all right guys so continue from our last lecture we were on the topic of heat transfer between different objects we had this example problem where you have a code coin dropped into a warm coin being dropped into a coat of beer and the heat transfer is between the two objects and eventually they're gonna reach the same final temperature okay one the coin by losing heat going down in temperature and the beer by absorbing the heat from the coin and um raise this temperature but eventually they reach the final same final temperature okay now for two objects okay this is how we handle it but what if in an isolated system we have more than two objects okay so here's our next example 50 grams of water at 20 degrees celsius is placed in a coffee cup 48 grams of water at 80 degrees celsius is then added to it the final temperature in the cup is measured at 80 oh no no 80. 47.8 degrees celsius calculate the heat capacity of the coffee cup know that cp specific heat of water is 4.18 joules per gram per degree celsius all right so now we have a question we have a ton of numerical conditions in it okay and now let's try to analyze how do we do this yeah so you have some cold water in a coffee cup so this is 20 degrees celsius 50 gram okay and then you had some hot water 48 grams 80 degrees celsius add it to it okay to the cup that contains the cold water now you have a lot of water together okay and we measured the temperature of the water now is 47.8 degrees celsius now so these are the conditions provided and we're looking for heat capacity c of the cup all right so first of all how many things are involved in this reaction in this process cold water obviously is your object one hot water is your object too okay when you mix them together of course the hot water is losing heat and the cold water is gaining heat that should be pretty obvious okay but since we're calculating the cup okay and let's be specific we're calculating heat capacity of this cup okay so the cup itself is object number three so you have three things that has involved in the heat exchange all right so for the object number one change of temperature is final temperature 47 minus 0.8 minus 20.0 for your object number two your change of temperature is 47.8 minus 80. what about your object number three well it says cold water is placed in the cup so delta t of the cup here is going to be the same initial and final temp final temperature is 48 7.8 for everybody the initial temperature of the cup should be the same with the cold water because cold water was in it at the beginning okay yeah so each of these three things head temp head temperature change and had absorbed or released heat during the process now how do we do this calculation now don't forget for all your heat change problems you're going to deal with in this class you have to have a isolated system or insulated system okay because only within in in insulated system you can stop worrying about okay what if i lose heat to the environment you know the environment in this case with the insulator system don't exist okay so all the heat transferred is between these three things q lost plus q gain equal to zero okay but you only have two things here in the equation no you only have two different scenarios in the equation okay they any objects you have in the system they either lose this heat or they gain the heat okay but do you have to limit yourself to one thing losing heat or one thing gaining heat no for our situation let's see who's losing heat hot water is definitely losing heat okay anything else losing heat you don't have anything hot anymore okay so who's gaining heat cold water is gaining heat that's for sure but look at the temperature change of the cup the cup has a higher final temperature than the initial temperature as well so the cup is also gaining heat okay so really instead of writing it as q loss plus q gain you go to zero you can also write q of the hot water plus q of the cold water plus q of the cup equal to zero so whether you have three objects four objects 120 objects as long you know as you are in an isolated system this equation stands heat of everybody add up to zero energy will not be created energy will not disappear for no reasons energy is only transferred between the objects that you have in the system all right so once you establish this now it's all about breaking each cue down to expressions using these conditions okay so what do we know about these objects hot water and cold water both water and we know cp of water so obviously this should be m of the hot water cp of water delta t of the hot water the cold water is mass of the cold water same cp delta t of the cold water and then for the cup okay we don't have a cp but we are looking for a c how do we represent q with c q equal to c delta t so c of the cup delta t of the cup and the other t of the cup just happened to be the same thing with the cold water so all together these things equal to zero now then it's about plugging your numbers in it's a long equation but everything should be in place mass of the hot water 48 grams times cp 4.18 joules per gram per degree celsius so grams cancel now delta t of the hot water 47.8 minus point zero degrees celsius degree cancel the degrees celsius cancel so you have jurors mass of the cold water 50.0 grams 4.18 joules gram