Warm- up • Get a device and please put your phone in the holder • Log on to eClass and check the agenda • Fill in the notes from the back as you go through the Nearpod Thermochemistry notes The chemistry of Heat & Energy Part 1 What is Energy? Energy is the ability to do work or produce heat. What is Energy? Energy is the ability to do work or produce heat. Energy exists in 3 forms • potential energy • kinetic energy • in the form of heat Potential Energy “stored” energy that has the potential to do work; this is also energy due to the position of an object Kinetic Energy • the energy of motion *Most energetic! • Which particles have the most kinetic energy? Law of Conservation of Energy • Energy cannot be created nor destroyed only rearranged (transferred). • Energy can be converted from one form to another Energy Heat (q) is energy that is in the process of flowing from a warmer object to a colder object Heat can flow in one of two directions: Exothermic To give off heat; energy is lost (-q) Endothermic To absorb heat; energy is added (+q) Examples of Exothermic Reactions Burning Fireworks Examples of Endothermic Processes melting Evaporating Temperature & Heat Temperature is related to the average kinetic energy of the particles in a substance. Cold Hot Measuring Temperature Temperature may be measured using Celsius or Kelvin. Celsius Kelvin Part 2 Measuring Heat • A unit for measuring heat is the calorie (cal) • calorie -the amount of heat needed to raise 1 g of water 1°Celsius 1 Calorie = 1000 cal A fruit and oatmeal bar contains 142 Calories. Convert this energy to calories. 142 Cal 1000 cal 1 Cal 1 Cal = 1000 cal =142,000 cal Joules The SI unit for measuring heat and energy is the Joule (J) 1 cal = 4.184 J Burning a peanut releases 30 Calories a) how many calories are released? 30 Cal 1000 cal = 30,000cal 1 Cal b) how many Joules are released? Burning a peanut releases 30 Calories a) how many calories are released? 30 Cal 1000 cal = 30,000cal 1 Cal b) how many Joules are released? 30000 cal 4.184 J 1 cal =125,520 J Part 3 Storing Heat The following slide shows a graph of different substances being heated for up to 60 minutes.. After 30 minutes, the metal temp is ~45oC. Storing Heat Different substances have different capacities for storing heat energy (Specific Heat) 0.6 J/g°C 4.184 J/g°C 1.0 J/g°C Specific Heat (c) The amount of heat needed to raise the temp of 1g of a substance by 1°C is called specific heat. Why does water have such a high specific heat? water metal • Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. • Metals have weak bonds and do not need as much energy to break them. How to calculate changes in heat energy Heat calculations Use the formula: q = mc△T q m c △T How to calculate changes in heat energy q = mc△T Heat lost or gained (J) q =_____________________________ mass of the substance (g) m =_____________________________ change in temperature (Tf – Ti) △T = ___________________________ specific heat of substance (j/g ◦C) C = ____________________________ Warm up-Practice Problem 1 on page 7 in the notes How much heat is transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm? q = m x c x △T specific heat of iron is 0.449 J/goC. Heat calculations Use the formula: q = mc△T q m c △T Practice Problem 1 How much heat is transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm? q = m x c x △T q = 5.00 x 0.449 x (100-20) q = 179.6 J specific heat of iron is 0.449 J/goC. Practice Problem 2 How many joules are required to lower the temperature of 100.0 g of iron from 75.0 oC to 25.0 oC? q = m x c x △T q = 100 x 0.449 x (25-75) q = - 2245 J specific heat of iron is 0.449 J/goC. Practice Problem 3 How many calories of energy are given off to lower the temperature of 100.0 g of iron from 150.0 oC to 35.0 oC? q = m c△T q = 100 x 0.449 x (35-150) q = -5163.5 J specific heat of iron is 0.449 J/goC. Solution: Problem 4 How much heat is transferred to a 10.