these are the answers to the AP Chemistry packet that covers topics 6.1 to 6.5 on this slide you can see unit six at a glance unit six is entitled thermochemistry in this video we will take a look at topic 6.1 endothermic and exothermic processes topic 6.2 energy diagrams topic 6.3 heat transfer and thermal equilibrium topic 6.4 heat capacity and calorimetry and topic 6.5 energy of phase changes in the video description area there is a link to the AP Chemistry course and exam description which we call the CED for short and there is also a link to the packet that accompanies this video topic 6.1 endothermic and exothermic processes here are some Essential Knowledge statements from the AP Chemistry course and exam description temperature changes in a system indicate energy changes energy changes in a system can be described as endothermic and exothermic processes such as the heating or cooling of a substance phase changes or chemical Transformations when a chemical reaction occurs the energy of the system either decreases exothermic reaction increases endothermic reaction or Remains the Same for exothermic reactions the energy lost by the reacting species the system is gained by the surroundings as heat transfer from or work done by the system likewise for endothermic reactions the system gains energy from the surroundings by heat transfer to or work done on the system the formation of a solution may be an exothermic or endothermic process depending on the relative strengths of inter molecular interparticle interactions before and after the dissolution process when studying energy changes and the flow of heat it is helpful to differentiate between the system which represents the portion of the universe that we choose to focus on and the surroundings which represents everything else when a chemical reaction occurs the system is often the portion of the universe where attractive forces between particles are broken and or formed question one a sample of water at room temperature is added to a beaker and placed on a hot plate the initial temperature of the water is recorded the hot plate is turned on and the temperature of the water is monitored over time would you describe the change that occurred in this experiment as an endothermic process or an exothermic process justify your answer by indicating the direction of heat flow use the terms system and surroundings in your answer so in this experiment the heat is being transferred from the hot plate to the sample of water in the beaker this is classified as an endothermic change heat is transferred from the hot plate surroundings to the sample of water the system now in question two we also have an increase in temperature but there is no hot plate involved question two two different solutions aquous hydrochloric acid HCL and aquous sodium hydroxide NaOH occupy separate beakers at room temperature the initial temperature of each solution is recorded and the solutions are combined in a styrofoam cup the temperature of the solution is monitored over time so as you can see the temperature is increasing on the graph would you describe the change that occurred in this experiment as an endothermic process or an exothermic process justify your answer by indicating the direction of heat flow use the terms system and surroundings in your answer so from the information in the graph we know that the temperature is increasing but in this experiment there is no hot plate or any external source of heat so the increase in temperature must be Rel related to energy changes from the chemical reaction between HCL and NaOH so the heat is being transferred from the system to the surroundings as this chemical reaction takes place this is an exothermic process heat is transferred from chemical changes occurring at the particle level the system to the water molecules that are present in the solution the surroundings in question three it says consider the process of dissolving an ionic solid such as pottassium hydroxide Koh or calcium chloride ca2 in water as represented by the equations above the particle diagram below represents the dissolution of an ionic solute in a polar solvent such as water the dissolution process Can Be Imagined as occurring in three steps in step one we start with a solid ionic compound and the ions are pulled apart from each other so they're separated so attractive forces between ions are broken in Step One in step two we see a collection of solvent molecules and then there are some empty spaces in between those solvent molecules so we're pulling apart some of the molecules of solvent and they're moving away from each other so attractive forces between the polar molecules of solvent are being broken so attractive force is broken in step one attractive Force broken in step two and in step three the ions of solute and the polar molecules of solvent they form attractive forces so attractive forces are formed between the ions and the polar molecules in step three so step one and two involves attractive forces that are broken step three involves attractive forces that are formed now in unit six it's very important for you to understand how to classify the change in attractive forces between particles as being either endothermic or exothermic let's take a look breaking attractive forces between particles is an endothermic process energy is absorbed forming attractive forces between particles is an exothermic process energy is released before I move on to topic 6.2 I just want to make some more General comments about endothermic and exothermic processes in chemistry two examples of experiments during which a change in temperature may be observed are the following a solute is dissolved in a solvent forming a solution like you saw in question three of this packet and a chemical reaction occurs forming a new substance like you saw in question two of this packet if temperature changes are detected in these types of experiments it is related to energy changes in the system and at the particle level energy changes are associated with attractive forces between particles so breaking attractive forces endothermic forming attractive forces exothermic information about breaking and forming chemical bonds is included in topic 6.7 that will be for another video topic 