 John, like most people in the STEM fields, has an unhealthy addiction to caffeine that he could totally quit whenever he wants to. It's just that he doesn't want to. Anyway, let's say he gets up in the morning and brews a fresh pot of coffee and pours 16 ounces into a well-insulated thermos to bring along to his 8am class. When it is poured, the coffee is about 94 degrees Celsius which is way too hot for John to drink, meaning he often goes into class uncaffeinated, much to the detriment of his students. He decides to add ice to the coffee, which quickly drops the temperature to a much more reasonable level. John's freezer keeps his ice cubes at about negative 5 degrees Celsius, and they come out of the tray in 1 inch by 1 inch by 2 inch chunks called cubes, despite not being cube-shaped. The latent heat diffusion of ice in one atmosphere is 334 kilojoules per kilogram, and the density and specific heat capacity of ice at negative 5 degrees Celsius are 917.5 kilograms per cubic meter, and 2.06 kilojoules per kilogram Kelvin, respectively. How many whole ice cubes should John add to ensure that his coffee reaches an equilibrium temperature as close to 60 degrees Celsius as possible? I will begin with a system diagram. I'm drawing my system as the inside surface of the thermos, and I'm including both the ice and the coffee in my system analysis. Note that you could analyze the ice as a separate system and the coffee as a separate system, which would involve two energy balances, but it's just as easy to group everything together. I can begin my list of assumptions by neglecting the air at the top of the thermos. So my system is only ice and coffee, not air. Two. I'm going to assume that no mass escapes around our system boundaries. Under normal circumstances, that would mean that I'm treating this as a closed system, but note that in a closed system, energy can cross the boundary of our system as heat and work. In our analysis, we are treating the system as being perfectly insulated, based on the fact that it's a well insulated thermos. Therefore, we are neglecting heat transfer through the container walls. Furthermore, there are no opportunities for work, so there are no opportunities for energy to cross the boundary of our system, either. Therefore, I'm treating this not as a closed system, but as an isolated system. Now that I've got my system defined, I can begin an energy balance on the system. I begin with delta E is equal to energy in minus energy out. For an isolated system, there are no opportunities for energy to cross the boundary, therefore E in and E out are both zero, and you're probably thinking, oh great, delta E is equal to zero, that's unhelpful. But remember that we are defining our system as both coffee and ice, meaning if I were to split the delta E into both components, delta E of the ice plus delta E of the coffee must equal zero. Therefore, delta E of the ice must equal delta E of our coffee just in opposite directions. Therefore I will add a negative sign on one of the two terms. My energy change is still delta U plus delta KE plus delta PE, and if I neglect changes in kinetic and potential energy for both the ice and the coffee, then I will be left with just delta U of the ice is equal to negative delta U of our coffee. We have ourselves a delta U term for both ice and coffee, which means that we have to consider how to evaluate it. The best option here would be to look up the internal energy at the beginning and end of the process for both substances, but that isn't really an option here because A, we don't really have good ice tables for situations like this, and B, we don't have coffee tables at all. And even if we were to assume that the coffee was close enough to water, so as to use the water tables for coffee properties, we still have to contend with the fact that we only have one independent intensive property for our coffee, that's temperature. So we could approximate the change in internal energy of the coffee by using the assumption that the specific heat capacity of water is constant across that range, but for the ice, we can't just use the specific heat capacity to approximate the entire internal energy all at once. We have to split it across the phase. And I think that this is easier to get your head wrapped around by picturing it on a graph. I'm a very visual learner, so I like to plot things wherever possible. So let's look at a plot of temperature versus internal energy. If you consider water starting in the solid phase, as we add energy to the water, we are increasing its sensible energy. The temperature of the ice will increase until eventually it reaches a point where it can contain no more sensible energy without changing phase. It becomes saturated with energy, at which point additional energy will go into the latent energy of the ice. This occurs at a constant temperature because no more energy can go into sensible energy until the ice has changed phase into liquid water. Then as we add energy, we can increase the sensible energy again, until eventually it reaches a point at which it is saturated with energy again, and we cannot add energy to the sensible energy of the water anymore. We have to change phase again. And once the phase has changed into the vapor phase, we can increase the sensible energy again. And on this graph, these changes in phase should really be horizontal lines. We'll try to draw those better. And on this graph, the region in the middle here is the liquid phase. The region on the left here is the solid phase. The region on the right is the vapor phase, or gas phase. And these horizontal segments represent the latent energy associated with that phase change. Remember that for the transition from liquid to vapor, that's called latent energy energy, interesting, latent energy of vaporization, and for the transition from solid to liquid, that is called the latent energy of fusion. And then these three segments here are sensible energy terms. So when we're evaluating an energy change within a singular phase, especially something like the liquid phase, or our temperature range is going to be relatively small, because remember, this horizontal segment between solid and liquid would be at about zero