equal Hot Object Cold Object Temperature Thermal Energy Where: Q1 = Heat lost by the 1st system Q2 = Heat gained by the 2nd system

23. Heat Transfer Calculations • Combining  • Gives us: Where: m1 = mass of system 1, in grams c1 = specific heat capacity of system 1, in J/(g.°C) ΔT1 = temperature change in system 1 (Tf-Ti1), in °C m2 = mass of system 1, in grams c2 = specific heat capacity of system 1, in J/(g.°C) ΔT1 = temperature change in system 2 (Tf-Ti2), in °C Special note: Tf is the same for both systems, but Ti is different

24. Sample Question • Calculate the mass of cold water at 10 °C that it would take to cool 10.0 g of hot (95°C) glass to 30°C. Formula: ̶ m1c1ΔT1=m2c2ΔT2 Data: m1 = 10g c1 =0.84 J/(g∙°C) Ti1 =95°C Tf=30°C c2 = 4.184 J/(g∙°C) Ti2 = 10 °C To find: ΔT1 ΔT2 m2 Step 1: Calculate ΔT1 : ΔT1 = Tf– Ti1 = 30 °C – 95 °C = -65 °C From Table Step 2: Calculate ΔT2 : ΔT2 = Tf–Ti2 = 30 °C – 10 °C = 20°C Step 3: calculate the mass of water (m2) = -65°C = +20 °C = 6.5 g Round to a sensible number of significant digits. = 6.5248565965 g Answer: It would take 6.5 g of water to cool the glass.

25. Finding the Final Temperature • Quite often questions will ask for the final temperature of the two systems. • This rearrangement of the previous formula is sometimes useful: Where: Ti1 = initial temperature of system 1 Ti2 = initial temperature of system 2 Tf= final temperature of both systems Everything else means the same as before. The derivation of this formula is shown on page 139 of your textbook, so I won’t repeat it here. You can actually use formulas given before to solve this type of problem, but this form is easier to input if you are working with a TI83 calculator or the equivalent.

26. Sample Question • A 500 g package of frozen strawberries at -4.0 °C has a specific heat capacity of 3.50 J/(g.°C). It is placed in 2.00 kg of warm (40.0°C) water in an insulated cooler to thaw. What is the final temperature of the strawberries and water? Formula: Data: m1 = 500 g c1 = 3.5 J/(g∙°C) Ti1 =-4.0°C m2 = 2 kg = 2000 g c2 = 4.19 J/(g∙°C) Ti2 = 40 °C To find: Tf Step 1: Calculate Tf: Tf = = = 32.398815 °C I did this in two steps to show the cancellation of units more clearly. You can do it in a single step if you have a scientific calculator, as shown on the following calculator display. = 32.4 °C Round to a sensible number of significant digits. Answer: The final temperature is 32.4 degrees celsius.

27. Personal Observation Personally, I have never bothered to memorize the formula on the previous slide. I just use the three simple formulas earlier in the chapter: Q=mcΔT, –Q1=Q2 and ΔT=Tf-Ti. Then I break my calculations into many small steps. It takes me a bit longer, but I don’t have to commit a really complex formula to memory, and I find I make fewer mistakes.

28. Read Section 3.4, pp. 138-140 • Do Questions 1 to 11 on page 141 • Also: • Chapter End Questions on page 145 to 146

29. Chapter 4 Enthalpy Change

30. Overview: Enthalpy is defined as the total energy of a system. In theory, enthalpy is the sum of the kinetic and potential energy that a system contains at a given pressure. In practice, it is impossible to measure the total enthalpy of a system, since there are too many variables involved. Instead, we concern ourselves with the energy changes that occur in a system. This gives us a way of finding how much potential energy changes to kinetic, or vice versa, during a chemical change. 4.1 Page 148 Enthalpy (H)

31. Enthalpy (H) • Enthalpy is the total energy of a system • It includes all the potential energy and all the kinetic energy of the system. • In practice, it is impossible to find these values exactly, so we never talk about total enthalpy. Where: H = total enthalpy of a system Ek = the kinetic energy of the system Ep= the potential energy of the system