6.7 is Bond enthalpies now on this particular slide I'm showing you something from an earlier video back in unit 2 this is topic 2.2 intramolecular force and potential energy and this Essential Knowledge statement says a graph of potential energy versus the distance between atoms is a useful representation for describing the interactions between atoms such graphs illustrate both the equilibrium Bond length and that will be along the x axis the separation between atoms at which the potential energy is the lowest and the bond energy which would be on the y- AIS the energy required to separate the atoms so this is from unit two but it's a good review related to this idea of breaking and forming bonds so this particular potential energy diagram you may remember from unit 2 we have potential energy on the y- axis and the distance between the nuclei of two atoms on the x axis now back in unit two we would use a diagram like this to analyze the properties of a bond and we talked about both Bond length which you can see along the x axis and the bond energy which you can see along the Y AIS I'm going to focus here on just the bond energy so looking at the minimum in potential energy it looks like the minimum in potential energy on this y AIS is somewhere between -400 and -500 now this diagram represents the bond between two atoms of hydrogen now if you look up the bond energy of the hydrogen hydrogen bond the value is 436 K per mole so what does that mean well we have a minimum in potential energy when the bond is formed and then we have a potential energy of zero when the two atoms are so far apart that there is no bond at all depending on which direction you're moving the 436 K per mole could either be absorbed or released so suppose you already have the bond if the bond is already formed then it would take 436 K of energy to break one mole of hydrogen hydrogen bonds so Breaking the Bond would be an endothermic process in which energy is absorbed but if I go in the other direction if I start with the atoms and they're separated from each other and the bond is formed 436 K of energy is released when one mole of hydrogen hydrogen bonds are formed so again the idea is that that could be the energy required to break the bond or that could be the energy released when the bond is formed okay so before I wrap up my discussion of topic 6.1 on this slide I'm just going to try to do my best to clarify some details that can be potentially confusing to students students often get confused about temperature changes and what that means for a process to be classified as either endothermic or exothermic so we're not talking about a hot plate or a bunson burner we're just talking about mixing chemicals together so either the process of forming a solution such as dissolving a solid in water or a chemical reaction that might be initiated by mixing chemicals together it is important to understand the direction of heat flow between the system and the surroundings and this information is summarized in the following table note that the thermometer is usually considered to be part of the surroundings and not the system so how does the temperature change during the proc process if the temperature decreases we say that heat flows from the surroundings into the system and since the thermometer is part of the surroundings the temperature goes down but we call it endothermic this can be exemplified by an instant cold pack so with an instant cold pack you would squeeze the bag to initiate this endothermic process heat flows from the surroundings to the system the temperature of the surroundings decreases and it feels cold and so the opposite of that process would be that over time during the course of the chemical reaction the temperature increases heat flows from the system into the surroundings so for example if you had instead of a cold pack you had an instant hot pack so you squeeze the bag to initiate an exothermic process in which heat flows from the system to the surroundings the temperature of the surroundings increases and it feels hot and then something that I will mention in topic 6.3 which is called heat transfer and thermal equilibrium you will study a different type of experiment so not a chemical reaction or the process of forming a solution but an experiment in which heat flows from a warmer object to a cooler object so one example of this experiment involves adding a hot piece of metal to a sample of water at room temperature and again that situation involves a difference in kinetic energy or temperature but it does not usually involve the breaking or forming of attractive forces between particles all right let's go ahead and move on to topic 6.2 which is called Energy diagrams here is an Essential Knowledge statement from the AP Chemistry course and exam description a physical or chemical process can be described with an energy diagram that shows the endothermic or exothermic nature of that process well hopefully these energy diagrams look familiar to you because in unit 5 kinetics topic 5.6 was on reaction energy profiles and they were essentially energy diagrams so it should be familiar to you all right here is question four in the packet the left diagram shown above represents the energy profile for a chemical reaction part a fill in the following information based on the left diagram what is the activation energy for the forward reaction so back in topic 5.6 reaction energy profile we learned that the energy difference between the reactants and the transition state so kind of like the top of the hill in the diagram is the activation energy for the forward reaction so to measure that on this particular diagram in question 4 it would be between 150 where the reactants start to 300 the top of the hill that's a difference of 150 kles so that is our activation energy 150 K according to the information in the diagram and the energy difference between the reactants and the products would be between 150 and 250 that's a difference of 100 so 250 minus 150 is 100 I'm going to call that difference positive 100 for a reason that I will explain later and is it endothermic or exothermic well since the products are at a higher level in terms of potential energy relative to the reactants this is an endothermic process