Presentation on theme: "Energy and Changes in Matter"— Presentation transcript:
1Energy and Changes in Matter Today’s ObjectivesDifferentiate between heat and temperature.Interpret the heating and cooling curves of various substances.
2Heat versus Temperature DefinitionEnergy Transfer between Substances, Related to the Total Kinetic EnergyA Measure of the Average Kinetic Energy of a SubstanceUnits of MeasurementJoules (J)Calories (cal)Celsius (oC)Kelvin (K)How to Measure?IndirectlyDirectly
3Match versus Ice Sculpture Which Has a Higher Temperature? A Match Which Has More Heat Energy? The Ice Sculpture
4Heat Heat always flows from high temperatures to low temperatures. Heat flows from the fire to the marshmallow.Heat always flows from high temperatures to low temperatures.
5Units of Measurement for Heat 1 cal = J 1000 cal = 1 Cal How many calories are in 54.0J?
6Heat and Physical Changes Add or Remove HeatChange in Kinetic Energy of the SubstanceChange in Potential Energy of the SubstanceChange in TemperatureChange in Phase
7Heating Curve Boiling Point Vaporization PE PE SolidLiquidGasVaporizationMeltingKEPEBoiling PointMeltingPointPEPotential Energy Changing with Phase ChangesKEKinetic Energy Changing with Temperature Changes
8Energy and Changes in Matter Today’s ObjectivePerform calculations related to heat and changes in temperature or phase.
9DT = Change in Temperature = Tfinal - Tinitial Calculating HeatTo calculate the amount of heat energy when temperature changes,Q = m Cp DTQ = heatm = massCp = specific heatDT = Change in Temperature = Tfinal - Tinitial
10What is Specific Heat (Cp)? Specific heat is the amount of energy required to change the temperature of a substance by 1oC. A substance with a high specific heat requires more energy to change its temperature than a substance with a low specific heat. Low Specific Heat = Good Conductors High Specific Heat = Good Insulators
11Examples Un-Popable Balloon Which balloon will pop first? One with Only AirOne with Air and WaterWhich substance is the best conductor?CopperGoldAir
12Q > 0 = heat is being absorbed. Example Problem #1How much energy is required to raise a 34.0g sample of copper metal from 20.0oC to 45.0oC? The specific heat of copper is Q = m Cp DT Q = (34.0g)(0.385 ) (45.0oC -20.0oC) = (34.0g)(0.385 ) (25.0oC ) = 327 JJg oC“to” – “from”Jg oCJg oCQ > 0 = heat is being absorbed.
13Q < 0 = heat is being released. Example Problem #2How much heat is released when a 3.20g sample of water is cooled from 83.0oC to 54.0oC? The specific heat of water is Q = m Cp DT Q = (3.20g)(4.18 ) (54.0oC -83.0oC) = (3.20g)(4.18 ) (-29.0oC ) = -388 JJg oC“to” – “from”Jg oCJg oCQ < 0 = heat is being released.
14Heat of Vaporization (Hv) Calculating HeatTo calculate the amount of heat energy when phase changes,Q = mHQ = heat m = massPhase ChangeHValue for HMeltingHeat of Fusion (Hf)+HfFreezing-HfVaporizationHeat of Vaporization (Hv)+HvCondensation-Hv
15Q > 0 = endothermic phase changes Example Problem #3Use heat of fusion!How much heat is needed to melt 56.2g of ice at 0oC? Q = mHf Q = (56.2g)(334 ) = 18,800 JJgQ > 0 = endothermic phase changes
16Example Problem #4How much heat is released when 120g of steam condenses? Q = mHV Q = (120g) (-2260 ) = -270,000 JUse a negative heat of vaporization!JgQ < 0 = exothermic phase changes
17Important Reminders When the Temperature Changes Q = mCpDT When the Phase ChangesQ = mHDT = Temperature You Are Going To – Temperature You are Coming FromPhase ChangeHValue for HMeltingHeat of Fusion (Hf)+HfFreezing-HfVaporizationHeat of Vaporization (Hv)+HvCondensation-Hv
18Energy and Changes in Matter Today’s ObjectiveDetermine experimentally the heat of fusion of water.
19Important Reminders about Your Experiment Make sure that you always have ice in your calorimeter at ALL times!Your constant temperature should be between -4oC and 4oC.If you spill the water, you will have to re-do the lab.This is especially important at the end when you are measuring the volume.Your volume will exceed 100mL!For your calculations, only use the accepted value for the heat of fusion when you are calculating percent error.
20Energy and Changes in Matter Today’s ObjectivesSolve problems related to heat and physical changes.