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm? q = m x c x △T specific heat of iron is 0.449 J/goC. q = 359.2 J Calorimetry and Specific Heat The amount of calories contained in a food sample can be determined using a calorimeter. What is a calorimeter? • Device used to measure the heat given off during chemical reactions The specific heat of liquid water is 4.184 0 J/g C We can use this known information to calculate the specific heat of other substances Assuming that all of the heat from a burning peanut is absorbed by 50ml (50g) of water in a calorimeter, calculate the amount of calories in the peanut if the water temp increased from 24˚C to 64˚C. (specific heat of water = 4.184 J/g˚C) q = m c△T 1 cal = 4.184J q = ? (this is the unknown in this question) m= c= ΔT= 1 cal 4.184 J Assuming that all of the heat from a burning peanut is absorbed by 50ml (50g) of water in a calorimeter, calculate the amount of calories in the peanut if the water temp increased from 24˚C to 64˚C. (specific heat of water = 4.184 J/g˚C) q = m c△T 1 cal = 4.184J q = ? (this is the unknown in this question) m = 50 g c= 4.184 J/g0C ΔT= (64 - 24) = 400C 8368 J 1 cal 4.184 J = 2000 cal THIS 33- Jayson 34- Emmanuel IS THE FRONT 1- Jeremiah 9- Aaron 17- Benjamin 25- Ajdin 2- Alex 10- Hau 18- Jorge 26- Jade 3- Mikai 11- Cheyla 19- Veronica 27- Dy-Mond 4- Ciara 12- Leala 20- Kyle 28- Camden 5- Joseph 13- Amira 21- Jenaisah 29- Rachel 6- Angelina 14- Rose 22- Arrington 30- Adrian 7- Brian 15- Robert 23- 31- McKayla’ 8- 16- 24- 32- THIS 33- Jamien 34- Jaylen IS THE FRONT 1- Hailey 9- Presley 17- Trey 25- Braden 2- Adbullah 10- Bomani 18- Braxton B. 26- Sofia H. 3- Juan 11- Bryan 19- Solomon 27- Liseth 4- Sofia O. 12- Blair 20- Mariah 28- Esthefany 5- Akeelah 13- Darren 21- Maya 29- Brianne 6- Ethan 14- Braxton C. 22- Alex 30- Kace 7- 15- 23- 31- 8- 16- 24- 32- THIS 33- Shileah 34- Jade IS THE FRONT 1- Johnnie 9- Jay 17- Edwin 25- Alex 2- Crysta 10- Christopher 18- Collin 26- Isaac 3- Silas 11- Lauryn 19- Journey 27- Asia 4- Abbie 12- Kaytlynne 20- Olivia 28- Amber 5- Nathan 13- Michelle 21- Nathalie 29- Carlos 6- Nalani 14- Ashley 22- Marc 30- Matthew 7- Bukunmi 15- Deborah 23- Harrison 31- Daniel 8- Maithee 16- Suraj 24-Jaelyn 32- Mekhi THIS 33- Taiba 34- Elizabeth IS THE FRONT 1- Desmond 9- Sean 17- Leyah 25- Heydi 2- Brandon 10- Maggy 18- DeAndre’ 26- Mia 3- Yasiry 11- Giancarlo 19- Adam 27- Xaria 4- Meer 12- Bella 20- Jewel 28- Carlito 5- Jazmine 13- Debbie 21- Sam 29- Omari 6- Emely 14- Dante 22- Rhianna 30- Kelton 7- 15- 23- Mighela 31- Roderick 8- 16- 24- 32- Katie THIS 33- 34- IS THE FRONT 1- Abby Lomas 9- Sekai 17- Corey 2- Hezekiah 10- Helen 18- Emmanuelle 26- Efren 3- Walt 11- Dontay 19- La’Aysa 27- Ja’Kyla 4- Abigail 12- Kathy 20- Aliya 28- Mariah 5- Nick 13- Will 21- Valza 29- Kele 6- Ava 14- Robert 22- Toby 30- America 7- Sean 15- 23- Kamari 31- Morgan 8- 16- 24- 32- 25- Arianna How much heat is in your snack •Purpose • To calculate how many calories a Cheeto has •Safety • Goggles • It’s fire, be careful •When you are ready for fire, raise your hand Clean up, clean up, everybody clean up •Empty the soda cans •Throw away any leftover Cheeto •Clean up any crumbs •Clean up any spills Calculations •Flaming Hot Cheeto Puffs •0.44 Cal/g •Regular Cheeto Puffs •0.44 Cal/g A fruit and oatmeal bar contains 142 Calories. Convert this energy to calories. 142 Cal 1000 cal 1 Cal 1 Cal = 1000 cal =142,000 cal Burning a peanut releases 30 Calories a) how many calories are released? 30 Cal 1000 cal = 30,000cal 1 Cal b) how many Joules are released? Burning a peanut releases 30 Calories a) how many calories are released? 