the potential energy of the products is higher than that of reactants therefore energy is absorbed so now we have to draw a diagram on the right and some things are going to be similar and some things are going to be different let's take a look so Part B says on the diagram on the right draw an energy profile for a different chemical reaction with the following features the potential energy of the reactants should be the same in each diagram so we'll start our reactants at an energy level of 150 K the activation energy for the forward reaction should be the same in each diagram so to go all the way up to the transition state that would be at 300 so we have an activation energy of 150 but now where the products are relative to the reactants that's going to be different so it says the absolute value of the energy difference between reactants and products should be the same in each diagram but the reaction represented by the diagram on the right should show the opposite change in energy in other words if the left diagram represents an endothermic reaction then the right diagram should represent an exotherm MC reaction so I had a change in energy for the diagram on the left of positive 100 and that's endothermic the change in energy between reactants and products on the diagram on the right is negative 100 so it's exothermic so as we compare these two diagrams on the left the change in energy between reactants and products was positive 100 and on the right the change in energy between products and reactants is negative 100 so endothermic on the left exothermic on the right now the word energy and enthalpy are very similar in AP Chemistry so topic 6.6 which will be presented in the next video says that the enthalpy change again enthalpy and energy are very similar the enthalpy change of a reaction gives the amount of heat energy released for negative values of enthalpy change or absorbed for positive values for the change in enthalpy so that's a future topic topic 6.6 all right our next Topic in this packet is topic 6.3 heat transfer and thermal equilibrium here are some Essential Knowledge statements from the AP Chemistry course and exam description particles in a warmer body have a greater average kinetic energy than those in a cooler body collisions between particles in thermal contact can result in the transfer of energy this process is called heat transfer heat exchange or the transfer of energy as heat eventually thermal equilibrium is reached as the particles continue to collide at thermal equilibrium the average kinetic energy of both bodies is the same and hence their temperatures are the same question five in a certain experiment a piece of metal is heated to a temperature of 100 C in a boiling water bath then the hot metal is quickly transferred to a sample of water at 20° cius in an insulated container part A describe the direction of heat flow at the moment that the sample of metal is added to the sample of water at 20° C well if the temperature of the metal is 100° C but the surrounding sample of water is only 20° C then heat flows from the warmer piece of metal at 100° C to the cooler water molecules around it at 20° Celsius Part B at what point during the experiment will the transfer of heat between the metal and the water be complete at some point these two samples will have the same temperature so heat transfer will be complete when thermal equilibrium is reached and the temperature of the metal and the water are the same value now let's move on to our next topic this is topic 6.4 heat capacity and calorimetry here are some Essential Knowledge statements from the AP Chemistry course and exam description the heating of a cool Body by a warmer body is an important form of energy transfer between two systems the amount of heat transferred between two bodies may be Quantified by the heat transfer equation Q equal m * C * delt T calorimetry experiments are used to measure the transfer of heat the first law of thermodynamics states that energy is conserved in chemical and physical processes the transfer of a given amount of thermal energy will not produce the same temperature change in equal masses of matter with differing specific heat capacities heating a system increases the energy of the system while cooling a system decreases the energy of the system the specific heat capacity of a substance and the molar heat capacity are both used in energy calculations chemical systems change their energy through three main processes Heating and Cooling phase Transitions and chemical reactions in calorimetry experiments involving dissolution temperature changes of the mixture within the calorimeter can be used to determine the direction of energy flow if the temperature of the mixture increases thermal energy is released by The dissolution process exothermic on the other hand if the temperature of the mixture decreases thermal energy is absorbed by the dissolution process endothermic the heat transfer equation is listed as follows Q equals m * C * delt T where Q represents the heat which is normally in Jewels but sometimes in units of kles M is the mass which is usually in units of grams C is the specific heat Capac capacity which is defined as the amount of heat required to raise the temperature of 1 G of a substance by 1° C or by 1 Kelvin again the specific heat capacity is also known as the specific heat delta T is the change in temperature and because the units of the specific heat are Jewels per gram degre C but sometimes Jews per gram Kelvin you might be confused about whether or not your temperature should be listed in units of degrees Celsius or kelvin so I just wanted to comment on the fact that the delta T the change in temperature has the same magnitude the same absolute value whether it's in degrees Celsius or kelvin because we're focusing on the change for example if the temperature changes from 20 Dees C to 25 de C that would also be a change of 5 Kelvin if you went from 29 kin to 298 Kelvin so the magnitude of that change would be five whether it was listed in degrees cels or kelvin here is a section of the AP Chemistry equations and constant sheet this section is entitled thermodynamics electrochemistry and you can see that the first equation listed in this section is the