32. Enthalpy Change (ΔH)AKA: Heat of Reaction • Enthalpy Change is the energy exchanged between a system and its surroundings during a physical change or chemical change. • In theory, the enthalpy change should equal the difference between the enthalpy of the products and the enthalpy of the reactants: Where: ΔH = the heat of reaction (enthalpy change) Hp = enthalpy of the products Hr= enthalpy of the reactants

33. Enthalpy Change in Practice • Although finding the actual totalenthalpy(H) of a system is nearly impossible, finding the enthalpy change(∆H) is quite easy to do by experiment. • Since nearly all the energy change of most reactions is in the form of heat (thermal energy), a calorimeter can find the amount of heat absorbed or released during a change. This will be your heat of reaction, from which we get the enthalpy change (∆H)

34. Overview: Chemical and physical changes can absorb or release energy. Endothermic changes absorb heat from their surroundings. Exothermic changes release heat into their surroundings. For chemical reactions its usually easy to tell. If materials become hotter, an exothermic reaction is occurring. If they get colder, it is an endothermic reaction. For physical changes, it is not quite so straight forward. 4.2 Endothermic and Exothermic

35. Endothermic and ExothermicDefinitions • An Endothermic process is a change that absorbs heat or other forms of energy from the surroundings. • Examples: Photosynthesis (absorbs light), evaporation (absorbs heat), dissolution of NH4NO3 (absorbs heat) • An Exothermic process is a change that releases energy into the surroundings. • Examples: Respiration (releases chemical energy), condensation (releases heat), dissolution of H2SO4 (releases heat)

36. Endothermic and Exothermic Physical Changes • Physical changes, like freezing, melting, boiling vaporization, condensation and dissolution can absorb or release energy. Heat Absorbed Heat Released

37. Gas Rapid vaporization is called “boiling”, Slow vaporization is “evaporation” Sublimation occurs when a material “evaporates” from a solid straight to a gas, like dry ice or iodine. Vaporization Exothermic Process Liquid Condensation Endothermic Process Solid Condensation / Deposition Sublimation Terminology associated with Change of Phase And the energy exchanges Liquid Solid Liquid Melting (fusion) Freezing (solidification) 37

38. The Staircase Analogyanother way to remember physical changes Imagine the states of matter to be like a staircase. Solids are lowest in energy, so they are on the bottom step. Gases are highest, so they are the top step. Gas Most energy To go from solid to liquid or gas means “climbing” the stairs. Using up energy to go up. These changes are endothermic. Liquid More energy Going from gas to liquid, or from liquid to solid gives you back energy, as you bounce down the stairs. These changes are exothermic Solid Low energy

39. Heat Curve of a Pure Substance If a pure, cold solid is slowly heated, its temperature will increase, until it starts to melt. 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 Then, even though you are still heating it, the temperature stays the same until all the solid is melted. The energy absorbed to melt it is called the heat of fusion gas Boiling Point liquid & gas ΔT(l) Temperature (°C) liquid = Melting Point Specific heat capacity of the liquid can be determined from the inverse slope of the curve solid & liquid ΔH(l) solid ΔHfus ΔHvap Heat of vaporization Heat of fusion 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Energy absorbed by substance (joules / gram)

40. Reversing a Heat Curve If instead of heating a cold, solid substance, we cool a hot gaseous substance, we get a similar heat curve, but reversed in appearance. Some of the terminology changes when we reverse a heat curve. 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 gas Condensation point Instead of a boiling point, we have a condensation point. Instead of heat of vaporization we have heat of condensation. liquid & gas Instead of a melting point we have a freezing point, and the heat change is called heat of solidification. liquid Temperature (°C) Freezing Point solid & liquid solid ΔHs ΔHcond Heat of solidification Heat of condensation 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Energy (joules / gram) released

41. Things You Must be Able to Find From the Heat Curve of a Pure Substance • The melting/freezing point temperature • From the vertical axis (1st plateau level) • The vaporization/condensation temperature • From the vertical axis (2nd plateau level) • The heat of fusion (+) or solidification (–) • From the horizontal axis (difference of 2 measurements) • The heat of vaporization (+) or condensation (–) • From the horizontal axis (difference of 2 measurements) • The specific heat capacity of the solid or liquid substance. • Inverse slope of a section of the curve.

