 You have a big old box containing nitrogen and an electric resistance heater. The box is sitting on a table in a very large 20 degrees celsius room. The nitrogen is initially at 25 degrees celsius and 200 kilopascals, and at that instant you measure the heat loss through the walls of the box to be 150 watts. Now you plug in the heater, that's a 120 volt wall outlet, and measure the current running through the heater to be 8.5 amps. You walk away, irresponsibly leaving the heater on for a really, really long time. I want to know first how hot does the nitrogen in the box get. Then I want to know what is the pressure of the nitrogen at that point, that is once it has reached its maximum temperature. Lastly I want to know if the big old box is a cube 100 centimeters on a side, how many grams of nitrogen are in that box. I will begin with a drawing. There, with that expert drawing in place we can start to talk about what's actually happening here. So we have a rate of electrical work entering our system. That rate of electrical work is going to increase the temperature of the nitrogen. The nitrogen is rejecting heat to its surroundings. Initially, at the beginning of the process, we have a temperature difference of 5 degrees celsius. Then, as the temperature of the nitrogen increases, the rate of electrical work entering will be the same, but the rate of heat rejection will increase because this rate of heat rejection is a function of delta T. That means that the higher the temperature of the nitrogen gets, the closer q dot out is to the rate of electrical power entering, and eventually the temperature of the nitrogen will level off. At that point, the rate of electrical work entering is going to be the same as the rate of heat transfer out. That means all of the energy coming in is going back out again. So that leads me to my first question. Is this a steady process or a transient process? You could argue that it's both depending on which side we care about. If the temperature of the nitrogen levels off, starting at 25 here, then the point at which it reaches maximum temperature, we could draw here, and this side of the analysis would be steady, this side would be transient. And our actual analysis, in order to answer the question, is going to be to look at this process and consider this point. So once we reach a point at which the rate of heat rejection is the same as the rate of electrical work in, that's the temperature that we actually want. So this is the answer to part A. Does that make sense? So we're analyzing the transient process, but we are actually using what represents steady state as the threshold between the transient part and the steady part. That's how we define our ending state point. By the way, that's what it means by a very, very long time. We are allowing the nitrogen to get as hot as it possibly can. Furthermore, the fact that the room is described as very large implies to us that there is a lot of air in the room. In reality, as the nitrogen is heating up and rejecting heat to the air around it, the temperature of the air is also going to be increasing. By treating the room as being very large, we are saying that there's so much air relative to the nitrogen that it can absorb all this heat and not change its temperature a significant amount. So, we are making the assumption that the temperature of the air in the room stays the same the entire time. Next question, is this an open system or a closed system? Right, I'm defining my system as the mass of nitrogen in the box, and I wasn't told enough information to deduce that there is any nitrogen entering or leaving the box, therefore I'm calling this a closed system. So, let's start to get into it and see what further assumptions we have to make. If we are defining our process as starting here and ending here, then our end state point is going to be at the transition from transient analysis to steady analysis. So, it's the very beginning of the point at which it's steady state. So, if I look at my energy balance and I simplify it for steady state analysis, I'm going to divide everything by dt, at which point I have dedt is equal to e.in minus e.out. And for steady state, nothing can change with respect to time. Therefore, this is zero. Therefore, the rate of energy entering must be the same as the rate of energy exiting. Then I can write that as q.in and work.in because I have a closed system, which means that energy can only cross the boundary as heat transfer or work. And I recognize that I have electrical work entering, but no opportunity for work out. And I have heat rejected, but no opportunity for heat transfer in. So, I'm calling this work in is equal to q.out on a rate basis. I will also point out while we're here that depending on how you define your system, you could actually call this a heat transfer in as well if you wanted to. And like you could say that all of the work, the electrical work is being converted into heat transfer. So, if you wrapped your system around the heating coil, then it's actually being converted from work to heat transfer before it enters your system. So, it's entering your system as heat transfer, but that doesn't actually matter over here. Whether you call it a q.in or a work.in, the fact that the energy entering is then leaving is the important part. Not how you note it. Anyway, this rate of work entering is going to be electrical work. And I recognize that for our purposes, we are calling that voltage times current because we are hand waving any power factor