heat transfer equation Q heat equals m mass time C specific heat capacity times delta T the change in temperature all right let's take a look at question six in question six it says the specific heat capacity of solid iron Fe and liquid water H2O are listed in the table above so we see that for solid iron it's 0.46 and for liquid water it's 4.18 and the units of specific heat or specific heat capacity are Jews per gam de C part A calculate the amount of heat that is required to raise the temperature of a pure sample of 25.0 G of liquid water from 20.0 De C to 75.0 De C include units in your answer so it's going to be the 4.18 for the specific heat of liquid water that I will use in this calculation along with the mass and the delta T I'm solving for Q which is the amount of heat required to make this change so the mass is 25.0 G the specific heat of water is 4.18 jewles per gam degre C and the difference in temperature between 75 and 20 is 55.0 De C when I multiply 25 by 4.18 by 55 on my calculator I get 574 47.5 looking at these three numbers that I used in the calculation since they all have three significant figures I'm going to round off my answer to three significant figures so 5,750 would be my numerical answer where there's no decimal at the end of that value now it says include units in your answer so the units would be jewels in this case so 5,750 Jews if I had decided to convert this from jewels to kles I would have divided by a th to go from Jews to kog you would multiply by 1 kog over 1,000 jewles so you also could have written 5.75 K in your answer for part A let's take a look at Part B in Part B now we're talking about the iron a pure sample of 125 G of solid iron so there's the specific heat capacity of solid iron absorb 5.32 K of heat the initial temperature of the solid iron sample is 21.0 Dees C calculate the final temperature of the solid iron again include units in your answer a common mistake that students would make is to not include units in their calculations so a student might write down that Q equals 5.32 because that is the amount of heat but it's supposed to be consistent with the units of specific heat capacity which are Jews per gram degre C and 5.32 is actually kog so be careful when you're setting these problems up to recognize the difference in units so 5.32 K is actually 5,320 Jew that's why in this calculation set up I have on the left side of the equation 5,320 jewles 125 gr represents the mass of the iron 0.46 Jew per g degre c is the specific heat of solid iron and then since the delta T is unknown that is what I'm solving for in Part B so I'm going to go ahead and do the math of 5,320 divided 125 divided 0.46 and so on my calculator I get 92.5 2 and since the 0.46 the specific heat of iron has only two significant figures I'm going to go ahead and round off this value of delta T to two significant figures I'll call that 93 I do have to include units in my answer so I'm going to say 93 degrees cels now be careful you've come this far watch out for the fact that we're not solving for the delta T in this problem and then stopping there it says calculate the final temperature of the solid iron so 93° C is not the final temperature of the iron but rather the change in temperature 21.0 de C is the initial temperature of the iron so 93 plus 21 equals 114 that is my final answer 114° C technically I could have converted this into units of Kelvin although I'm not really sure why I would do that so if I had written my answer as 387 Kelvin that would also be technically correct but there was no need to convert from Celsius to Kelvin in this problem now let's move on to part C of this question it says calculate the molar heat capacity of liquid water in units of jewles per mole degre C so we already know from the table that the specific heat capacity of liquid water is 4.18 and the units are Jew per G de c what does this quantity tell us well it tells us that it takes 4.18 jewles of heat to raise the temperature of 1 G of liquid water by 1° C but the molar heat capacity of liquid water would be how much heat is needed to raise the temperature of one mole of liquid water by 1° C so what I have to do in my calculation is I have to get rid of the unit of gram and replace that with a unit of mole now in my setup I have 4.18 jewels on the top and then gram and degree celsius on the bottom so in my conversion factor I don't want to change the jewels and I don't want to change the degrees Celsius but I'd like to get rid of grams and replace it with moles since GRS is already on the bottom I'm going to put the unit of grams on top so that grams will cancel out and then I'll put the unit of mole on the bottom because that's what I want in my final answer so Jews per mole degrees Celsius are the desired units and on the periodic table I can calculate the molar mass of H2O so one mole of H2O has a molar mass of 18.016 G from the information on the periodic table now when I do this math I will be doing 4.18 times 18.016 to get my answer 75368 that's a lot of significant figures I'm going to round this off to three significant figures so 75.3 and the units are jewles per mole degre c one of the Essential Knowledge statements from topic 6.4 is that the specific heat capacity of a substance and the molar heat capacity are both used in energy calculations in a certain experiment a sample of solid iron and a sample of liquid water each absorb a certain amount of heat data from the experiment is listed in the table above Part D A student makes the claim that the final temperature of the liquid water will be the same as the final temperature of the solid iron because each sample has the same mass that's 100.0 G and absorbed the same quantity of heat which is 1.00 K do you agree or disagree with the student's claim justify your answer with a calculation to support your choice now if you are watching this video and following along with the calculations in this packet you may already have an idea about whether you should agree or disagree with the students's claim but I'm not going to start with those words I'm going to start with the calculation and then I think it'll be obvious at the end of the calculations whether we should agree or disagree with the students claim so the calculation