42. Heat Curve of a Mixture The heat curve of a mixture of substances is less precise than that of a pure substance. The mixture “melts” over a range of temperatures, and each component “boils” at a different temperature. 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 Fractional Distillation #3 gas Fractional Distillation #2 Fractional Distillation #1 Boiling range We can use the different boiling temperatures of the mixtures components to separate them by capturing the vapours separately—a process called fractional distillation. Temperature (°C) liquid Melting Range slushy The mixture melts more gradually than pure matter. solid 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Energy absorbed by substance (joules / gram)

43. Sample ProblemHeat curve of “Imaginarium” 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 Use the heat curve to determine: The melting point of imaginarium. The boiling point of imaginarium The specific heat capacity of liquid imaginarium the heat of fusion of imaginarium the heat of vaporization of imaginarium ΔT(l) = (70-(-20)) =90℃ Temperature (°C) ΔH(l) = (900-500) =400J/g == 4.44 J/g℃ ΔHfus= (500–300)= 200 J/g ΔHvap=(1300-900)=400 J/g Heat of fusion Heat of vaporization 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 Energy absorbed by substance (joules / gram) *Heat of fusion and vaporization are given in J/g. To convert to the more standard form ( j/mol) we could multiply this answer by the molar mass of imaginarium (if we knew it). Answers: –20°C +70°C 4.44 J/g°C 200 J/g* 400 J/g*

44. Endothermic and Exothermic Chemical Reactions • During a chemical reaction, • energy is absorbed as the old bonds in the reactants are broken. This phase of the reaction is endothermic • Energy is released as new bonds form in the products. This phase is exothermic. • If more energy is absorbed than is released, the overall reaction is endothermic. If more energy is released than is absorbed, then the overall reaction is exothermic.

45. Visualization of a Chemical Change Energy Absorbed Energy Released H H H H H O H H H O O O H O H O H H O O H H O H H O H H H H Products 2H2O atoms 4H + 2O Reactants 2H2 + O2 ① ② Bond breaking Bond forming In the chemical reaction visualized here, hydrogen burns with oxygen to form water. This happens in two stages. ① The collisions of hydrogen (H2) and oxygen (O2) molecules break the molecules into atoms, absorbing a little energy. ② The oxygen and hydrogen atoms then form new bonds, making water molecules and releasing large amounts of energy. The overall reaction is exothermic, but the initial stage is endothermic. Be aware that this visualization is a bit of a simplification, as explained in the next slide

46. In reality, the process of reaction is a bit more complicated. Molecules do not instantly jump apart into atoms… they tend to clump into intermediate complexes, as we shall see in the next chapter. However, the mathematics of energy is the same for the simple visualization shown on the last slide as it is for the actual reaction pathway, so we’ll use the simple version for now.

47. Simple Enthalpy Diagrams • Simple enthalpy diagrams show the difference between the enthalpy of the reactants and the products. • In an endothermic reaction, the enthalpy increases (ΔH is positive) • In an exothermic reaction, the enthalpy decreases (ΔH is negative) Endothermic Reaction Exothermic Reaction

48. Assignments • Read sections 4.1 and 4.2 • Do questions 1 to 6 on page 155

49. Overview: Energy Balance is the sum of the energy released when old bonds are broken (a positive number) and the energy absorbed as new bonds are formed (a negative number). The energy balance can tell us what the enthalpy change is during a reaction. Energy balance can also be calculated from bond energies, many of which have been found experimentally and recorded in charts. 4.3 Energy Balance

50. p. 156 Bond Energies • The text book contains tables of bond energies, such as the one on page 419, or the small one here (from page 156). • Each bond energy can represent the energy required to break a bond, or the energy released if a bond is formed (per mole). • The energy absorbed while breaking bonds is represented by a positive number, and the energy released by a negative number, • so breaking the bond between two hydrogen atoms would require +436 kJ of energy per mole of atoms. • If you formed new H—H bonds, you would release energy. (–436 kJ/mol) Also see p. 419