here because it's all just turning into heat transfer. So, I'm taking 120 volts multiplied by 8.5 amps. And when I get out, it's going to be treated as a quantity in watts. So, if I pop up the calculator and I take 120 times 8.5, I get 1020 watts. And this is all once maximum temp has been reached. Therefore, q9 out is equal to 1020 watts as well. Get rid of this so I have more space to work. Okay. So, see if you follow this logic. I'm saying heat transfer out is a linear function of delta t. So, whatever the heat transfer out is at 5 degrees of delta t, which was 150 watts. Yep. And then at 10 degrees of delta t, it would be 300. Right? Because double the delta t means double the heat transfer out. And 20 would be 600. So, what would the delta t have to be for the rate of heat rejection to be 1020 watts? Why don't you pause the video and try that on your own? So, I'm saying 5 over 150 is equal to question mark over 1020. Therefore, question mark is equal to 5 times 1020 divided by 150. Therefore, question mark is going to be 5 times this proportion, which is 34. So, does that mean my temperature for part A is 34 degrees Celsius? No, it does not. That's a temperature difference between the nitrogen and the surroundings that represents maximum temperature. That is the temperature at which the rate of heat rejection is the same as the rate of electrical work in, which is the point at which the temperature of the nitrogen is going to stop increasing. It's the point at which it levels off. That's a delta t between the temperature of the nitrogen and the ambient temperature. Therefore, my temperature of my nitrogen at maximum temperature is 54 degrees Celsius. And since I'm defining my process as starting at the initial conditions, ending at the point at which we transition from transient analysis to steady analysis, I'm going to call that state 2. Therefore, I'm describing my answer as t2. Now, a common question I would get at this point would be, hold up, John. How come it's 54 and not 59? Did the temperature of the nitrogen start at 25 degrees Celsius? And every other time in the past, delta t has been t2 minus t1. t1 is 25, so 25 plus 34 is going to be 59, 54. You're right, but we're not describing the temperature change across the process. We're describing the temperature difference between the nitrogen and the ambient conditions. So we have to add this temperature, not the initial temperature of the nitrogen. Okay, then part B. I want to know what the pressure of the nitrogen is once it has reached maximum temperature. For this, I'm going to be treating the nitrogen as an ideal gas, so I will add that to my assumptions. I'm not sure why I'm writing the word assumptions. Let's try that again. Nitrogen is ideal. Then if the nitrogen is ideal, I can write this as pressure times volume is equal to the mass of the nitrogen times the specific gas constant of nitrogen times temperature. I can relate the pressure and the volume of the nitrogen to the mass, the specific gas constant and the temperature. And then I recognize that for this process, I have a rigid container. Therefore, the volume is constant. It doesn't actually say that the cube is rigid, but we were given no indication as to a change in volume or anything that could indicate a change in volume. Therefore, I will actually write that as an assumption as well. And since I'm running out of assumption room, I'm going to move my energy balance. So instead of writing volume as constant, I'm going to be a little bit lazy and write this as delta V is equal to zero. Then I have a closed system, which means that I'm assuming that we are analyzing the same mass of nitrogen the entire time. Therefore, mass is also constant. And then the specific gas constant by definition is going to be constant as long as what we're analyzing is still nitrogen. Therefore, I could say the proportion of pressure over temperature, which is equal to mass times specific gas constant divided by volume is itself constant. Then I could say P1 over T1 is equal to P2 over T2. Therefore, P2 is going to equal P1 multiplied by T2 over T1. Then I was described a pressure at state one of 200 kilopascals. I know the initial temperature is 25 degrees Celsius. So I can populate those down here. 200 kilopascals. And then I'm multiplying by 25 plus, excuse me, that's the initial temperature, 25 plus 273.15 in order to convert that from a relative temperature to an absolute temperature. And then I'm saying T2 is 54 plus 273.15. And then calculator, you can do that for us. 200 times 54 plus 273.15 divided by 25 plus 273.15 yields 219.453 kilopascals. Therefore, the highest pressure experienced by the nitrogen in the box will be 219.453 kilopascals. That is the pressure corresponding to the highest temperature experienced by the nitrogen. By the way, another common question here is how do we know that's an absolute pressure and not a gauge pressure? We really don't. We don't know explicitly, but the rule of thumb we use is if we don't have enough information to determine otherwise, the default is assuming that it's an absolute pressure. So all pressures are assumed or treated as being an absolute pressure unless there's something that indicates otherwise. That indication would likely involve a description of the gauge. So I could say a gauge on the side of the box reads 200 kilopascals, or I could say that a manometer indicates that a pressure difference across the height of the fluids is going to be 200 kilopascals, that sort of context glue. The fact that we have nothing like that indicates to us that we're treating this as an absolute pressure. So the absolute pressure at the beginning was 200 absolute pressure at the end is 219.45. We could write that down as an assumption if we wanted to, that would generally be good practice. If you're not sure whether or not you should list something as an assumption, it's probably important too. But because that's such a the commonplace analysis, again, I'm writing the word assumption. Why am I doing that? I can't do two things at once. I can't talk and speak. Talk and speak. I can't talk and write. That's absolute pressure. 