 Imagine for the moment that I asked you to determine how much heat was required to increase the temperature of 350 milliliters of water from 68 degrees Celsius to 90 degrees Celsius. Well, the way that we would set this up would be to establish a system. I could define my system as the mass of the water. I could probably get away with assuming that the change in mass is negligibly small. At which point I would have heat entering my system and increasing the temperature. And if I were to make some sweeping assumptions like the density of our water doesn't change, therefore as the temperature changes the volume doesn't change there's no work opportunities. And if I were to neglect any heat transfer in the hour direction and just call it Q in the thing that I'm solving for and if I were to neglect any changes in kinetic potential energy my energy balance would simplify down to Q in is equal to mass which if I assumed was constant I could factor out times U2 minus U1. The mass we could determine because we know the volume and we can look up the density of water at 68 or 90 degrees Celsius. So the question really becomes how do we determine delta U? And for that we have a couple of different options. The one that we know already is going to be the best option so I'm going to call this options for evaluating delta U and delta H. Option one is to look them up and subtract them and this is what we've done so far. We have two state points one is at 68 degrees Celsius the other is at 90 degrees Celsius. We could presumably look up an internal energy the problem with that is we don't actually have enough information to perform to look up properly. We don't know another independent intensive property which would be required to fix the state. We just have a temperature so we could probably get away with you know let me start a new page we could probably get away with assuming something. I mean if we were to assume say that the pressure was I don't know let's call it 100 kilopascals that's basically one atmosphere right? I mean it's close enough I would say. We have a temperature at day one then of 68 degrees Celsius and a pressure at day one of oh about 100 kilopascals. Okay with that information assumed I can look up E1. All I would have to do there is find the intersection of 68 degrees Celsius and 100 kilopascals in my tables and that process is going to begin by determining the phase at day one which I could do by looking up the saturation temperature corresponding to my pressure and comparing my temperature to that or I can look up the saturation pressure corresponding to my temperature and compare my pressure to that. In this case I will look up the saturation temperature corresponding to 100 kilopascals and compare 68 to that temperature. That saturation temperature at 100 kilopascals is going to be t-sat at one bar which if I jump into our tables is going to be 99.63 degrees Celsius. My temperature at day one is less than 99.63 degrees Celsius therefore it must be a compressed liquid. So if I peruse my compressed liquid tables surely I will find a pressure sub table corresponding to one bar and oh alas I don't have one. I mean I have a pressure sub table corresponding to 25 bar and 50 bar and 75 bar but not one bar which means the bestest way to get a property here would be to interpolate to interpolate between 25 bar and our saturation condition because remember this is all describing water our compressed liquid tables and our saturated liquid vapor mixture tables we're just splitting them into phases so the saturation condition that is the saturated liquid property at 68 degrees Celsius is going to be on one side of my data point and the property at 25 bar and 68 degrees Celsius is going to be on the other side of my data point. So my interpolation would just be something like our pressure which is one bar minus p sat at 68 degrees Celsius divided by our pressure on the lowest sub table which is 25 bar minus p sat at 68 degrees Celsius that proportion then would be applied for linear interpolation to the proportion of the way we are between those internal energies so u1 minus uf at 68 degrees Celsius divided by u at 25 bar 68 degrees Celsius minus uf at 68 degrees Celsius well I know one I know 25 but I don't know anything else so we go into our tables first let's try to find the saturation pressure corresponding to 68 degrees Celsius well if I jump back into our saturation tables I want the one ordered by temperature and if I find 68 degrees Celsius I can see that the saturation pressure is oh man I don't happen to have a row for 68 degrees Celsius so I have to interpolate so that linear interpolation is going to be 68 not what I wanted let's try that again 68 minus 65 divided by 70 minus 65 the proportion of the way we are between 65 and 70 and we're saying that proportion is the same as the proportion of the way we are between the pressures so x minus 0.2503 divided by 0.3119 minus 0.2503 and we are solving that relationship for x we get 0.287 so we're saying that the saturation pressure at 68 degrees Celsius is about 0.28726 bar 0.28726 bar I mean approximately because we used linear interpolation then u at 25 bar and 68 degrees Celsius surely that's just going to be an easy look up in my compressed liquid tables I jump back to our compressed liquid tables I can see that at 25 bar I have a nice row for 68 degrees so oh man I don't so interpolation number two coming on up we are going to take 68 minus 40 divided by 80 minus 40 is equal to the internal energy at 68 degrees Celsius which I'm calling x for now minus the internal energy at 25 bar and 40 degrees Celsius 167.25 divided by the internal energy at 80 degrees