degrees celsius times 47.8 minus 20 degrees celsius degrees celsius cancel c of the cup plus c of the cup is what we wanted delta t of the cup is 47.8 minus 20.0 degrees celsius the total of that equal to zero so essentially this is a number this is also a number and those two numbers together plus c cup let's say that's our x okay times another number equal to zero that's a pretty simple algebraic situation that you can solve okay rearrange the equation you find that your c of the cup the heat capacity of the cup 23.4 joules per degree celsius okay because these numbers combine to give you a jewel unit and x times degrees celsius equal to that and the unit of the sea should be jewelers per degree celsius which is what we know okay heat capacity have unit of joules or calorie per degree celsius so good after you take your notes after you watch the video now pretty good chance or opportunity for you to later on when you study the notes cover up my solution read the question again think about what we did here and try to solve this question on your own and can you finish this calculation set up this calculation finish it and get this result okay now so really at this point if this thermal property is provided it doesn't really matter okay how many different objects you have in a given insulated system okay within this system everybody's total heat eventually okay equal to zero and you can plug in your numbers make sure your final temperature uh initial temperature your different uh several properties are in the right place okay and you should be able to set up a equation starting from this original equation of q last q loss plus q getting at zero and do your calculation all right so so far okay everything that we did here okay on heat exchange whether it's heat transfer between two different objects or the heat change of a given individual object okay whether we use q loss plus cooling equal to zero or we use the equation of q equal to m c p delta t and c m delta t or um c dot t okay one thing you should notice that is that there's no chemistry here if you're dealing with one object obviously you don't have a chemical reaction and when if you're work when we're doing dealing with two or three different objects okay we're not focusing on chemical reaction this this quite this example is water and water okay obviously no reaction last example was a coin and beer obviously no chemical reaction so why do you stick a whole bunch of heat information where no chemistry is involved in a chemistry course no because the reason of that is that what we learn here provide us with a way of trying to to find out what kind of energy is involved in chemistry and that is a huge subcategory of chemistry okay so after we learned the heat transfer the next thing we're going to be talking about here really involves some chemistry and we call it called this part thermal chemistry okay now we have talked about this as we start the energy um section okay we said that a very very important aspect of chemistry huge reason why people study chemistry is the energy involved in chemical reaction okay now you think about the energy resources that people are using right now okay if you're using fuel any kind of fuel fossil fuel whether it's coal whether it's uh petroleum okay at the end of the day what do we use fuel for we burn it but we're supposed to get energy out of it burning gasoline in your car eventually providing enough energy to push your pistons and make your car move okay and that is all chemistry energy okay that entire process of gasoline burning and releasing energy to the environment is what we study in thermal chemistry okay and if you think about batteries okay we talked about it earlier that okay batteries battery reactions are all redox reactions okay why can redox reaction be used as batteries number one of course is because redox reactions always involve um electron transfer okay and with the electron transfer if you make the half reactions happen on different electrodes and lead the electrons out through wires and go back to the other electrode where it needs electrons you've got a full circle closed circuit and that's going to give you electricity okay but the other reason is that all chemical reactions involve some form of energy okay yeah and redox reactions okay involve energy loss or energy gain okay and whatever reaction that involves energy loss okay eventually we want to try to see if we can get that energy and use it okay if you're wondering well you told us all redox reactions can be made into batteries and what if i have a redox reaction that's absorbing energy well if the redox reaction you're studying is absorbing energy then the reverse reaction of it is releasing energy so you can always find an opportunity fork for reaction that provide energy to the environment and try to use the energy involved okay so energy involved in chemical reaction is the reason that we study thermochemical chemistry all chemical reactions involved energy change some absorb energy from its environment some release energy to its environment okay now and how much energy is involved in a chemical reaction when i burn this much of gasoline okay how much energy is going to be produced during this process of burning if i use copper zinc reaction single replacement reaction as a battery if