21Example #1Determine the identity of a substance that requires 89.1J to raise 53.8g sample from 18.0oC to 22.3oC. Q = m Cp DT Cp = = Cp = = 0.385mDT(53.8g)(22.3oC-18.0oC)89.1JmDTQ(53.8g)(4.3oC)89.1JJg oCCopper
22Example #2Determine the amount of heat energy required to change 4.50g of ice from -15.0oC to water at 56.0oC. This problem is different because we are changing phase AND temperature.
23Heating Curve of Water Step 1: Ice at -15oC Ice at 0oC Q=mCpDT Step 2: Ice at 0oC Water at 0oC Q=mHfStep 3: Water at 0oC Water at 56oC Q=mCpDT
24Step 1: Ice at -15oC Ice at 0oC Q=mCpDT Example #2 - ContinuedStep 1: Ice at -15oC Ice at 0oC Q=mCpDTQ = m Cp DT Q = (4.5g)(2.05 )(0oC oC) Q = (4.5g)(2.02 )(15oC) Q = 138 JJg oCJg oC
25Step 2: Ice at 0oC Water at 0oC Q=mHf Example #2 - ContinuedStep 2: Ice at 0oC Water at 0oC Q=mHfQ = m Hf Q = (4.5g)(334 ) Q = JJg
26Step 3: Water at 0oC Water at 56oC Q=mCpDT Example #2 - ContinuedStep 3: Water at 0oC Water at 56oC Q=mCpDTQ = m Cp DT Q = (4.5g)(4.18 )(56oC - 0oC) Q = (4.5g)(4.18 )(56oC) Q = 1050 JJg oCJg oC
27Example #2 - ContinuedDetermine the amount of heat energy required to change 4.50g of ice from -15.0oC to water at 56.0oC. Calculate the total amount of heat. Step 1 Q = 138 J Step 2 Q = J Step 3 Q = 1050J103, 000J
28Example #3How much energy is released when 4.00g of steam at 110.0oC is cooled and condensed to form water at 90.0oC? Step 1: Steam at 110oC to 100oC Step 2: Steam Condenses to Water Step 3: Water at 100oC to 90oC
29Example #3 - ContinuedStep 1: Steam at 110oC to 100oC Q = mCpDT Q = (4.00g)(2.02 )(100oC - 110oC) Q = J Step 2: Steam Condenses to Water Q = mHv Q = (4.00g)(-2260 ) Q = JJg oCJg
30Example #3 - ContinuedStep 3: Water at 100oC to 90oC Q = mCpDT Q = (4.00g)(4.18 )(90oC - 100oC) Q = -167 J Total Energy Released Q = J J J Q = JJg oC
31Thermochemistry the study of the heat changes during chemical reactions, and the effects on chemical and physical processes.Today’s ObjectivesDifferentiate between exothermic and endothermic reactionsInterpret the potential energy diagrams for chemical reactions.
32Heat and Chemical Reactions In chemical reactions two things occur,The original chemical bonds are broken. This requires energy.New chemical bonds are formed. This releases energy.Because the energy it takes to break the bonds and the energy that is released when the new bonds are formed are not always equal, heat is either absorbed or released in chemical reactions.
33Heat and Chemical Reactions Endothermic ReactionsExothermic ReactionsAbsorb EnergyFeel cold to the touchHeat is treated as a reactant.Release EnergyFeel warm to the touchHeat is treated as a product.
34Potential Energy Diagram for an Endothermic Reaction (kJ)PE of ProductsHeat of Reaction(DH)PE ofReactantsPEReactants < PEProductsDH = Products – Reactants = 350kJ – 200kJ = 150 kJ
35Potential Energy Diagram for an Exothermic Reaction (kJ)PE ofReactantsHeat of Reaction(DH)PE of ProductsPEReactants > PEProductsDH = Products – Reactants = 15kJ – 40kJ = -25kJ
36Law of Conservation of Energy Any chemical or physical process does not create or destroy energy.Energy Released or Absorbed by the SystemEnergy Absorbed or Released by the Surroundings=System = What You are StudyingSurroundings = Everything Else/Generally Water
37CalorimetryHeat must be calculated indirectly. In other words, scientists use experimental means to calculate the heat energy associated with a process. These experimental means are generally referred to as calorimetry.
38Tips for Solving Calorimetry Problems Successfully Calculate how much heat is gained or lost by the water.Assume that the amount of heat gained or lost by the water is equal to the amount of heat lost or gained by the other substance or reaction.Solve for the missing value for the other substance or reaction.