degrees Celsius, if this is at a pressure of approximately one atmosphere. And this horizontal segment up here is going to be at about 100 degrees Celsius. So the temperature change of the liquid water is going to be relatively small, so it's probably reasonable to assume that the specific heat capacity of the liquid water is constant across that range. For example, if you consider the initial condition of our coffee, we are going to be right about here. I can call that state point C, let's say, and what we are ending at is 60 degrees, which is going to be right about here, which I will call X for now. C is at 94 degrees Celsius and X is at 60 degrees Celsius, meaning that the change in temperature is only 34 degrees Celsius. It would be reasonable to assume constant specific heats across that range to approximate the delta U from C up to X. That difference in delta U is going to be what we plug in for our change in energy of our coffee, but for our ice, we have a slightly more complicated situation, because the ice begins down here at negative five. If I call that state point I, then going from I up to X is not just a matter of plugging in the specific heat capacity for that temperature range. I have to analyze from I up to here, and then I have to use the latent heat diffusion to get across the phase change. And then once I have changed phase entirely, I can jump up to X using the specific heat capacity of our water. So I'm going to be using the specific heat capacity of the ice to get from I up to zero degrees Celsius, then the latent heat diffusion to get across the phase change, and then the specific heat capacity of liquid water to get from the right side of the latent energy of fusion here up to state point X. So if I were to list the assumption of constant specific heats, then I can evaluate my delta U by using C delta T. Note that if this graph were drawn to scale and we had equal masses of coffee and ice, then determining our equilibrium temperature is just going to be a matter of splitting the horizontal distance between I and C, because the amount of energy that leaves the coffee is the same as the amount of energy that goes into the ice. If I had three times as much coffee as I had ice by mass, then that means that I would meet one quarter of the way over from the coffee. Since I don't know my masses or at least my relative proportion of masses, I have to use the position of the equilibrium point to determine the relative proportion of the masses. Meaning I am likely going to end up in a situation where I'm describing the energy change on a mass proportion. Anyway, starting with my delta U of my ice, I can get from I up to zero using the specific heat capacity of our solid water, which is ice. By all right, that is mass of ice times the heat capacity of the ice times the relevant temperature change, which is zero degrees Celsius minus negative five degrees Celsius. Then I add to that our latent energy of fusion. Note that the latent energy given to us in the problem statement describes the energy on a per unit mass basis. So in order to describe a total change in energy, I have to multiply by mass. And then once we've changed phase, we can use the specific heat capacity of water to get from zero degrees Celsius up to X. Note that our mass is still the same. So I'm still going to be calling it the mass of ice. And then this temperature would be Tx minus zero degrees Celsius. I guess to be consistent here, I will change negative five to Ti. So that's the change in internal energy of the ice going from I up to X. And that's equal to the change in internal energy from C to X. Remember that we describe the delta as n minus beginning, which would be mass of water times the specific heat capacity of the water times Tx minus Tc. But because of the negative out front, I'm going to actually write that as mass of water times the heat capacity of water times Tc minus Tx. So my energy balance leaves us with an equation which relates the mass of the ice and the specific energy change of the ice. Note that it's still just the specific energy change. It just happens to be written across three terms to the mass of water and the specific energy change of the water. Again, water here referring to coffee. So if I wanted to, I could write the mass of ice divided by the mass of water is equal to C water times delta T divided by C ice times delta T plus latent heat diffusion plus C water times delta T. The mass proportion is equal to the energy proportion. The same logic holds, we're just writing it a little bit differently. Speaking of writing this differently, I want the mass of ice for now. So I can solve this for mass of ice. And then we can start to think about what we're going to be plugging in here. I have enough information to determine the mass of the water. Can look up the specific heat capacity of water. I know Tc, I know Tx. I know C ice because I was given C ice. I know zero, I know Ti. I know the latent heat diffusion because I was given the latent heat diffusion. I can look up the C water and I know Tx and zero. So to be able to continue here, I have to look up the specific heat of water and I have to determine the mass of water that represents the coffee. So the specific heat capacity of water will come from table A19. So if we jump into our property tables, we can see on table A19 a specific heat capacity for liquid water as a function of temperature that's going to be our first term here. Furthermore, remember that because we're talking about ice and water as incompressible substances, we are describing their specific heat capacity as a single term regardless of Cp or Cv. The best course of action here would be to use the specific heat capacity that was halfway across our range or at least approximately halfway across our range. In this case, because we're talking about two different ranges, I could use one value for C down to X and then a different value for zero up to X. But since I'm already approximating, I'm just going to use a single value for all of water and I'm going to use the property that is halfway between C and zero degrees Celsius. So halfway between 94 and zero is going to be 47. 