30 Cal 1000 cal = 30,000cal 1 Cal b) how many Joules are released? 30000 cal 4.184 J 1 cal =125,520 J Practice Problem 3 How many joules of energy are given off to lower the temperature of 100.0 g of iron from 150.0 oC to 35.0 oC? q = m c△T specific heat of iron is 0.449 J/goC. Practice Problem 3 How many calories of energy are given off to lower the temperature of 100.0 g of iron from 150.0 oC to 35.0 oC? q = m c△T q = 100 x 0.449 x (35-150) q = -5163.5 J specific heat of iron is 0.449 J/goC. Problem 4 How much heat is transferred to a 10.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm? q = m x c x △T specific heat of iron is 0.449 J/goC. q= Solution: Problem 4 How much heat is transferred to a 10.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm? q = m x c x △T specific heat of iron is 0.449 J/goC. q = 359.2 J Warm up • • • Finish/ turn in How much heat is in your snack? Lab Finish/ turn in WORKSHEET: Energy Units Conversions from Tuesday USE YOUR NOTES!!! Today • • Notes on Specific Heat Calculations Specific Heat Calculations google form ● You can determine the specific heat of an unknown metal by placing a heated piece of the metal in water in a calorimeter. ● Once you have measured the final temperature of both the metal and the water, you can work out the specific heat of the metal. q metal = 30.5 x c x -61.9 q water = 50.0 x 4.184 x 9.1 q metal = q water 1) A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. How much heat was given off by the lead? What is the specific heat of lead? First, find the q for water q = m c△T Hot Lead 1) A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. How much heat was given off by the lead? What is the specific heat of lead? First, find the q for water q = m c△T Hot Lead q = 100 x 4.184 x (34.4- 25.3) = 1) Then, use that (q) to calculate the (c) of Lead.A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. *Law of conservation of energy: heat given off by lead = heat transferred to water q = m c△T q = 100 x 4.184 x (34.4- 25.3) = 3807.4 J Hot Lead -3807.4 = 45.0 x c x (34.4 - 100) =c 1) Then, use that (q) to calculate the (c) of Lead.A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. *Law of conservation of energy: heat given off by lead = heat transferred to water q = m c△T q = 100 x 4.184 x (34.4- 25.3) = 3807.4 J Hot Lead -3807.4 = 45.0 x c x (34.4 - 100) 1.29 J/g0C =c 1) Then, use that (q) to calculate the (c) of Lead.A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. *Law of conservation of energy: heat given off by lead = heat transferred to water q = m c△T -3807.4 = 45.0 x c x (34.4 - 100) -3807.4 = -2952 x c Hot Lead -2952 -2952 1.29 J/g0C =c A 17.9 g sample of unknown metal was heated to 48.31 oC. It was then added to 28.05 g of water in an insulated cup. The water temperature rose from 21.04 oC to 23.98oC. What is the specific heat of the metal in J/goC? q = m c△T First find the heat given off from the metal to the water q = 28.05 x 4.184 x (23.98- 21.04) = 345 J Second, find the specific heat (c) of the metal. -345 = 17.9 x c x (23.98- 48.31) 0.792 J/g°C = c Practice The temperature of a piece of copper with a mass of 95.4g increases from 25oC to 48 oC when the metal absorbs 849J of heat. What is the specific heat of copper in J/goC? q = m c△T 849 = 95.4 x c x (48-25) 849 = c = 849 (95.4 x 23) 2194.2 0.387 J/g°C = c Remember: (-q) = Exothermic = releases energy (+q) = Endothermic = absorbs energy Today • Grab a device • Go to google classroom for the google form • Follow the instructions carefully Warm up • Matter is made up of particles that could be in the solid, liquid or gas phase. Read the descriptions and identify the phases of matter (what phase is 1, 2, and 3)? Today • • Heat and Phase Changes Notes Phase Changes and Heating Curves google form Heat & Phase Changes You will want a calculator. (click here) Remember: (-q) = Exothermic = releases energy (+q) = Endothermic = absorbs energy Kinetic Theory of Matter An explanation of how particles in matter behave. There are three basic assumptions to the kinetic theory: 1. All matter is composed of small particles (atoms, molecules, and ions). 