 A student living in a 3 meter by 5 meter by 2 meter dormitory room turns on her 10 watt fan before she leaves the room on a hot summer day, hoping that the room will be cooler when she comes back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat transferred through the walls and the windows, determine the temperature in the room when she comes back 10 hours later. Assume the specific heat capacity is constant, and assume the room to be at 100 kPa and 20 degrees Celsius in the morning when she leaves. I will begin with a system diagram indicating that it's perfectly insulated with our fuzzy line, which show that it's adiabatic. I have a fan that is blowing air around, and that fan is causing energy to enter the room. And depending on how you define your system, that energy could appear as a couple of different things. In my case, I'm going to define my system as the volume of the room, and I'm going to assume that no air enters or exits that, so that's both a control mass and a control volume. And I'm going to qualify the energy entering as an electrical work. So I'm going to call it a work in. Then I'm going to establish some assumptions. We were told to assume the heat capacity of air is constant. That doesn't need to go in my assumptions list. I was told to assume that the initial temperature and pressure were 20 degrees Celsius and 100 kPa, so I don't need to list those in my assumptions list. But I will assume that the only thing the student has in her dormitory room is air. So room contents are just air. Even the fan takes up zero volume and contributes zero mass. Somehow. Two, I'm going to say that the volume of the room doesn't change. So as the temperature changes, I'm assuming that the expansion or contraction of the air is not, you know, deforming the walls in any way. They are relatively rigid, or at least rigid enough to support that minor change of pressure. I am assuming that the, I am simplifying it a little bit by treating it as a closed system. So I'm assuming that no mass of air escapes under the door or around the gaps in the door or around gaps in the window or through any vents in the ceiling or anything, just the same mass of air the whole time. Then I was again told to neglect any heat transfer. So I don't need to list adiabatic as one of my assumptions. I don't have to account for heat transfer through the walls and windows. All I have is a perfectly sealed and perfectly insulated room and that energy is entering in the form of electrical work. So I guess closed system is kind of implied by all the doors and windows are tightly closed, but we listed it anyway. Now, if I set up an energy balance on this room, I can say, first of all, that I am treating this as a closed system. Therefore, E n could be Q in or work in and E out could be Q out or work out. Then I have a transient process. Hold on, let me pose that question to you. Hey guys, shouldn't I treat this as steady state or not? No, I should not. I can't neglect the effects of time because we have a duration. The presence of a duration strongly hints to us that we are analyzing a transient process. Therefore, because it's transient, I'm going to include the left hand side as delta U plus delta KE plus delta PE. Then I can begin to simplify. I was told to treat this as perfectly insulated. I can reasonably neglect any work out. The only work out that really would occur would be boundary work, I guess, if we were, like, deforming the walls and changing the volume of the air. But I'm not. Then I'm going to neglect changes in kinetic and potential energy of the air as well. Now, I know what you're thinking. You're thinking, John, but isn't the fan changing the kinetic energy of the air? And you're right, it is. But all of that change of kinetic energy is happening internally. It's on the inside of our system. The system as a whole, the average kinetic energy of the mass of the air in the room, is not changing. So that means we have delta U is equal to, excuse me, that's a big U. Delta big U is equal to the work entering. And I could write work in as specific internal energy if I were to incorporate mass by saying M2U2 minus M1 times U1. But because I've assumed it's a closed system, the mass is going to come out. And I'm left with mass times specific internal energy at state two minus the specific internal energy at state one. I have a delta U, which means I'm faced with a decision. Do I use option one and look them up or do I use option three? Well, we've been told to assume the specific heat capacity of the air is constant. So we are going to be deploying option three. The question really becomes, do we use Cb or Cp? Well, remember, variety of gases like this air, which by the way, I could write explicitly as an assumption. It doesn't matter if I have a constant pressure or a constant volume process. Cp is defined as dH dt, regardless of if it's constant volume or constant pressure. And Cv is defined as du dt. Therefore, which one I use is just dependent on if I have a delta U or a delta H. So we can write this as the integral of du is equal to the integral of Cv dt, which because Cv is constant, comes out of the interval and we're