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1 Energy SECTION 1: THERMOCHEMISTRY 2 Thermochemistry Thermochemistry – thermo = energy/heat Energy changes occur during reactions and phases changes Many reactions release energy making the surrounding hotter. Combustion gives off energy as heat and light. Some reactions absorb energy. Their surroundings feel colder. This gas tube gets cold as liquid under pressure changes to gas. Energy is the ability to do work or transfer heat. Energy used to cause an object that has mass to move is called work. Energy used to cause the temperature of an object to rise is called heat. (energy to make molecules move) Thermochemistry Forms of Energy Kinetic energy is energy an object possesses by virtue of its motion: 1 Ek = mv2 2 Notice energy is a function of an object’s velocity (speed) and its mass Kinetic Energy is gained by an object when it is accelerating Thermochemistry © 2012 Pearson Education, Inc. Which has more kinetic energy: A slow moving locomotive, A fastball, Or a bullet in flight? Which would do more damage if it ran into you? That’s Kinetic energy! Thermochemistry © 2012 Pearson Education, Inc. 6 Forms of energy Kinetic Mechanical – motion of object Electricity – electric current flow of electrons Light (radiant) – light waves Chemical Kinetic: Heat – molecular motion (speed energy – indicated by temperature) Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. The energy always comes from attraction (like gravity in cartoon above) Or repulsion (the coils of a compressed spring) Thermochemistry 8 Potential Mechanical – ex: compressed spring Electric (“potential”)? Stored in a battery, static electricity Chemical energy (potential) when chemical bonds and IM forces are formed or broken. 9 Energy conversions - energy is transferred and always changing forms Ex: In a car: gasoline’s energy (chemical PE) is converted into motion (mechanical KE) Light bulb: electrical energy converted into light (and heat – thermal) energy Easy bake oven heats with a light bulb. (fire hazard!) 10 Law of conservation of energy Energy lost = Energy gained Efficiency: most conversions Lose some energy as waste heat Typical Light bulb = 10% efficient 10% light 90% heat Compact fluorescent = 60 to 100% efficient LED lights: 5 x less energy use than compact fluorescent Light emitted in all directions makes them less efficient 11 Heat (kinetic energy) – two definitions 1) Total motion energy of all molecules in a sample Depends on Temperature (average motion energy of a molecule) And Sample size (number of molecules present) Which has more heat in it? Lake in summer Boiling water (Hint: Which has more molecules?) This type of heat energy can not be measured easily 2) Heat is the Energy transferred (lost or gained) when a sample undergoes a change in form (PE) Which way does it flow? Ex: Ice is placed in warm water. Water: temp decreases, KE is lost Heat is transferred from the water to melt the ice Ice: solid becomes liquid PE is gained Does water gain “Cold” from the ice? Ice is a heat sucker! 12 13 1 scoop of NaOH in 50 mL of H2O Temp changes o From 100C to 200C Δtemp = 10 C 1 scoop of NaOH 2 scoop of NaOH in 100 mL of H2O in 100 mL of H2O Temp changes o From 100C to 150C Temp changes o From 100C to 200C Δtemp = 5 C Δtemp = 10 C Can you explain the difference? Hint 1 : Potential energy is released when bonds form between molecules Hint 2: Kinetic energy is absorbed by the water in the surrounding solution How do the three trials compare in terms of heat? Why did trial 2 cause half the temperature change as trial 1? Why did trial 3 cause twice the temperature change as tial 2? 14 Energy flow Kinetic energy always flows “down hill” (higher to lower KE) Hotter substances lose heat to cooler substances Energy flows from higher temp to lower temp – until both samples at same final temperature. Energy Transformations: Potential energy (change in form) 15 energy changes form: is converted into Kinetic energy (change in Temp) Endothermic changes – absorbing heat MELTING ICE (COLD) Phase change Gains PE heat flows in WARM WATER temp decreases Loses KE Endothermic Changes - reactions or phase changes, which absorb heat (KE PE) Evidence The Surroundings get colder Energy Transformations: 16 energy changes form: is converted into Potential energy (change in form) Kinetic energy (change in Temp) Endothermic changes Boiling (vaporizing) Phase change Gains PE Ex: boiling/vaporizing heat added from outside heat flows in frost CO2 Tank WARM metal temp decreases Loses KE Expanding vaporizing CO2 absorbs heat from the surroundings. frost forms on the cold metal Energy Transformations: Potential energy (change in form) 17 energy changes form: is converted into Kinetic energy (change in Temp) Endothermic changes Dissolving Phase change Gains PE Ex: Also: Dissolving of most salts (ex: cold pack) heat flows in WARM water temp decreases Loses KE Salt crystals break apart as they dissolve - absorbing heat from the surrounding pack making it cold. Energy Transformations: Potential energy (change in form) 18 energy changes form: is converted into Kinetic energy (change in Temp) Exothermic changes – release heat Ex: BURNING (combustion) Chemical change Loses PE heat flows out Exothermic Changes reactions or phase changes, which release heat (PE while KE ) AIR temp increases gains KE * Evidence – surroundings get warmer Energy Transformations: Potential energy (change in form) 19 energy changes form: is converted into Kinetic energy (change in Temp) Exothermic changes – release heat Ex: Freezing Physical change Loses PE Ex: Freezing (solidifying) – heat removed to freeze a liquid heat flows out colder AIR temp increases gains KE Solid ice has less PE than liquid water Energy Transformations: Potential energy (change in form) 20 energy changes form: is converted into Kinetic energy (change in Temp) Exothermic changes – release heat Ex: Dissolving Chemical change Loses PE heat flows out Ex: Dissolving NaOH – water gains the heat water temp increases gains KE Water forms bonds with the dissolved salt releasing energy Water gets hot! Energy Transformations: Potential energy (change in form) 21 energy changes form: is converted into Kinetic energy (change in Temp) Law of conservation of energy – Energy lost by one substance = energy gained by other substance 22 GAS (highest PE) Energy absorbed LIQUID Endo-thermic Energy Released Exo-thermic SOLID (lowest PE) 23 Measuring Energy Section 2A Calorimetry 24 Measuring energy : Heat Capacity Why does sand get hot while water stays cool? Water has a greater heat capacity! …than sand. 25 Measuring energy: Heat Capacity Is the Energy required to raise the temperature of a substance (make molecules speed up) The same amount of heat is given to two different compounds Ethanol’s temperature rises to 62oC While water’s only rises to 40oC Which has a greater heat capacity? (ability to absorb and hold heat) 26 Measuring energy: Heat Capacity Is the Energy required to raise the temperature of a substance (make molecules speed up) The same amount of heat is given to two different compounds Water’s heat capacity is higher. Ethanol’s heat capacity is lower. It takes more heat to raise its temperature It takes less heat to raise its temperature Which has a greater heat capacity? (ability to absorb and hold heat) 27 Measuring energy: Heat Capacity Is the Energy required to raise the temperature of a substance (make molecules speed up) Its also the heat that must be removed to cool a substance. Heat capacity depends on the Intermolecular (sticky) forces Water has stronger hydrogen bonding Ethanol’s sticky forces are weaker 28 Specific heat capacity (C) – is a physical constant It’s the Heat required to raise 1.0 grams of a substance by just 1o C Water is commonly the standard 1 calorie = heat to raise 1 gram of water by 1oc 1000 calories = 1 kilocalorie = 1 “Calorie” (notice the capitol C?) 1 Joule (J) – is the metric unit of energy 4.18 joules = 1 calorie Water is the Standard! Example problem 1: How many calories of heat are needed to raise 10 grams of water by 10 oc? the heat capacity of water (specific heat) : …is 1 calorie per gram per 0c so 1 calorie x 10 g water x 10oc = 100 calories Formula: Q = C m ∆t Heat = (specific heat capacity) X (mass) X ( change in temp oc) 29 30 Ex 2: What is the specific heat C of a substance which absorbs 20. J of heat to raise 5 grams of the substance by 2 oC? Q = C m ∆t Substitute: (20 joules) = C (5.0 g)(2.0 oC) and solve: 20. J (5.0 g)(2.0 oC) =C C = 2.0 joules / goC Are the IM forces in this substance stronger or weaker than in liquid water? 