2. These particles are in constant, random motion. 3. These particles are colliding with each other and the walls of their container. Kinetic Theory-Solids • Particles of a solid are tightly packed vibrating in a fixed position. • Solids have a definite shape and volume. • Solids do not change shape (volume) without outside forces Kinetic Theory-Liquids • Particles of liquids are not as tightly packed as solids, they are able to to slide over one another. • Liquids have indefinite shape and definite volume • Liquids will spread out, but will not fill any container Kinetic Theory-Gases • Particles of gases are very far apart and move freely. • Gases do not have a definite volume or shape. • Gases will completely fill any container Warm up Matter is made up of particles that could be in the solid, liquid or gas phase. Read the descriptions and identify the phases of matter (what phase is 1, 2, and 2- Liquid;3)?3- Gas • 1- Solid; Plasma Endothermic PHASE CHANGES Description of Phase Change Solid to liquid Liquid to gas Solid to gas Term for Phase Change Heat Movement During Phase Change Melting Heat goes into the solid as it melts. (endothermic) Evaporation Heat goes into the liquid as it vaporizes (endo) Sublimation Heat goes into the solid as it sublimates (endo) Exothermic PHASE CHANGES Description of Phase Change Term for Phase Change Condensation Gas to liquid Liquid to solid Freezing Heat Movement During Phase Change Heat leaves the gas as it condenses (exo) Heat leaves the liquid as it freezes (exothermic) What is happening -As temperature increases/ heat is added, intermolecular forces are being broken Heating Curves A heating curve shows the temperature change as thermal energy, or heat, is added to a substance Heating curve video Heating curve of water (click here) Which region(s) represent kinetic energy (temperature) of the substance increasing?1, 3 and 5 Which region(s) represent kinetic energy of the substance staying the same?___________ Does the graph represent an endothermic or exothermic process? 2 and 4 represent potential energy increasing. There is no increase in temperature Which region(s) represent kinetic energy (temperature) of the substance increasing?1, 3 and 5 Which region(s) represent kinetic energy of the substance staying the same?2 and 4 Does the graph represent an endothermic or exothermic process? Which region(s) represent kinetic energy (temperature) of the substance increasing?1, 3 and 5 Which region(s) represent kinetic energy of the substance staying the same?2 and 4 Does the graph represent an endothermic or exothermic process? Heating Curve for Water 1. What is happening to water at each letter? A. ____________ B. ____________ C. ____________ D. ____________ E. ____________ Heating Curve for Water 1. What is happening to water at each letter? (ice) A. Solid ____________ B. Melting ____________ (water) C.Liquid ____________ D. ____________ Boiling E. Gas ____________ (steam) Assignment • Is on google classroom • Phase Changes and Heating Curves google form Heating Curves To calculate the energy changes as a substance warms or cools the equation q=mCΔT is used. Ex: Calculate the heat required to pass through region A-B if 20g of ice was being heated. (specific heat ice = 2.09J/g°C) q = mc∆T = 20 x 2.09 x ( 0 - -25) = 1045J Calculate the heat required to pass through region C-D if 20 g of water was being heated. (specific heat