200 kilopascals is absolute. Okay. Then part C. If the big old box is a cube 100 centimeters on a side, how much nitrogen is in the box? Well, for that, I'm going to use the same ideal gas law. Except this time, I don't have the luxury of describing a proportion. So I'm going to have to calculate the specific gas constant. Remember that the specific gas constant of a substance is going to be the universal gas constant divided by molar mass of that substance. Therefore, the mass is going to be pressure times volume divided by specific gas constant times temperature, which I'm going to write as pressure times volume divided by universal gas constant divided by molar mass divided by temperature. And I could plug in either state one properties or state two properties. It doesn't actually matter. I could plug in P one, V one and T one, or I could plug in P two, V two and T two. Generally speaking, you should use as few of the values that you calculated as possible. So as to try to limit the effect of any errors that you may have made, or compounding rounding errors or the uncertainty inherent in making a calculation with a value that you have some uncertainty in anyway, I mean the 200 kilopascals is going to be plus or minus some value because no pressure measurement is actually perfect. So I'm going to use P one, V one and T one just to try to limit uncertainty. So I'm going to plug in the pressure at state one, which was 200 kilopascals. And then I'm multiplying by the volume of the cube. The volume of a cube is going to be the side length multiplied by the side length multiplied by the side length. That's side length cubed. So 100 centimeters cubed. Then I divide by the universal gas constant, which I'm going to get from the inside of the front cover of my textbook. That's the calculator. It's not what I want. I want this. And the inside of the front cover indicates that the universal gas constant in metric units is 8.314 kilojoules per kilomole kelvin. And then we need the molar mass of nitrogen. And for that, I will go back to my textbook this time navigating all the way back to the appendix, specifically appendix one, which is going to have metric units. And I want table A1. I'm going to find diatomic nitrogen. And I see that its molar mass is 28.01 kilograms per kilomole. Then lastly, I'm dividing by temperature, which is 25 initially. Yes. Yes. So 25 plus 273.15 kelvin. Then in order to get an answer in grams, I'm going to have to convert this mass dimension into grams. So I can say one kilogram is 1000 grams. And then in order to get everything else to cancel, I'm likely going to have to break apart kilopascals, kilojoules, and convert cubic centimeters into cubic meters, but one thing at a time kelvin cancels kelvin, kilograms, cancels kilograms, kilomoles, cancels kilomoles. So I have kilopascals times cubic centimeters times grams divided by kilojoules. So I will break the kilojoule into its components, which instead of going back into joules and the Newton meters, I'm going to write instead as 1000, excuse me, one kilonewton times one meter. Similarly for kilopascals, I'm going to write one kilopascal as a kilonewton per square meter, recognizing that the kilonewtons will cancel, meaning I don't have to cancel to 1000s and save a little bit of space. Kilopascals, cancels kilopascals, kilodjoules, cancels kilodjoules. Then I'm left with cubic meters in the denominator and cubic centimeters in the numerator, which means that the last conversion here is one meter, contains 100 centimeters, and then I cube everything. One cube is boring. Cubic centimeters are going to cancel cubic centimeters, and meters and square meters cancels cubic meters. And then I want enough room to be able to actually write an answer, so I will bring this all up a little bit. And then let's put our calculator to work here. So first up we have the wake-up button, and then 200 times, I'll add a leading parentheses, 200 times 100 cubed times 28.01 times 1000, 1001 John, it's not the same. And then I divide by 8.314 times the quantity 25 plus 273.15 times 100 cubed. Note that if I had been paying a little bit more attention, I could have just canceled the 100 cubed. But here we are. So I have 2259 grams of nitrogen, and that's that problem. I will point out that analyses like what we did up here, where we were making a deduction based on how we understood the actual mechanisms to work, I think is the hardest part for people wrapping their head around thermodynamics for the first time. It's not just a matter of applying the same algorithm every time. You have to use the understanding of what actually affects the energy to deduce what's going to happen, which means that you have to have a little bit of a deeper understanding of what's happening. I would encourage you to get into the habit of asking yourself why you're making the steps that you're making instead of just how to proceed.