for iron involves the specific heat capacity of iron so 0.46 jewles per gram degre Celsius and because the amount of heat is in kog I have to be careful I have to watch out for those units like I mentioned earlier so one kogle is actually a th000 jewles q equals m time C * delta T we're solving for delta T in this calculation so I have a th000 jewles of heat I have 100.0 G of iron the specific heat of iron is 0.46 jewles per gam degre celsus and since I don't know the value of delta T that's what I'm solving for so the delta T would equal 1,00 divided by 100 divided by 0.46 and on my calculator I get a value of 2.74 so I'll say approximately because I'm looking at you know two significant figures I'll say 22 degrees C for my delta T and when I add that to the initial temperature of the sample which was 25° C 22 plus 25 would be a final temperature of 47 now before I switch over to the calculation for water take a look at the difference in the C calculation I'm still going to have 1,000 Jew that's not going to change I'm still going to have 100.0 G that's not going to change but on the bottom instead of 0.46 I'm going to have 4.18 so you can already imagine that when you divide by a larger value then your final answer for the delta T is going to be a smaller value let's take a look so for water I have 1,000 jewles of heat I have 100.0 G but now I have 4.18 Jew per g degre c for my specific heat of water and when I do this math my delta T works out to be 2.39 so approximately 2.4 de Cel and when I add that to 25 the initial temperature of the sample my final temperature of the water is 27.4 De C so now I can say with confidence and supporting my answer with a calculation that I disagree with the students claim that the final temperature of the water is not going to be the same as the final temperature of the iron and that's because there are different values for specific heat capacity so that changes the final temperature now you may have already processed it in this way in this particular equation we have q which is the heat we have M which is the mass and were constant in this experiment but as you can see the spe the specific heat multiplied by the change in temperature represents an inverse relationship so the iron had a smaller value for the specific heat and a larger delta T and then vice versa the water had a higher value for a specific heat and a smaller delta T so again an inverse relationship between C and delta T Part D of question six in this packet is a good example of this Essential Knowledge statement from topic 6.4 which says the transfer of a given amount of thermal energy will not produce the same temperature change in equal masses of matter with differing specific heat capacities all right let's move on to question seven in the packet a calorimeter is a device used to record the flow of heat during a physical or chemical process a simple version of this piece of equipment involves a styrofoam coffee cup and a thermometer sometimes a lid is used on the cup to minimize the loss of heat to the surroundings in question seven it says a pure sample of solid copper with a mass of 100.0 G is placed in a boiling water bath at 100 0° C for several minutes now if this scenario of adding a hot piece of metal to a cooler sample of water if that sounds familiar I did mention earlier in this video for question five of the packet when I was talking about topic 6.3 heat transfer and thermal equilibrium a very similar experiment to what we're talking about right here in question seven so the hot piece of copper is transferred to a sample of water which is at room temperature and the initial temperature of the water in the calorimeter is 22.0 De C eventually thermal equilibrium is reached when the temperature of the mixture of copper and water reaches a maximum value the final temperature of the mixture is 28.7 De C assume that no heat is lost to the container or the surroundings outside the container and the specific heat capacity of liquid water as was mentioned in question six of this packet the specific heat capacity of liquid water is 4.18 jewles per gam degre celi all right part A of this question the magnitude of delta T for the liquid water in this experiment because it was initially at 22.0 De C before the metal was added and it was at 28.7 De C when it reached thermal equilibrium with the metal that difference 28 7- 22 is 6.7 de C calculate the magnitude of delta T for the copper which began at a temperature of 100.0 De C in the boiling water bath and then ended at 28.7 De C when it reached thermal equilibrium with the cooler sample of water so that difference between 100 and 28.7 is 71.3 De C now now you might be wondering about whether or not it makes a difference if I refer to the final temperature minus the initial temperature is that the way I should calculate it every single time is it tfal minus t initial but what I did by using the word magnitude I was referring to the absolute value so the S of delta T is ignored here I'm just focusing on the absolute value of the difference and not necessarily worrying about whether the change in temperature is positive or negative so in Part B of this question I think because we have a difference in delta T A student makes the following claim that the amount of heat energy lost by the copper is greater than the amount of heat energy gained by the water so if you're just focusing on the delta T then you might assume that somehow the copper lost more heat because its delta T was greater than that of the water so we clearly have to make a choice about do we agree or disagree with this student's claim one of the Essential Knowledge statements from topic 6.4 says that the first law of thermodynamics which is also known as the law of conservation of energy states that energy is conserved in chemical and physical processes so the correct answer to question seven part B is the following I disagree with the student's claim so when the hot piece of metal is added to the calorimeter with the cooler sample of water there is heat transfer but according to the first law of thermodynamics also referred to as the law of conservation of energy the amount of heat energy lost by the metal in this experiment is equal to the amount of heat energy gained by