liquid water = 4.184 J/g°C) q = mc∆T = 20 x 4.184 x ( 100 - 0) = 8368J Calculate the heat required to pass through region E-F if 20 g of steam was being heated. (specific heat water vapor= 2.02 J/g°C) q = mc∆T = 20 x 2.02 x ( 125 - 100) = 1263J Enthalpy and Hess’ Law you will want a calculator (click here) System & Surroundings • Scientists study heat changes in systems and surroundings. • A system is the part of the universe we are studying. • The surroundings are everything else outside of the system • System + surroundings = the universe System and Surroundings Beaker & solution Table & air Heat is Exchanged Between the System & the Surroundings Energy out Energy in Enthalpy (H) • Heat transferred from a system to its surroundings during a chemical reaction at constant pressure is called Enthalpy (H) . • Can’t measure H, only ΔH • At constant pressure, ΔH = q = mCΔT What is Enthalpy? Enthalpy- a measure of heat-flow within a close system -can only be measured as a change, not a state of being. -noted as ΔH ΔH = q ΔH = mcΔT **this only works in a closed system Measuring Enthalpy (ΔH) If heat leaves system and goes into surroundings, and it feels hot to the touch ΔH = - (exothermic) Measuring Enthalpy (ΔH) If heat goes from surroundings into system and it feels cool to the touch ΔH = + (endothermic) Heat of Formation Values (ΔHf ) • Every substance has an energy associated with it called ΔHf (ENTHALPY OF FORMATION). • REMEMBER to make sure the substance you look up also have the correct states of matter. Enthalpies of Reaction • All reactions have some ΔH associated with it 2H2(g) + O2(g) → 2H2O(l) ΔH = - 572 kJ • How did we calculate that ΔH? ΔHrxn = (sum of ΔHproducts) - (sum of the ΔHreactants) Enthalpies of Reaction • How do we know the ΔH’s of the products and reactants? ΔHrxn = (sum of ΔHproducts) - (sum of the ΔHreactants) First: elements have a ΔH of zero (0). Second: we use a table of known values for compounds Enthalpies of Reaction • All reactions have some ΔH associated with it 2H2(g) + O2(g) → 2H2O(l) • ΔH H2 = 0 kJ/mol • ΔH O2 = 0 kJ/mol • ΔH H2O = -285.83 kJ/mol ΔH = - 572 kJ ΔHrxn = (sum of ΔHproducts) - (sum of the ΔHreactants) BUT!! It is 2H2O, so: • • ΔH = (2*-285.83) - (2*0 + 0) ΔH = -572 kJ Hess’ Law Hess’s Law: If you add two or more chemical equations to get an overall equation, then you can also add the heat changes (ΔHs) to get the overall heat change. Hess Law Example • 3A + 3B 🡪 4C + D • 2A + 2B 🡪 2C • A + B 🡪 2C + D • 3A + 3B 🡪 4C + D ΔH = ???? ΔH = -594kJ ΔH = 198kJ ΔH = -396kJ • To calculate the heat of reaction we add all the Enthalpy changes (ΔH) for all of the reactions. How to add chemical equations: 1. If two identical substances are on opposite sides of the arrow, they will cancel (reduce). 2. If two identical substances are on same side of the arrow, add the coefficients together. 3. Keep substances on the same side of the arrow in the final equation. Example: Add the following equations. C + O2 🡪 CO2 CO2 🡪 CO + ½ O2 Example: Add the following equations. C + O2 🡪 CO2 CO2 🡪 CO + ½ O2 Example: Add the following equations. C + O2 🡪 CO2 CO2 🡪 CO + ½ O2 C + ½O2 🡪 CO Example: Add the following equations. 2 Cu + O2 🡪 2 CuO 4 Cu + O2 🡪 2 Cu2O 6Cu + 2O2 🡪 2CuO + 2Cu2O Hess’ Law • Basically: if we can add the reactions together, then we can add the ΔH’s • A + 2B → C + D • B → 2C + D • A + 3B → 3C + 2D • ΔH= -20 kJ ΔH= -45 kJ ΔH = ? Practice • A + 2B → C + D • B → 2C + D • A + 3B → 3C + 2D H= -20 kJ H= -45 kJ ΔH = -65 kJ Practice • 2A + 5B → C + D • A→C+B • 3A + 4B → 2C + D H= -237 kJ H= 200 kJ ΔH = -37 kJ C2H4 🡪 2 C + 2 H2 2 C + 3 H2 🡪 C2H6 ΔH= 152kJ ΔH= -85kJ C2H4 + H2 → C2H6 ΔH = 67 kJ End of material -ticket-out-of-the-door -Asynchronous assignment tomorrow -Test for Unit 5- Thermo/Kinetics is May 6th