Celsius and 25 bar which is 334.29 minus the internal energy at 40 degrees Celsius and 25 bar which is 167.25 and I can devalue of 284.178 284.178 okay so we are two thirds of the way through these intermediate stepping stone properties the third one then is going to be uf at 68 degrees Celsius I'm sure that'll be a nice look up and not at all an interpolation if we jump back to our saturation tables this would be temperature again and I find 68 degrees Celsius and I look for uf at 68 degrees Celsius and I see that we still don't have a row corresponding to 68 degrees Celsius in our saturation tables so I'm going to be using the same temperature proportion to interpolate that I used for the pressure so this is going to be 68 minus 65 divided by 70 minus 65 and that's equal to the thing that I'm looking for minus 272.02 divided by 292.95 minus 272.02 and I get 284.578 okay 284.578 kilojoules per kilogram okay now that I've done those three interpolations can actually get to u1 and I get here we go 1 minus 0.28726 divided by 25 minus 0.28726 and then I'm taking that equal to x minus 284.578 divided by 284.178 minus 284.578 and we get 284.566 okay so u1 is about 284.566 kilojoules per kilogram cool one interpolation done for the next one I'm going to be taking the condition at I believe it was 90 degrees Celsius yeah 90 and we're assuming it's still 100 kilopascals about so t2 is 90 degrees Celsius and p2 is about one bar still 100 kilopascals so u2 then going to be coming from our compressed liquid tables because it still has not yet reached the saturation temperature therefore it must be in the compressed liquid region so I'm going to have to set up the same interpolation that I did last time this time using one bar minus p sat at 90 degrees Celsius divided by 25 bar minus p sat at 90 degrees Celsius and I'm saying that's equal to u2 minus uf at 90 divided by the internal energy at 90 degrees Celsius and 25 bar minus uf at 90 degrees Celsius in this interpolation I know 1 I know 25 p sat at 90 I can get from my saturation tables if I jump over to the saturation tables by temperature scroll on down to 90 fortunately for me I have a pressure it is 0.7014 so I can populate that 0.7014 bar then the internal energy at 90 and 25 bar going to come from our compressed liquid tables 25 bar and 90 degrees Celsius man I don't have a row for that either so we're going to have to interpolate again that's going to be 90 minus 80 divided by 100 minus 80 and that proportion is equal to our internal energy which is what we're looking for minus the value at 80 334.29 divided by the value at 100 which is 418.24 minus the value at 80 which is 334.29 and with that I can finally get this number 376.265 and I can populate that 376.265 kilojoules per kilogram all I need now is uf at 90 degrees Celsius I'll go back to my tables navigate by temperature find 90 degrees Celsius thank fortune that I should have a row for 90 and I have 376.85 376.85 and with that I can perform the interpolation so if I set up another interpolation here this is one minus 0.7014 divided by 25 minus 0.7014 and I'm setting that equal to the value that we're looking for minus 376.85 divided by 376.265 minus 376.85 and I get a value of 376.843 so my delta u with option one is u2 minus u1 which is going to be 376.843 minus 284.8566 and we get 92.277 kilojoules per kilogram okay we got an answer how confident are we in that answer well I'm pretty confident in our math I mean we did like 10 interpolations but I'm confident that we did them correctly but think about the assumptions that we made to get to that number we assumed a pressure we assumed that the pressure was the same we assumed that we could use linear interpolation to approximate the values that means that we have n uncertainty in our result and that uncertainty may or may not be relevant to what we're doing depending on what we're doing but this number we might need a more accurate number for example if we were talking about how much energy was required to pasteurize milk that is a very precise temperature that we're trying to hit we would probably want to be accurate but the fact that this problem statement didn't give me much information implies that the precision of the result isn't super important I mean plus or minus a couple of percent is probably acceptable error so since we're already approximating by assuming a pressure and using linear interpolation it might be okay if we assume just a little bit more error in order to get a result that's close enough so the first approximation method that we will use is saying that when we're in the compressed liquid region comma and below the lowest pressure sub table we can probably get away with assuming that our properties are basically the same as the saturated liquid property at that temperature so essentially what we're doing here is neglecting the effects of pressure within the compressed liquid region close to the saturated liquid line and that logic is based on the observation that pressure doesn't have nearly as much effect on the properties in the compressed liquid region as does the temperature I mean if we jump over into our compressed liquid tables and get rid of the calculator for a second if I were to compare the internal energy at 25 bar and 80 degrees Celsius to the internal energy at 50 bar which is twice the pressure and 80 degrees Celsius they change by what a third of a percent maybe actually less than that half of a third of a percent like a sixth of a percent but if we compare 25 bar and 80 degrees Celsius to 25 bar and 100 degrees Celsius which is a proportion of temperature change that I can't do in my head I mean let's actually try that out 