this much of copper zinc is used during the process well how much energy is going to be released as electricity now that you can see is the obvious important factor when it comes to study the energy side of chemical reactions okay now first of all the amount of energy involved in a chemical reaction okay we have a new concept to describe it we call it enthalpy we use capital letter h to represent it okay like potential energy okay enthalpy is a state function which means you can only you don't have absolute value of enthalpy for a given chemical or given chemical reaction what you can only study is during a process of change how much energy is gained or lost okay so although we have the definition of enthalpy as energy involved in chemical reaction when we measure that what we measure is the change so we call it delta h okay and when you whenever you think about delta edge delta h to a reaction okay is very much the same as q okay the heat the heat change to a object okay an object could lose heat and you would have a negative q value an object could have could gain heat and you would have a positive value same thing for the reaction a reaction could be absorbing energy from the environment okay and then you would say the delta h for that reaction is positive because you're getting energy okay if a chemical reaction releases energy to the environment you will see that the delta edge being labeled as a negative value because you're losing energy during the reaction okay yeah and every chemical reaction is a process like this okay the reason of that very simply if we label different chemicals potential energy on a reaction progress chart you would see okay originally if you have some chemicals let's say the reactants that have lower potential energy and later on you made some products and if those products just happen to have higher potential energy okay then the energy difference between the reactants and products okay is going to be a delta h and in this case because your reaction have to absorb energy from the environment to lift your reactants to products and this is what this would be a process where you absorb energy we say the delta h is positive for this reaction okay or you could have a bunch of reactants that has a higher potential energy and then it turns into products that has lower potential energy okay and in that case you can say that my chemical reaction okay the other edge difference between the reactants and products is a negative value and we're releasing energy to the environment okay energy releasing process so burning gasoline okay or you know discharging a battery all falls into this category and we call it exothermic reaction so energy releasing process is exothermic okay energy absorbing process okay like photosynthesis of green plants okay like the charging process of a rechargeable battery okay absorbs energy from the environment from the environment we call it endothermic reaction okay and now absorbing energy endothermic all right so of course now you can see in chemical reactions the amount of energy released or absorbed during a chemical reaction is always proportional to the amount of the reactants used the more reactants you use the more energy it releases or absorbs okay and now when you put the amount of energy okay absorbed or released during a chemical reaction that information together with the balanced chemical equation for example we know that burning of natural gas methane with oxygen gas produce carbon dioxide and okay now so we know this is the balanced chemical equation okay but if i put a delta h together with it and we say delta h naught or dot edge standard is negative 127 kilojoules now these two piece of information together makes this equation a thermal chemical equation okay and this means that if you have now the little circle here not or the standard okay as we read it means that if you have this many moles of stuff one mole of ch4 reacting with two moles of oxygen and ended up producing one mole of carbon dioxide and two moles of water then you will be release releasing negative sign means and exothermic releasing energy to the environment you will be releasing 127 kilojoules of heat okay to the environment okay so with a thermochemical equations we can do a lot of things for example if you want to if you use let's say um natural gas okay to heat up water in a pot okay knowing how much water you need to heat up to what temperature okay you can eventually calculate and of course if you want to include the uh the pot okay you cannot you would also need to know some thermal properties of the pot but use if we have those data okay we can actually figure out okay to heat up this pot of water from let's say room temperature 20 degrees celsius to boiling 100 degrees celsius okay how much heat do i need and once you know how much heat you need you would be able to figure out using a thermochemical equation how much natural gas i actually need to burn to produce that much heat to heat up my water so a lot of the commonly known chemical equations at this point people already know okay their thermochemical equation and means the energy involved in those chemical reactions and if you search a chemistry index okay and for our textbook if you search the appendices where it has thermochemical reaction data okay you can find a