left with delta U is equal to Cv delta T. Therefore, my work is going to be mass times the Cv of air times delta T. Delta T is what we are looking for because I want T2. And on the electrical work side, I don't actually know the electrical work, but I do know the rate at which the work is entering and a duration. So instead of running work, it might be more convenient for me to write electrical power and duration. I recognize that electrical power is going to be the magnitude of work accomplished over the process divided by the duration. Therefore, I can say work is equal to electrical power times duration. So in this equation, I know T1 delta T and power so far. I'm looking for T2, which means I need to come up with the Cv of air and the mass of the air. For the mass of the air, I'm going to use our ideal gas law. I know you've grown very dependent on the property tables, and that's all well and good. But for ideal air, we don't need a property table to tell us what the mass is. Because we've assumed it's an ideal gas, we can treat it as an ideal gas for the purposes of calculating a mass. The mass of the air, we could write as the pressure times the volume divided by the specific gas constant times temperature. Which pressure and which temperature to use doesn't really matter as long as I use them consistently. So if I plug in state one for pressure, I have to plug in state one for temperature as well. Because I don't know T2, I'm not going to use state two properties, I'm just going to use state one properties. The volume of the room is going to be 3 meters by 5 meters by 2 meters, because we assumed that the only thing in this room is air. So we don't have to subtract any volume dedicated to furniture or people or the fan or whatever. So I have electrical power multiplied by duration is equal to E1 times the volume of the room divided by specific gas constant of air times T1 times CV times T2 minus T1. So next is going to be calculating a couple of these intermediate quantities. And again, I'm going to do this all symbolically because I prefer to only perform the arithmetic once. So instead of plugging in the specific gas constant for air, I'm going to recognize that it is the universal gas constant divided by the molar mass for air. The universal gas constant comes from the inside of the front cover over our textbook. That's 8.314 kilojoules per kilomole Kelvin in the case of the metric unit system. For the molar mass of air, I'm going to be using table A1, which is going to show us that our molar mass is 28.97. Then the volume of the room is going to be 2 meters times 5 meters times 3 meters, pressure we know, temperature we know, CV of air, we can look up. For looking up the specific heat capacity of ideal gases, we have two tables to refer to. We can look at table A20 and table A21. A21 is the heat capacity as a function of temperature using line fit data for a variety of ideal gases. If we needed one of these gases specifically, we could use this if we had to. But for most of these substances, for most of these substances we're going to analyze, we have table A20, which gives us the specific heat capacity, CP, the specific heat capacity, CV, as well as k. By the way, k is just the proportion of CP over CV, which is useful in certain circumstances. We can formalize that if you like by recognizing that k is CP over CV. And while we're here relating these quantities to one another, let me take you aside and talk about our H value. H we define as U plus pressure times specific volume. And if I wanted to, I could plug in RT for ideal gases and write this as H is equal to U plus R times T. Because PV is equal to RT for ideal gases, again referring to the specific volume, not the total volume. And if I were to take the derivative of all three of these quantities with respect to temperature, I would have DH DT is equal to DU DT plus R, meaning for ideal gases, CP, CV and R are all related to one another. We often write R as CP minus CV. So if you're in a circumstance where you know CP and you know the specific gas constant, but you don't have CV, you can calculate it this way. Anyway, on to our actual lookup. We are looking for the CV of air. We are assuming that it's constant. And the best thing to do here would be to look up the CV of air halfway between T1 and T2. Well, if I go into this table, I can see I have CV of air at a variety of temperatures here. However, I don't know T2. Because I don't know T2, I can't actually use halfway in the middle. So the best thing to do here is to use the temperature at day one and go a little bit over it if I have to commit one way or the other, because presumably the temperature is increasing. So the temperature at day one was 20 degrees Celsius, which my calculator can help us convert into Kelvin. I take 20 plus 273.15 and I get 293.15 probably 293.15. So because I'm just grabbing a value that's close to halfway in the middle, we're going to commit up instead of committing down. I mean, it's closer anyway. So we're using the value at 300 Kelvin, which is 0.718 kilojoules per kilogram Kelvin. And that again came from table 820. So now I think we know everything. We know