100 plus 273.15 divided by 80 plus 273.15 that's a difference of five percent changing the temperature five percent instead of doubling the pressure we are affecting the internal energy way more I mean that's a difference of like 80 kilojoules per kilogram so much more substantial is the temperatures effect on an internal energy than the pressures that we assume that the pressure has no effect that's the simplification that we are making in this shortcut and when we apply this shortcut then all we have to do to come up with delta u here is to say u1 is oh about the same as uf at 68 degrees Celsius and u2 is about the same as uf at 90 degrees Celsius so uf at those two temperatures is something that we can come up with relatively easily if I jump back into my saturation tables by temperature and find 68 and 90 I mean I still have to interpolate for the uf value at 68 degrees and just in the interest of comparing the amount of time spent here let's recompute that again so I'm saying this is 68 minus 65 divided by 70 minus 65 and we're calling that about the same as x minus 272.02 divided by 292.95 minus 272.02 and we get a uf value at 68 degrees Celsius of 284.578 and for the value at 90 we don't have to interpolate we can just look at it and we see that it is 376.85 376.85 now let's calculate a delta u that would be u2 which is 376.85 minus 284.578 and we get 92.272 so just for comparison here we did one interpolation instead of what like eight and we got a number that was within .005 kilojoules per kilogram do you think that's acceptable error I mean let's actually try computing that shall we that would be 92.272 minus 92.277 divided by 92.277 we have introduced an error of 0.005 percent I would argue that that uncertainty is within and the uncertainty that we had in this result in the first place because we performed a bunch of interpolations all of them were linear interpolations and we assumed a pressure don't forget that that assumption of pressure is going to have a huge result on our result if we were trying to increase the temperature of water in Colorado where the ambient pressure is going to be much lower than sea level or if we were looking at a situation like Amsterdam where perhaps it's higher than sea level that pressure is going to have a result on our answer so we can't even be that confident in this answer being the correct one the correct one being in heavy quotation marks there because we can only be as confident in our answer as we are about the information that we have in order to be able to build it that's why it's so important to list your assumptions when you're working through these problems anyway we incurred just a tiny bit of error in order to have an answer in two minutes instead of 20 minutes I would argue that that's acceptable error but wait we're not done let's go another step further let's look at how internal energy varies as a function of temperature if I were to plot that out to scale which I have done for you we can see that the internal energy of water plotted as a function of temperature looks like this so where are we on this spot oh we're we're way down here we're talking about a difference in internal energy between this temperature and this temperature now what observation can you make about that relationship it looks very nearly linear right so one of the assumptions that we can make when we are talking about approximating this result is that that relationship is in fact linear and in order to be able to describe that a little bit better let's introduce a concept and that concept is heat capacity when we are talking about heat capacity what we are describing is the amount of energy required to raise the temperature of a substance by one degree and just like how we describe specific internal energy and specific enthalpy a little bit more usefully than total internal energy and total enthalpy it is often more convenient for us to describe a specific heat capacity and specific heat capacity is heat capacity divided by mass which is the energy required to raise the temperature of a unit mass of the substance by one degree and the variable we use to abbreviate heat capacity is c so if i were to start a table here i can say heat capacity can describe a couple of different things for liquids and i should add that this is incompressible liquids and solids and the way that we define heat capacity for incompressible liquids and solids is with the derivative of internal energy with respect to temperature and when it's an uppercase value like this that's total heat capacity is the derivative of total internal energy with respect to temperature and specific heat capacity which would be in lower case c would be the derivative of specific internal energy with respect to temperature again uppercase is total heat capacity which we almost never use lower case c is specific heat capacity which we are going to use all the time and the reason it is so useful is because we can describe del u as c del t so if we were able to integrate we could determine a delta u that delta u would be the integral of c as a function of temperature with respect to temperature so option two on our list of shortcuts which is also option two on our list of options would be to determine how the specific heat capacity varies as a function of temperature and integrate there are lines of best fit based on empirical data out there for a lot of different substances you can approximate the specific heat capacity at any temperature by connecting dots that we've established based on recordings of how the temperature changes as you are adding or decreasing energy but it gets even better if we assume specific heat capacity is not a