General comments
I thought the lecture went well, although I wish I was clued-in enough to notice the strange occurence with the coffee cup experiment (see below). It seems that limiting the active elements of the lecture to 2 PRS questions and 1 turn-to-your-partner problem is about right. It leaves plenty of time for me work work things out on the board.
Responses to 'Muddiest Part of the Lecture Cards'
(62 respondents)
1) I suspect some foul play in the coffee cup experiment. Ideally it should have taken 3 minutes and 20 seconds and it took 2 minutes and 57 seconds. How can the actual time be less than the ideal (no heat loss) estimate>. (14 students) Why didn't somebody mention this in class!!!??? I transposed the two numbers in my head, so it never dawned on me that something was amiss. So how did this happen? My guess is that the thermometer was too close to the heater. In a more careful experiment, one would use a small mixer to make sure the temperature was everywhere the same. In short, what went wrong was that the system was not in thermodynamic equilibrium (one temperature, measured one place did not define the state of the system)! It could also be that even though the heater was rated at 125W that it was delivering more power, though this is less likely. Chris wrote up a nice discussion of the experiment that fills in some of the details.
2) Otto Cycle Questions (15 students) Why can we leave out a step in the Otto Cycle? We don't leave out a step. We replace two strokes of the process (piston moves up and pushes the hot exhaust gases out, then piston moves down and pulls in a fresh mixture of fuel and air) with a single constant volume cooling process. The animation shows the path the actual process takes--you will note that the exhaust phase does not produce any work and it doesn't expend any fuel energy so it is common to simplify it as we have done in class. Does the process 3-4 have anything to do with heat? Yes it does. It represents the waste heat that is expelled to the atmosphere. I am not sure how you got engine efficiency in the engine process--wouldn't you just measure how much gas you use? Yes and no. Efficiency is defined as work output per fuel burn. So I need two quantities. We did use the fuel burned [cv(T3-T2) is Joules of fuel burned], but we also needed to find the work. Why is the expansion adiabatic and not isothermal? The time it takes for the piston to move from the top of the cylinder to the bottom is much shorter than the amount of time it takes to transfer appreciable heat from the walls, so adiabatic is a good assumption. Why is w3-4 negative--the piston is expanding, isn't it doing work on the surroundings? Yes it is. I think I had the signs correct on the board, but maybe I made a mistake. They are correct in the notes. How can the engine cycle be quasi-static? When is something not quasi-static (e.g. instantaneous change)? (1 student). Read this and this. It depends on what the system you consider is and how long it takes to come to equilibrium. For example, it is fair to say that the teacup we measured in class what not in thermal equilibrium when taken as a whole, but individual chunks of water within the teacup were in equilibrium.
3) I am a little confused on cv and cp (6 students). First read this again. Now let me add a few of the more subtle points. The quantities u, h, and T are all properties of the system and thus are functions of the thermodynamic state of the system (not the manner in which you got there). So the expressions du=cvdT and dh=cpdT are valid at any point in any ideal gas process (it doesn't have to be constant volume or constant pressure) as long as it is possible to define the state of the system. (That is, you must at least be able to assume that it is in quasi-thermodynamic equilibrium at the point you are considering.)
4) I am still a bit confused as to why cv<cp (even after your explanation) (4 students). First, it is only true for a gas. For solids and most liquids cp is approximately equal to cv. The difference is that gases are compressible--that is, they change their specific volume relatively easily. So when we add heat to a gas it will expand unless we constrain it. The expansion takes some of the energy away from internal energy to do work on the surroundings. When we constrain the volume of the gas and do not allow it to expand, it therefore takes less heat addition to increase its temperature by 1K. so cv<cp.
5) I am confused as to when you have to use g in relating the states of a system before and after a change. Why does p1v1/p2v2=RT1/RT2 not always work? (1 student) At any point in a process with a thermally perfect gas where the thermodynamic state can be defined pv=RT. However, often you will be given all the information at one state (say state 1) and only one thermodynamic variable at a second state, but you will be told that the process between the two states is quasi-static and adiabatic. Then you can use the fact that pv^g is constant for a q-s adiabatic process to get the second property you need to fully define state 2.
6) Still having trouble with adiabatic versus isothermal when making a p-v diagram. (1 student) Just remember, the two processes are very different. The shape of the two curves are similar on a p-v diagram. However, the adiabatic curves are steeper. Typically which one is relevant for a particular engine will be specified in the problem statements.
7) Is a gas fuel mixture an ideal gas? What can be approximated as an ideal gas? (1 student) Gas-fuel mixtures do behave as ideal gases. It is important however to take account of the change in cv and cp with temperature when doing combustion problems.However, in this class we will assume that cp and cv are constant to simplify the analysis. A complete discussion of the limitations of the ideal gas law is contained in S, B, and VW.
8) Not completely clear on how to use the First Law to solve-- how/where to plug in which values (1 student). This is one of the primary objectives of the course. But it will take practice. All of the homeworks from here on out and most of the discussions in class will focus on this topic (i.e. you have many more opportunities for practice on the way).
9) No mud (17 students). Good.