the water and now that we know this now that we know that the Heat lost by the metal is equal to the heat gained by the water we can set up a calculation in part C of this question that is based on that fact so use this information from the experiment we have all the data we need as far as temperature and Mass to calculate the value for the specific heat capacity of copper so Q equals m * C * delta T and the heat that is lost by the copper is equal to the heat gained by the water now again as I mentioned earlier in the video that the magnitude of the delta T for the copper is 71.3 Dees C if you were doing T final minus t initial then you might consider that to be a negative number so I'm not using negative signs at all in this setup I'm focusing on the magnitude of heat on both sides of the equation which is why you don't see any negative signs anywhere so the heat lost by the copper is equal to the heat gained by the water and I look on the left and the right and I see 100.0 G of copper 100.0 grams of water so I'm going to go ahead and cross those off to make my math easier the specific heat of the copper which is symbolized by lowercase C I don't know what that value is the specific heat of water was given as 4.18 jewles per gram degre celsus and obviously from part A I have my delt t for the copper and my delta T for the water so when I do this math to solve for C the specific heat of the copper I get 71.3 * C on the left side of the equation and 4.18 * 6.7 gives me a value of 28.0 on the right side of the equation so now I'm going to go ahead and just divide both sides by 71.3 and So my answer on my calculator is 0.39 three and my units of specific heat are the same on both sides of the equations that would be Jewels per gam degrees celsi so that's my answer to part C 0.393 jewles per gram degre Celsius that's the specific heat capacity of the copper now as I transition from question seven to question 8 you'll notice that these two experiments question seven and question 8 are very very similar to each other there is a slight difference and it has to do with what specific quantity that we are calculating in question 8 but let me go ahead and read the details of question 8 the specific heat capacity of solid aluminum is 0.91 Jew per gam degre C and the specific heat capacity of liquid water is 4.18 Jew per gram degree celsius a pure sample of solid aluminum with a mass of 100.0 grams is placed in a boiling water bath at 100.0 degrees C for several minutes then the hot aluminum sample is quickly transferred to a sample of 100.0 G of liquid water in a coffee cup calorimeter the initial temperature of the liquid water in the calorimeter is 22.0 De C eventually thermal equilibrium is reached when the temperature of the mixture of solid aluminum and liquid water reaches a maximum value assume that no heat is lost to the container or the surroundings outside the container so here comes the difference in question seven what we did not know what we had to calculate was the specific heat capacity of the copper metal but in this question question 8 what we do not know is the final temperature of the mixture of the metal and the water so calculate the final temperature of the mixture when thermal equilibrium is reached we're going to use the same principle which is based on the law of conservation of energy also known as the first law of thermodynamics and that is that the heat lost by the aluminum in this experiment is equal to the heat gained by the water in this experiment now there's a problem with the setup right now I have delta T on the left and I have delta T on the right but these two values of delta T are not the same the magnitude of delta T for the aluminum should be greater than the magnitude of delta T for the water so what I have to do is come up with a value for the final temperature now based on the information given to us in this problem we know that the initial temperature of the hot sample of aluminum was 100° C and we know that the initial temperature of the water The Coffee Cup calorimeter before the aluminum was added is 22° C so the final temperature of the mixture of aluminum and water when thermal equilibrium is reached is a number that we don't know so I'm going to call this x and x represents a number that falls in between 22 and 100 so I'm going to write for my setup in this equation the delta T for aluminum and the delta delta T for water in terms of X but if you're wondering how do you know whether to write the delta T as 100 - x or x - 100 or 22 - x or x - 22 here's what I've done on both the left and the right side of the equation I've written the value of delta T in terms of X but I have done it in such a way so that the value of delta T is positive on both sides of the equations so I choose to focus in this setup for this calculation on the magnitude of delta T and I want to avoid any issues with adding a negative sign to the equation for the loss of heat so because X is a number that falls between 22 and 100 100 - x is a positive value and x - 22 is also a positive value so I hope that makes sense I'm trying to focus more on the change in terms of a positive value and not get confused by adding a negative sign to one side of the equation okay the next thing I'm going to do in solving this problem is what I had done in question seven of the packet where I recognize that there's 100.0 G of aluminum and 100.0 gram of water that's the same value so I'm just crossing them off on both sides of the equation to simplify the numbers in the calculations now I have .91 * 100 - x = 4.18 * X -22 the next step in the calculation is to multiply and distribute so I have on the left 91 - 0.91 x and on the right I have 4.8x minus 9196 I'm now going to combine like terms and simplify and when I'm all done and I finished off my calculation my answer for X which is the final temperature of the mixture when thermal equilibrium is reached it works out to be 35.9 de C okay well hopefully my explanation made sense for how to calculate the final temperature in question 8 now it is time to move on to our next topic this is topic 6.5 energy of phase changes here are some Essential Knowledge statements from the AP Chemistry course and exam description energy must be transferred to a system