lot of chemical reactions that already comes with the delta edge value all right so now how did we get these values how do we get the delta h values for a given chemical reaction now now you're going to need the stuff that you learned earlier in the heat part okay if i can make a chemical reaction happen okay within an insulated system okay and if my reaction release heat then stuff in my system would absorb those heat if my reaction is absorbing heat then the stuff in my system is going to provide those heat now then i can calculate delta h because now your q loss plus q gain could be written in a different way okay the other edge plus q equal to 0 in a insulated system okay because again now although this time energy is coming from your chemical reaction okay or energy is provided to your chemical reaction but energy is still conserved in the insulated system so again what happens in the inside of the system stays in the insulated system okay the total of f all the energy exchange eventually equal to zero and how do we do that calculation how do we deal with that process let's see that from our next example okay so 50 grams of 1.0 molar naoh solution at 20 degrees celsius is mixed with 50. let's use milliliters since we have molar milliliter of 1.0 molar hcl solution also at 20 degrees celsius in a coffee cup okay final temperature in the cup is measured at 26.5 degrees celsius if the coffee cup has heat capacity c equal to 23.4 joules per degree celsius and all solutions have density equal to 1.0 grams per milliliter and cp 4.18 4.184 joules per gram per degrees celsius calculate delta edge of the neutralization reaction happened all right so now let's take a look at this so you have this looks like a very complicated question okay because you have so many different numerical conditions volume concentration mass temperature heat capacity specific heat capacity and where do i start well first of all you're calculating the delta edge for the neutralization reaction okay so let's write that reaction out you have naoh reacting with hcl producing you already know what it produces if you don't you're you're really in trouble water and salt okay and i had a chemical reaction okay now what happened i have 50 milliliter of one molar 50 milliliter in one mole now you don't need me to tell you that this obviously is not a limiting reagent situation right you have two reactants reacting with each other in the one one-to-one molar ratio and you have the same volume and same concentration so you have the exact same amount okay this is a stoichiometric amount for both reactants so at the end of the reaction both of your sodium hydroxide is gone and and your hydrochloric acid is gone okay so this is this one's definitely not about stoichiometry now all right so what else do we know now we know that both of these two things have 20 degrees celsius temperature so so this is a difference from what we had earlier in the insulated system and no chemical reaction insulate the system situation because in that situation you always need to have a hard object code object and you know and don't forget this time you also have a coffee cup okay now it had because it it's given you a heat capacity which means this guy can absorb and release heat in your system okay and since the reaction was happening within the cup and originally both of your two reactants is at 20 degrees celsius so you can be sure that your cup starts at 20 degrees celsius as well okay now without the chemical reaction if this is just water instead of two chemical solutions in this cup do you expect heat exchange the answer is obviously no everybody is at the same temperature nobody's cold nobody's hot nobody's losing heat nobody's gaining heat okay but in this situation you actually have a final temperature 26.5 degrees celsius what does that tell you your solution got hotter why if there's no heat transfer during the process how can my solution get harder well if they're just simply objects that has nothing to do with each other okay you won't have a hotter solution the factor matter is we admit that there is a chemical reaction going on and chemical reaction as we know now always involve energy so now let me ask you this question is this chemical reaction endothermic or exothermic okay and you look at this you go well the final temperature is higher the initial temperature is lower that means the solution absorbs heat that is a endothermic reaction and if that's how you think i'd have to tell you that no you're wrong okay what do you mean i'm wrong solution absorb the heat right yes absolutely the solution absorbed heat okay but who provides the heat if the solution gained heat in the insulated system somebody lost heat right who lost heat think about that for a second your solution both of your solution are now in the cup the solution is gaining heat the cup is gaining heat who lost heat you can't tell me that nobody lost heat okay because then you are going to be the first person to rewrite okay the law of energy conservation no so how you're going to explain this situation if you say well it's not endothermic well think about this that equation that we had earlier we said in insulated system where a chemical reaction happens we can always have delta h plus q equal to zero okay for every object in the system the