function of temperature at which point it comes out of the integral so if i jump back to this plot of internal energy and temperature for water we see that that relationship especially in the liquid region i mean here let me zoom in on that liquid region that relationship is perhaps a slight curve i mean maybe the slope of that line is a function of temperature but i would argue that it probably is close enough to being aligned that we could get away with calling it a line in most circumstances so if we assume if we assume the specific heat capacity is not a function of temperature or we could say if we assume specific heat capacity is constant then it comes out of the integral at which point i just have the integral of u is equal to c times the integral of delta t delta u is equal to c delta t and the specific heat capacity for a substance can be looked up where do we look it up you ask where do we look up anything in our tables in our tables specifically if we jump over to table a 19 we can see the specific heat capacity that would be indicated here as a cp instead of just a c for reasons that will make a little bit more sense later on specific heat capacity here for a variety of substances including but not limited to water at different temperatures is going to be oh i don't know 4.211 or 4.179 or 4.182 they do change as a function of temperature but not by much if we're willing to accept a little bit of error we could assume that it's constant it doesn't change as a function of temperature and just multiply this number by our delta t now the question becomes which value do we use the best thing to do here would be to use the value that is closest to halfway between our temperatures so halfway between 68 and 90 is going to be 79 right so 79 degrees Celsius plus 273.15 is 352.15 so yeah we could interpolate between 4.195 and 4.22 but why we are already approximating it would be pointless for us to try to be that precise about our approximation does that make sense we are incurring the error the uncertainty by assuming that it doesn't change as a function of temperature so using 4.195 five instead of 4.195 it's probably going to be unnecessary precision like we're already approximating we might as well approximate we wanted to be precise we would be precise anyway so then if i'm using 4.195 kilojoules per kilogram kelvin and i multiply by a temperature difference in kelvin which would be 90 minus 68 i get 92.29 now we got in our previous iterations 92.277 and 92.272 so again let's calculate a percent error here 92.29 minus 92.277 about by 92.277 do you think that's close enough and the answer is yeah it is for this circumstances probably just fine because we weren't that confident in this number anyway the circumstances of the problem are going to be what you use to infer how important precision is for now for our exams i will probably hint very heavy-handedly what direction i'm looking for you to go in in the real world it would depend on the circumstances of your problem what is your safety factor how precise do you need to be in this calculation what other things are you assuming and how much error are you incurring by making those assumptions anyway so assuming heat capacity is constant is our third option for evaluating delta u and delta h for our purposes in this class we are only going to be using options one or three we are not going to be using option two at all and that's because we don't have good data for how specific heat capacity of water changes as a function of temperature other than just using these data points and trying to connect them so if we're going to approximate we're going to approximate and if we're going to not approximate we are going to try to not approximate those are our two directions does that make sense so for our purposes i'm only expecting you to use options one and three and which one is better or worse in which circumstances is something that you'll get the hang of the only hard and fast rules that you need to know are option three becomes less accurate the further apart your temperatures are and you can never evaluate delta u or delta h using option three across a phase change because doing so neglects the latent energy associated with that phase change i mean if i jump back to this plot here this vertical distance right here that's the latent energy associated with going from liquid water to vapor water that latent energy would be neglected if you looked up the slope of this line and then used this temperature difference between here and here right because he would just he would tell you this vertical distance was your change in internal energy and you would neglect this huge latent energy associated with the change in phase also while we're here let me just point out look at how much bigger the latent energy is going from liquid to vapor then the sensible energy to go from zero degree Celsius to 100 degree Celsius that energy required to take liquid water at a constant pressure from zero degree Celsius to its boiling point freezing point all the way to its boiling point is what like a fifth of the energy required to change the phase once it gets to the boiling point that's why when you put water on the stove and you bring it to a boil it takes way longer to actually all boil than it does to get to boiling in the first place it's because of the fact that this is so much bigger than that sensible energy change anyway those hard and fast rules are all you need to know for now you'll get the hang of whether or not to assume constant specific heats based on the circumstances as we go forward before we go forward though let me pad out our definitions of our specific heat capacity a