to cause a substance to melt so from solid to liquid or boil from liquid to gas the energy of the system therefore increases as the system undergo a solid to liquid or liquid to gas phase transition likewise a system releases energy when it freezes from liquid to solid or condenses from gas to liquid the energy of the system decreases as the system undergoes a liquid to solid or gas to liquid phase transition the temperature of a pure substance remains constant during a phase change the energy absorbed during a phase change is equal to the energy released during a complementary phase change in the opposite direction for example the molar enthalpy of condensation from gas to liquid of a substance is equal to the negative of its molar enthalpy of vaporization from liquid to gas similarly the molar enthalpy of fusion can be used to calculate the energy absorbed when melting from solid to liquid or the energy released when freezing from liquid to solid all right this is question nine in the packet in question n it says the diagram shown at right represents a heating curve for a pure substance the experiment begins with a sample of a pure solid heat is added at a constant rate and the temperature of the substance is monitored over time so in the diagram we see heat added on the x-axis and we see the temperature on the Y AIS we have five different curve segments and we have to describe the change that occurs during each part of the experiment so the experiment begins with a sample of a pure solid that means that during segment AB we have a solid and we can clearly see that the temperature is inre inreasing so the solid is heated and its temperature increases but notice that what happens during segment BC is that although heat is being added there is no change in temperature during line segment BC and that's consistent with the information that I had mentioned earlier that the temperature of a pure substance remains constant during a phase change so what's happening during segment BC is that the solid melts now for the next segment CD we have all liquid and the liquid is heated and its temperature increases we come to the next flat portion of the diagram a plateau where heat is being added but the temperature is not changing so the next phase change would be going from a liquid to a gas you could call that evaporation I will call that that the substance is reaching it its boiling point so the liquid is boiling changing from a liquid to a gas during this portion de and of course the last portion would be now we have all gas so the gas is heated and its temperature increases so we've answered the details of question nine let's take a look at question 10 which is talking about the attractive forces and whether it's classified as endothermic or exothermic so the phase changes are listed as melting solid to liquid evaporation liquid to gas sublimation solid to gas in terms of a particle diagram view when you go from a solid to a liquid some of the attractive forces between the particles are broken when you go from a liquid to a gas essentially all of the attractive forces between the particles are broken and from a solid all the way to a gas clearly attractive forces are broken so attractive forces are broken during melting evaporation or sublimation if you were to reverse that process attractive forces would be formed as you change from a liquid to a solid or a gas to a liquid or from a gas all the way to a solid with freezing condensation and deposition attractive forces are formed as I mentioned earlier in this video it takes energy to break attractive forces so breaking attract RVE forces is an endothermic process and then the reverse of that when you form attractive forces energy is released so it's exothermic when attractive forces are formed something else to mention about phase changes is that there's a term referring to the molar heat of fusion so what exactly is the molar heat of fusion well it's defined as the amount of heat required to convert one mole of a pure substance from a solid to a liquid at its melting point so molar heat of fusion refers to the melting point and the energy involved in that process and then the molar heat of vaporization is referring to the boiling point that's defined as the amount of heat required to convert one mole of a pure substance from a liquid to a gas at its boiling point and as mentioned in the Essential Knowledge statements you could reverse that process and go from either a gas to a liquid or from a liquid to a solid and the amount of energy would be the same it would just be released instead of absorbed okay let's take a look at our next question in the packet this is question 11 so in question 11 the most important piece of advice I can give you is to read the question carefully if you don't read this question carefully you might miss out on important details about the quantity of heat involved or about the direction of the flow of heat so so in question 11 it says the following information about CH4 which is known as methane is listed in the table above which of the following statements accurately describes the net flow of thermal energy when 32.8 G of CH4 is converted from a gas to a liquid at 112 Kelvin so in the table above we have the boiling point of CH4 which is 112 Kelvin and we have the molar heat of vaporization for CH4 which is 8.17 KJ per mole so at first glance you focus on the word vaporization which is going from a liquid to a gas you understand that liquid to gas is an endothermic process and that thermal energy would flow from the surroundings to the sample in an endothermic process but as I said read the question carefully because in fact they're asking about the transition not from a liquid to a gas but rather in the opposite direction from a gas to a liquid so that would be not an endothermic process but rather an exothermic process and so as I had mentioned earlier in one of the Essential Knowledge statements for topic 6.5 a system releases energy when it freezes or condenses now that we know we're looking for the transition from a gas to a liquid which is an exothermic process the thermal energy should flow from the sample to the surroundings so the correct answer is either a or C now we have to focus on the amount or the magnitude