two solutions and the cup okay if we view them object as objects all three of them absorb heat so the q and those are the only three objects in the system so the q is absolutely positive and for you to make the energy conserved for you to make the total equal to zero what does that tell you about your delta h if this is positive your delta h must be negative and if your delta h is negative what kind of chemical reaction do you have of course you have an exosomic reaction okay and now let's explain this i put the two chemical solutions together in the cup the chemicals within them start to react going through this neutralization reaction okay and during the process the reaction because it's exothermic is going to release a whole bunch of chemical energy okay and that chemical energy is then absorbed by the solution and the cup so although no other objects are providing energy or losing heat okay still the chemical reaction itself provide energy to the solution and to the cup and that's why the temperature of the cup increase all right so now how do i do this calculation now obviously you want the other edge so that stays in the solution in the equation okay now what else can i do well let's think about a cue okay although these are the two these are two separate solutions before you mix them okay but once you mix them okay they're one big solution okay the good thing is when we're trying to figure out the cue when we're viewing the solution as an object i don't have to worry too much about different chemicals concentrations moles all that stuff okay what i need to what i can view this big body of solution i can view it as oh this is 50 milliliter plus 50 milliliters so it's a total of 100 milliliter okay but i have cp for the solution so i really don't need milliliter but this thing is 1.0 grams per milliliter oh okay so i really just have 100 grams of total solution so i can do q of the solution and obviously i can do q of the cup okay so put them together to total equal to zero delta h is what we want which means q and q solution and q of the cup should both be numbers well q of the solution is m c p delta t and q of the cup is c delta t okay so let's see delta h should equal to mass of the total solution we just calculated it 100 grams times 4.184 joules per gram per degree celsius times delta t 26.5 minus 20 degrees celsius plus c double t 23.4 joules degrees celsius times 26.5 minus 20 same temperature difference for the cub degrees celsius and all that equal to zero that is a number that is also a number so of course i can figure out my delta h and the calculation gave me my delta h in this this case is negative two eight seven zero jurors okay or right in a more scientific way negative two point eight seven times ten to the third jurors or you could write it as negative two point eight seven kilojoules because kilojoules is ten to the third yours okay now we talked about this is an exothermic reaction and we ended up getting a negative result for that edge which is perfectly sound perfectly reasonable and by figuring out how much heat in an insulated system that is absorbed or released okay while a chemical reaction is happening in the system we can do always do calculations like this and figure out and actually now much vast majority of the thermal chemical reaction data that people have nowadays okay the the original ones okay are done this way okay we make a chemical reaction happen in the container okay usually in with the existence of some solutions we calculate we measure temperature change we calculate how much heat energy is transferred from the reaction to the things in your container okay and with the thermochemical data of the container of the solution then we can calculate and find out for this much of the chemical reacting with each other what is the delta h involved this process of using okay objects heat and insulated system to find out the delta h involved in a chemical reaction this method is called calorimetry okay calorimetry is a big branch of thermochemistry this is how we find chemical some chemical thermochemical data okay experimentally and this cup that we use okay now in a accurately set up thermochem calorimetry experiment we use a much better insulator a closed container with its heat capacity carefully measured okay previously known okay and this cup is acting as what we call okay so chemical reactions that edge could be measured through calorimetry by making the chemical reaction happen in the calorimeter yeah but once we got these regular delta h values of the given amount of chemical okay how do we turn this into a standard value that we can put together with the chemical equation because right now you cannot put oh delta h of this equal to negative 2.87 kilojoules okay you can't do that and expect this to be a thermochemical equation people are going to to be confused because you used a little bit of chemical okay if you change the amount of chemical if you use 100 milliliter of the solution if you use like a 3.0 molar solution then your value of delta energy is going to change and when you have equations thermochemical equation like this format that number can change okay so in our next lecture we're going to discuss how to turn the delta h value into the standard delta h value that's it for today and i'll see you on friday