47 plus 273.15 is going to be about 325 Kelvin. I could determine a closer value, but again, we are already approximating. So being extra precise on those sorts of approximations is a little bit moot. So at 325 Kelvin, I have a specific heat capacity of water of 4.182. Again, we are incurring error by treating the coffee as just water anyway. So being hyper specific about using 4.181 versus 4.182, etc. is going to imply that we are more confident in our result than we actually are. So 4.182 has per table a 19 and then our mass of coffee, which is mass of water, is going to be calculated by taking our density of water times the volume of our water. The density of water will come from table a 19 as well. Because we're already assuming the properties of a saturated liquid at our temperature, which is what this table is, by the way, saturated liquids. Therefore, 325 Kelvin has a density of 987.1 kilograms per cubic meter. And the volume of water I know because that was given as 16 ounces, which is 473.2 milliliters. At this point, I have everything I need to be able to compute an answer. And before I start plugging stuff in, I will open up a new page. So the mass of ice will be equal to 987.1 kilograms per cubic meter times 400 and 73.2 milliliters multiplied by 4.182 kilojoules per kilogram Kelvin multiplied by 94 minus 60. Again, remember that a change in degrees Celsius is going to be the same as a change in Kelvin. So I'm just going to write Kelvin here. Because if you take 94 minus 60, you get 34 and if you take 94 plus 273.15 minus 60 plus 273.15, you still end up with 34. In this relationship, Kelvin cancels Kelvin and milliliters will cancel cubic meters if I convert. A cubic meter is 1000 liters and a liter is 1000 milliliters, meaning liters are going to cancel liters, milliliters are going to cancel milliliters, cubic meters are going to cancel cubic meters, leaving me with kilojoules in the numerator. Then in the denominator, I'm going to begin with the specific heat capacity of the ice multiplied by the temperature change of the ice. The specific heat capacity of the ice was 2.06 kilojoules per kilogram Kelvin and the change in temperature of the ice was zero minus negative five, which I am surely going to write incorrectly on my calculator. So I'm just going to write this as zero plus five to hopefully avoid that issue. Kelvin cancels Kelvin. Then I add to that our latent heat of fusion of ice, which was 334 kilojoules per kilogram of ice. Then we add to that the specific heat capacity of water. We are still using 4.182 kilojoules per kilogram Kelvin multiplied by 60 minus zero Kelvin. Kelvin will cancel Kelvin again, meaning that I have kilojoules in the numerator and kilojoules per kilogram in the denominator. The kilojoules in the numerator will cancel the kilojoules in the denominator, meaning that my answer will be in kilograms. So if we pop up the calculator, we can begin this computation. I'm taking 987.1 times 473.2 times 4.182 times the quantity 94 minus 60 divided by the quantity 1000 times 1000. And then that entire quantity is divided by 2.06 times five plus 334 plus 4.182 times 60. And we get 0.1110222. And let me just scan through this to make sure that it looks like the division went correctly, which is good. So we get a mass of ice of 0.111022 kilograms. Now is that the answer that the question wants? No, I did not ask for mass of ice required. I asked for a number of ice cubes. So I need to determine how much mass there is in an ice cube that is one inch by one inch by two inches of ice that has a density of 917.5 kilograms per cubic meter, which means that I will take the density of ice times the volume of a cube, which will be 917.5 kilograms per cubic meter multiplied by one inch by one inch by two inches. And then in order to get kilograms, I need cubic inches and cubic meters to cancel, which means that I'm going to refer to my conversion factor sheet. And on my conversion factor sheet under volume, I see a conversion from cubic meters to cubic feet and cubic inches to cubic centimeters. So I will have to take this in two steps. I guess I could also convert from inches to feet and feet to meters. And I don't happen to have inches to meters directly, so six of one, half a dozen of the other. Two steps here, I will begin with conversion from cubic inches to cubic centimeters. That's 16.387, 16.387 cubic centimeters per one cubic inch. And then there are 100 centimeters in one meter, which one cubed yields cubic centimeters and cubic meters, giving me kilograms. So 917.5 times one times one times two times 16.387 times one cube, which is one divided by 100 cubed. And I get 0.030048. Then to determine number of cubes, I will take the mass of ice required divided by the mass of each individual cube. So 0.11122 divided by 0.030048 yields 3.69. Now is that the answer to the question? No, it is still not, because I did not ask for a fractionary answer. I asked for how many whole ice cubes should John add to ensure that he reaches as close to 60 as possible. Therefore, I will say four. John needs to add four ice cubes to make his coffee drinkable and as close to 60 degrees Celsius as possible. And I will point out here that all of this data is empirical. I actually measured the temperature of my coffee when it comes out of my coffee maker, and it is 94 degrees Celsius. My freezer does maintain about negative five degrees Celsius, and my ice cubes from my ice cube tray are one inch by one inch by two inches. And my ideal coffee temperature, equilibrium temperature, was determined by taking a series of tests, wherein I left a thermometer in my coffee and drank it at a couple of points until I found a good compromise between warm enough to still feel warm, but cool enough so as to not burn my tongue or slow down the ingestion of caffeine to any noticeable degree. I want to be able to drink my coffee and drink it fast, not just sip it slowly over time. If you want to come up with a calculation for yourself, you may want to determine an equilibrium temperature. Maybe take some measurements of your freezer and your coffee maker and try this on your own. Or maybe you guys can just drink coffee directly out of the coffee maker like some sort of Targaryen descendant. I don't understand it, but there we are.