little bit more for incompressible liquids and solids there is only one heat capacity it is c and the book calls it cp and again the reason for that will make a little bit more sense in just a second here when we are talking about substances solids and liquids do not make up all of the substances that we analyze instead we have to also define a value for gases and for gases we do not just use one specific heat capacity we use two a heat capacity add a constant pressure and a heat capacity add a constant volume and the reason that we use those two values is to establish kind of a border case i mean if you imagine that you were heating up a rigid container the amount of energy would take to increase the energy of this rigid container would be delta u because it's at a constant volume now if you compare that to the amount of energy it would take to heat up a piston cylinder arrangement that was allowed to move up and down in order to keep the pressure constant the heat coming in would go into the delta u but it would also go into the boundary work and if we defined the boundary work as the boundary work of an isobaric process then the boundary work in this particular instance would be pressure times change in volume so if we added together the total change in internal energy plus pressure times change in volume then we would end up with change in total enthalpy and if we were describing the specific heat capacity associated with that then we would not need to describe that on a specific basis therefore the relevant parameter would be specific enthalpy does that make sense when you are heating up a process with a constant volume all of your energy goes into delta u or more of your energy goes into delta u whereas if you are heating up a process at a constant pressure some of the energy goes into the delta u some of it goes into boundary work so it appears to take more energy to increase the temperature because you're not getting all of a return on your investment does that make sense these are the two border cases we establish one for internal energy one for a change in enthalpy and those are based on analyzing a constant volume process and a constant pressure process and that's why we add this p and v to indicate that we are analyzing a constant pressure process and a constant volume process so the specific heat capacity for constant pressure is defined as the partial derivative of specific enthalpy with respect to temperature but only when the pressure is constant and the specific heat capacity for constant volume is defined as the partial derivative of internal energy with respect to temperature but only when the volume is constant that's gases and specifically that's real gases if we make the simplification for ideal gases this gets a little bit easier yet that's based on the fact that for an ideal gas we say internal energy is only a function of temperature pressure doesn't affect it it is only a function of temperature there is empirical data to reinforce that but for now we're just saying that specific internal energy is only a function of temperature and because for an ideal gas specific enthalpy is specific internal energy plus pressure time specific volume and we can make the substitution that pressure times specific volume is equal to r times t because of the ideal gas law pressure times total volume is equal to mass times specific gas constant times temperature which we could write as pressure times specific volume because we're replacing volume over mass with specific volume is equal to specific gas constant and times temperature. Now all of a sudden, we have a function of temperature, we have a constant, and we have a function of temperature. Therefore, specific enthalpy is also only a function of temperature. So for ideal gases, specific internal energy and specific enthalpy are only functions of temperature. As a side note, you may have noticed that in your property tables, you have properties for ideal air in the same way that you have properties for water. But this table is much more straightforward. There's only temperature. The reason that you are looking up specific enthalpy and specific internal energy is a function of temperature is because they're only functions of temperature. You don't have to worry about what the pressure is. There's no pressure sub-tables. We're not splitting it into different phases. It's just ideal gases. Specific enthalpy and specific internal energy is a function of temperature. Anyway, because specific enthalpy and specific internal energy are only functions of temperature, then, for ideal gases, specifically, this requirement is not necessary. And this changes from a partial derivative to a regular derivative. So it doesn't matter if the pressure is constant or if the volume is constant. Cp and Cv have to do with enthalpy or temperature. Again, for an ideal gas, Cp is defined as dH dt, Cv is defined as du dt. So if you're in a situation with an ideal gas where you have assumed constant specific heats and you need to substitute Cp delta t or Cv delta t, which one you use is not dependent on if you have a volume or pressure that is constant. It is dependent entirely on if you have a delta H or a delta U. Right somewhere into your brain, H goes to P, U goes to V. The way that I would encourage that is by remembering Harry Potter and Ultraviolet. H goes to P, U goes to V. These are our definitions of specific heat capacity. Depending on what our phase is, and if we've assumed that a gas is ideal or not, we may be able to use two different specific heat capacities and they might just be a function of temperature, or we might just have one specific heat capacity to rule them all.