31 Ex 2: What is the specific heat C of a substance which absorbs 20. J of heat to raise 5 grams of the substance by 2 oC? Q = C m ∆t Substitute: (20 joules) = C (5.0 g)(2.0 oC) and solve: 20. J (5.0 g)(2.0 oC) =C C = 2.0 joules / goC Weaker, water’s specific heat is higher: 4.18 Joules/goC 32 Reference Table Formulas Heat q= mC∆t C for water Try one: Ex 3: how many joules are absorbed when 300. grams of water changes from 250C to 390C? Write the formula: Substitute values: Solve: Ans: 18000 Joules? 33 34 Ex 4: 100 grams of water at 250C loses 4200 joules of energy. What is its final temperature? Write the formula: Q = C m ∆t ∆t = Q Cm Substitute values: (4200 j )= (4.2 j/g0c) (100 g) ∆t Solve: ∆t Remember: equation only allows you to calculate ∆t = (4200 j) (4.2 j/g0c) (100 g) ∆t = 100c Final temp = 250C – 100C = 150C Original water temperature decreases by 10oC 35 Try these: 1. What quantity of heat is absorbed by 35 grams of water if its temperature increases by 100C? 1463 Joules 2. What is the specific heat of a substance which absorbs 2680 joules to change 85g from 260C to 470C? 1.50 Joules per gram per o Celsius 3. What is the final temperature when 35 grams of water at 150C absorbs 4630 j ? Δtemp = 32oC, final temp = 47oC 36 Calorimetry: Calculating heat of a change Suppose that a sample of wax is melted in boiling water How much heat is required to melt the wax? This is called the heat of fusion (heat to melt) Ex: 1 Heat of fusion of a solid (Hf) How much heat is absorbed per gram, if 10 grams of a solid melts in 300 grams of water, causing the water temperature to decrease by 500C? 37 Calorimetry: Calculating heat of a change Ex: 1 Heat of fusion of a solid (Hf) How much heat is absorbed per gram, if 10 grams of a solid melts in 300 grams of water, causing the water temperature to decrease by 500C? Step 1: how much heat did the water lose? (heat gained by the solid to melt) For the water heat lost: Q = C m ∆t For the water heat lost: Q = (4.18Jg-oc-)(300 g)(50oC) For the water heat lost: Q = 62700 Joules 38 Calorimetry: Calculating heat of a change Ex: 1 Heat of fusion of a solid (Hf) How much heat is absorbed per gram, if 10 grams of a solid melts in 300 grams of water, causing the water temperature to decrease by 500C? Step 2: how much heat did the solid gain per gram? (heat lost by the water) Heat gained by the wax IS the heat lost from the water Q = 62700 Joules lost by water = 62700 Joules gained by the wax Heat gained per gram of wax is 62700 J of heat 10 grams of wax 39 Calorimetry: Calculating heat of a change Energy released during the system (a reaction) is absorbed by the surrounding water Find the heat absorbed by the water to find the heat of reaction Measuring energy (Calorimetry) – Water is used as the standard > Heat lost or gained during a change will be exchanged with the water > Calculate the heat exchanged by the water to find out about the change 40 Ex: 2 Calculate the heat of solution for KNO3 in kJ/mole, if 10.1 grams of KNO3 is dissolved in 100g of water causing its temperature to decrease by 8.3 0C? Step one: how much heat did the water lose? (heat gained by the chemical) For the water heat lost: Q = C m ∆t For the water heat lost: Q = (4.18Jg-oc-)(100 g)(8.3oC) For the water heat lost: Q = 3500 Joules For the water heat lost: Q = 3.5 kiloJoules 41 Step two: If the sample had a mass of 10.1 grams, and Calculate the heat of solution (heat of dissolving) of the KNO3 (kJ/mole) Aka: How much heat (kJ) did each mole release? (hint 2: KNO3 = 101 g/mole 39 + 14 + 3(16) Q = 3.5 kiloJoules 0.100 moles Q = 35 kJ/mole 42 Heat of Reaction (∆H) – Reference Table I Heat of reaction – heat lost or gained for molar quantities indicated in balanced equation Ex: combustion of 2 moles of methanol releases 1452 kilojoules of heat Combustion reactions synthesis reactions Dissolving (Heat of Solution) 43 Try one: Calculate Specific heat capacity of a metal A 10 gram piece of hot metal at 1500C, is dropped into a calorimeter containing 50. g of water? The water started at 250C, and increased to 350C. Step one: how much heat did the water gain? (heat lost by the metal) Hint: Q = C m ∆t for the water Ans. 2090 Joules gained by the water 44 Try one: Calculate Specific heat capacity of a metal A 10 gram piece of hot metal at 1500C, is dropped into a calorimeter containing 50. g of water? The water started at 250C, and increased to 350C. Step two: If the metal had a mass of 10. grams, and started at 150. 