of heat so 32.0 G of CH4 does that represent one mole or something else let's take a look at the periodic table carbon has a atomic mass of approximately 12 hydrogen has an atomic mass of approximately 1 so for CH4 we're talking about a molar mass of approximately 16 and we have 32 G of CH4 which means we have two moles and so because we have two moles 8.17 is the amount of heat released when one mole of methane or CH4 changes from a gas to a liquid but since two moles of methane are listed in this quantity the answer is not a but rather C 16.34% two different substances is listed in the tables above part A calculate the quantity of heat that must be absorbed to increase the temperature of a pure sample of 0.375 G of solid C7 h63 from 25° C to the melting point of 159 de C and melt the solid completely include units in your answer answer so here is a generic heating curve and I want to highlight in red the fact that not only are we heating the solid we are also melting the solid completely so there are two parts of this calculation one part is to raise the temperature of the solid from 25° C to 159° CS but then we also have to melt the solid completely at its melting point of 159° C so the first part of this calculation involves heating the solid from 25° C to 159° C for this calculation we're going to use the specific heat capacity of the solid which is listed in the table as 1.17 jewles per G degre Celsius we'll use the heat transfer equation which is q = m * * C * delta T and in this example the delta T is 159 minus 25 that's a difference of 134° C so here's my setup m is the mass which is 0.375 G the specific heat capacity is 1.17 jewles per G degree celsius and the delta T is 134° C if I do this math I get 58.79 25 and my units are Jewels I will round this off to three significant figures so I'm going to say 58.8 jewles now I also have to melt the solid completely at 159° C but during a phase change there is no change in temperature so I cannot use qals MC delta T because the delta T is zero so how I do this is focus on the information in the table for the molar heat affusion notice that it says 27.1 kles per mole that means it takes 27.1 K to melt one mole of this substance at its melting point let's first convert from grams into moles and then we can use the molar heat of fusion so on the periodic table C7 H6 O3 has a molar mass of 13818 G per mole I will set up a conversion to go from G to moles so 0.375 gtimes 1 mole of this substance over its molar mass 13818 and now I can use 27.1 K per mole as a conversion factor and so when I do this math 37 5 / 13818 and then time 27.1 I get 0.07 35784 again I will round off to three significant figures so 0 0736 kles now it's very important that you focus not only on the fact that there are two steps in this calculation heating the solid and then melting the solid you also have to focus on the following we have to add these two quantities together but right now the units don't match we have 58.8 jewels and then we have 0.736 kles so Focus really closely on your units when doing these calculations so I'm going to convert kles to jewels and the way you do that is to multiply by a th000 so there are 1,000 jewels in one kle so so 0.736 K is equal to when I multiply by a th000 73.6 jewles now when I add these two quantities together my final answer and I have to include units in my answer I have 132 jewles but you could have also gotten 0.132 k for your final answer either one but again very important to pay attention to the fact that we had two steps in the calculation and we had to focus on a difference in units between jewels and kogs all right moving on to Part B of this question we have the same kind of situation let's take a look calculate the quantity of heat that must be absorbed to increase the temperature of a pure sample of 0.375 G of c2h6o liquid from 25 de C to the boiling point of 78° c and then evaporate the liquid completely so again just like in part A there are two parts of this question first we have to heat the liquid from 25° C to its boiling point of 78° c and then we have to evaporate to boil the liquid completely using a different part of the calculation so we have to not only heat the liquid but also boil the liquid completely so just like in part A this is a two-part calculation so in order to heat the liquid from 25° C to 78° C I will use the specific heat capacity of this substance and that's 2.46 jewles per gam degre c q equals m c delta T and my delta T is the difference between 78° C and 25° C that's a difference of 53 degrees C so m is the mass which is 0.375 g c is the specific heat capacity which is 2.46 jewles per gram degre Celsius and my delta T is 53 de C when I do this math I get 48.890717 I'm going to round this off to three significant figures so I'm going to say 48.9 Jew for the next part which is to turn this liquid into a gas by boiling it completely at its boiling point I have to use the fact that the molar heat of vaporization is in units of kog per mole there is no delta T when you're changing from a liquid to a gas at the boiling point so I can't use MC Delta t i I've got to convert from grams to moles so using information on the periodic table the molar mass of c2h6 works out to be 46.0 68 G per mole I'll start with grams 0375 G of c2h6o I'll set up a conversion factor to go from grams to moles so times 1 mole over 46.0 68 G now I can use the molar heat of vaporization as a conversion factor so times 42.3 K over 1 mole and when I do this math 375 divided 46.0 68 time 42.3 I get 0.344 328 I'll round this off to three significant figures so 0.344 and again my units are kles so before I can add these two quantities of heat together to get my final answer I've got to make a decision about how to make sure that the units match up so like I did in part A I'm going to convert kilog into Jewels so when I multiply by a th 0.344 K is equal to 344 Jew now I can add these two quantities of heat together so my final answer is either 393 Jews or you also could have gotten this correct with an answer of 0.393 K all right well that is the last part of question 12 it's the last question in the packet so this is the end of the video I hope that you found these answers and explanations helpful thanks for watching