duration, we know power, we know pressure, we know temperature, we know temperature, we can write the specific gas constant as two quantities that we know, and we know the volume in terms of three quantities, which means at this point, solving for T2 is just a matter of doing some algebra and then some arithmetic. So if I do the algebra in my head and hope for the best, I'm going to write this as T2 is equal to and then that would be power times duration times universal gas constant times temperature one divided by molar mass of air times E1 times volume of the room times CV of air quantity plus T1. And if I draw a big horizontal line, I can start plugging in numbers. T2 is equal to, first we had the power of the fan, which was 10 watts, and then the duration, which was 10 hours, and then the universal gas constant, which was 8.314 kilojoules per kilomole kelvin. And then T1, which was 1T plus 273.15 kelvin. And then we are divided by the molar mass of air, 28.97 kilograms per kilomole. And then P1, which is 100 kilopascals, the volume of the room, which was two meters by five meters by three meters. And then multiplying by the specific capacity for air on a constant volume basis 0.718 kilojoules per kilogram kelvin. And we want to be able to add this to 20 degrees celsius, which means that we need a quantity in kelvin. So I will recognize that so far kelvin cancels kelvin, which is a great start. And kilograms cancels kilograms, kilomoles cancel kilomoles, kilojoules, and kilojoules I can cancel. So to get rid of the remaining hours watts, cubic meters and kilopascals, I will break those into their components. So first up, a kilopascal is 1000 Pascals. A Pascal is a Newton per square meter. And a watt is a joule per second. And then a joule is a Newton times a meter. So watts cancels watts, Newton cancels newtons, cubic meters cancels meters, meters and meters. And then this kilopascal cancels this kilopascal, leaving me with seconds in the denominator, hours in the numerator, which I can fix one hour is 60 minutes. And then one minute is 60 seconds. Hour cancels hour. Minute cancels minute. Seconds cancel seconds, leaving me with kelvin, to which I'm adding 20 degrees Celsius. At this point, a lot of you are thinking, but John, don't we have to convert Celsius to kelvin before we calculate an answer? But remember that when we're talking about a temperature difference, a temperature difference is the same in kelvin as it is in degrees Celsius. And what we're computing here is actually a delta T. So if we come up with an answer in kelvin, like 10 kelvin, a temperature difference of 10 kelvin is the same as 10 degrees Celsius. Because if we were talking about the difference between say 40 and 50 degrees Celsius, the difference is 10. And if I add 273.15 to both 40 and 50, the difference is still 10. Anyway, I'm going to want our good friend, the calculator and calculator is going to take 10 times 10 times 8.314 times the quantity 20 plus 273.15 times 60 times 60. And then we are going to divide by 28.97 times 100 times two times five times three times 0.718. 0.718 times 1000. And we are probably going to get a syntax error. Okay, too many parentheses. And I get 14.0607. 14.0607 plus 20 is 34. So T2, assuming constant specific heats, is 34.0607. And that's technically the answer to the question, but I don't want to be done yet. The problem told us to assume constant specific heats. And that's fine. We computed an answer like we were supposed to. But what if that hadn't been there? What if we wanted to try working through the problem with just option one? Well, for that, we wouldn't have substituted in cv delta t for delta u. Instead, our approach would be used to want to look up u1 and use this relationship to calculate u2 and then use u2 to look up t2. So in the interest of character building, let's try that, shall we? I'm going to grab out this. We're going to pull this over to a new sheet. We're going to establish an alternate universe where we had u2 minus u1 here instead of cv delta t. Then I'm still using the ideal gas law, but I still plug in pressure times volume divided by specific gas constant times temperature. And I'm still plugging in 10 watts times 10 hours on the left hand side. The difference here is that I'm now using t1 to look up u1 using this equation to get to u2 and then using u2 to look up d2. Because for ideal gases, internal energy is only a function of temperature. So I don't need temperature and pressure to look up u1, I just need temperature. So let's start there. At state one, we had a temperature of 20 degrees Celsius. Let's look up u1 corresponding to that temperature. For that, I'm going to go into my property tables. And what I want is specifically table 822. Table 822 is the equivalent of our steam tables except for air. And remember that it's only one dimensional. We only have temperature driving our lookup. So I am going to take the temperature and I'm going to look up an internal energy. And for that, I have to convert 20 degrees Celsius to Kelvin, which remember is 293.15. So if I were to take 293.15 minus 290 divided by 295 minus 290, then I could apply that same proportion to the