0C, Calculate the specific heat (C) of the metal? Hint: Q = 2090 Joules Use Q = C m ∆t for the metal, find C Ans: 1.8 J/gram ? 45 Section 2B Heats of Reaction 46 Heat of Reaction in Equations Exothermic – reactions which release or lose energy - ∆H is negative Ex: ∆H = CH4 + 2 O2 CO2 + H2O – 890.4 kJ (For 1 mole CH4 890.4 kJ is released) In the equation: CH4 + 2 O2 CO2 + H2O + 890.4 kJ (energy is a product, written on right ) 47 Heat of Reaction in Equations Exothermic – reactions which release or lose energy - ∆H is negative Ex: 2CO + O2 2 CO2 + 566 kJ ∆H = – 566.0 kJ (Energy released for 2 moles CO) 48 Endothermic Reactions which absorb or gain energy - ∆H is positive Ex: dissolving of potassium nitrate: KNO3(s) K+(aq) + NO3–(aq) KNO3(s) + ∆H = +34.89 kJ 34.89 kJ K+(aq) + NO3–(aq) (Energy is a reactant – must be absorbed, written on left ) Try a few: 1. According to reference table I, how much heat is released by the combustion of 2.0 moles of propane C3H8 in a calorimeter? 2. Will the water in the calorimeter get hotter or colder? How do you know? 3. When KNO3 is dissolved in water the temperature of the water decreases. Is this change exothermic or endothermic? How could you tell by looking at reference table I? 4. Using reference table I, write the thermochemical equation for the synthesis of 2 moles of water vapor. 49 50 5. What will be the temperature change in 100. grams of water if 0.100 moles of KNO3 are dissolved in it? Step one: Find heat absorbed by the 0.100 moles of chemical during dissolving This is kiloJoules per mole Step two: Since this heat is absorbed from the water, find the temp change in the water. (Recall: Q is the heat absorbed in step one!) Q = C m ∆t ∆t = Q Cm ∆t = 8.35 0C 6. 8.0 grams of methane undergoes combustion in a calorimeter containing 5000. grams of water. The reaction proceeds according to the equation as shown on reference table I. CH4 + O2 CO2 + H2O + 890.4 kJ 1. What will happen to the temperature of water in the calorimeter? Explain. 2. How much heat is released by the complete combustion of the 8.0 grams of methane? 3. What will be the change in temperature of the water in the calorimeter? 51 52 53 54 Section 2C Energy of Phase Changes Energy in Phase changes Q=CmΔT Energy Gained to melt Energy Gained to boil 55 56 Heat of fusion (melting) / heat of vaporization (boiling) Heat of fusion – Heat absorbed to melt a solid (Joules per gram) Q = Hf m Q = heat absorbed Hf = heat of fusion (joules/g) m= mass of sample for water (Ice) Hf = 334 J/g (see ref. Tbl. B) Ex1: How much heat is absorbed to melt 100 grams of ice? Q = Hf m Q = (334J/g) (100 g) = 33400 Joules = 33.4 kJ 57 Ex2: What is the heat of fusion of a solid if a 50 gram sample absorbs 10 000 joules to melt? Q = Hf m Hf = 10 000 J 50 g Hf = 200 J/g and Hf = Q m Heat of vaporization – Heat absorbed to boil water (or lost as a gas condenses) For water Hv = 2260 Joules per gram Ex3: What mass of water can be vaporized by the addition of 100 Kilojoules of heat? Q = Hv m where Q is the heat absorbed Hv is the heat of vaporization (2260 J/gram) m is the mass of the water in grams Step 1: convert Kilojoules to joules 100 kJ x 1000 J = 1 KJ Step 2: setup and calculate Q = Hv m and m= m = 100 000 J 2260 J/g m= 44.2 g 100 000 Joules _Q_ Hv 58 Learning check: show work 1. How much heat is absorbed when 20. grams of ice melts in a calorimeter? 6680 joules 2. If the calorimeter contains 100. grams of water initially at 250C, what is the final temperature of the water once all the ice is melted? 9 oC 3. What is the heat of vaporization for a liquid if it takes 5000 joules of energy to vaporize 2.5 grams? 2000 J/g 4. Aluminum oxide decomposes according to the balanced equation 2 Al2O3 4 Al + 3 O2 According to reference table I, what is the ∆H for the decomposition of 1.0 mole of Al2O3? +1680 kJ 59 60 61 62 63 64 In terms of energy flow, explain why the temperature of the water in the calorimeter used in the investigation increases. Show a correct numerical setup for calculating the number of joules of heat gained by the water in the investigation. In the investigation shown, the change in heat of the copper is greater than the change in heat of the water. What error could account for this apparent violation of the law of conservation of energy? [do not use human error as part of the answer]