specific internal energy, which would be between 206.91 and 210.49. So let's try that interpolation, shall we? I'm going to take 293.15 minus 290 divided by 295 minus 290. And I set that equal to what I'm looking for, which is x minus the value at 290, which was 206.91. And I divide by 210.49 minus 206.91. Then I can get an x value of 209.165. And then from that internal energy, we can compute a u2 value. Again, I'm going to use algebra. So this is going to be u2 is equal to power times duration times universal gas constant times u1 divided by molar mass air times p1 times v1. And I'm adding u1 to that quantity to get u2. So we are taking 10 watts multiplied by 10 hours. And then we are multiplying by the universal gas constant 8.314 kilojoules per kilomole Kelvin. And we are multiplying that by t1, which was 20 plus 273.15 Kelvin. And we are dividing that by our molar mass of air, which is still 28.97 kilograms per kilomole. And then I am multiplying that denominator by p1, which was 100 kilopascals. And multiplying again by two meters times five meters times three meters. And then we are going to convert some units, which means we probably need more space. I will scoot this over to the left. And so far kilomoles cancels kilomoles and pretty much nothing else happens. So I'm going to break a kilopascal into a kilonewton time, excuse me, kilonewton per square meter and kill the joule until a kilonewton times a meter. Kilonewtons cancels kilonewtons. Kilonewtons cancels kilonewtons, kilopascals cancels kilopascals. And then Kelvin cancels Kelvin. Forgot about that. Then we don't change page. We have meters, meters and meters in the denominator, which cancels square meters and meters in the numerator. And I have hours, watts and kilograms left to contend with. So the time inside of the power and the time are going to take care of themselves. And what I want is an answer in kilojoules per kilogram, because I want to add that to our value of u1. So I'm going to take a watt and write that as a joule per second and then convert 1000 joules to 1 kilojoule. And then I have time appearing in the numerator and time appearing in the denominator. So instead of running this as 60 and 60 again, I'm just going to write there are 3,600 seconds in an hour. Normally I would advocate against trying to do mental math when you're doing unit conversions. Might as well do it anyway, but I'm kind of running out of space. And that's just something that I'm assuming you guys can take for granted at this point. 60 minutes in one hour and 60 seconds in one minute becomes 3600 seconds in an hour. So this will leave me in kilojoules per kilogram because I have kilojoules over here, kilograms right here. So I take 10 times 10 times 8.314 times the quantity 273.15 plus 20 times 3600. And I divide by 28.97 times 100 times two times five times three times two times three. It's not the same thing. And I multiply that by 1000. And I get a quantity that is 10. And I add that to my 209 number. And I get a U2 value of 219.261 219.261. And that's a quantity in kilojoules per kilogram, which means I can grab T2 now because I have a one-dimensional lookup. Internal energy is only a function of temperature for ideal gases, which means I can look up a temperature corresponding to an internal energy as well. If I jump into our tables and I see 219.261 occurs between 217.67 and 221.25, meaning our temperature is going to be between 305 and 310 Kelvin. So I can interpolate for a temperature, which I will do by taking 219.261 minus 217.67 divided by 221.25 minus 217.67. And I'm setting that equal to x minus 305 divided by 310 minus 305. And I get a temperature of 307.222. And what I wanted was a temperature in degrees Celsius. So we take that quantity and we subtract 273.15. Then I would get... Great. Thank you, calculator. 34.072. So from this, which should produce the most accurate answer, I get 34.072. Compare that to what we got when we approximated the answer, which was 34.061. Do you think that difference of a 100th of a degree Celsius is significant? No, certainly not. Because in this analysis, we are introducing so much uncertainty as a result of things like assuming the room is perfectly insulated. The heat transfer is going to have so much more effect on this answer than 0.01 degrees Celsius does. It's that sort of the breakdown of what's contributing to your error that allows you to make deductions about which shortcuts, which approximations are appropriate for the problem in front of you. For now as a general rule, it's going to be pretty straightforward if you have some obvious reason that you can't assume constant specific heats, including but not limited to the fact that I say don't in that problem, then you shouldn't assume the specific heat capacity is constant. But in other circumstances where maybe you're evaluating the temperature change of air and it's relatively small temperature change and there's no phase changes or anything, it's probably pretty convenient and probably relatively realistic. Don't forget though that you need to list that as an assumption. You have to be aware of the uncertainty you are introducing. Anyway, that's this problem.