 A closed system of mass 10 kilograms undergoes a process where there is an energy transfer by work from the system of 0.147 kilojoules per kilogram. An elevation decrease of 50 meters and an increase in velocity from 15 meters per second to 30 meters per second. The specific internal energy decreases by 5 kilojoules per kilogram and the acceleration of gravity is measured to be 9.7 meters per second squared. Determine the heat transfer occurring during the process in kilojoules and indicate direction. So first, let's parse out what we know. The mass is 10 kilograms and it would be reasonable for us to treat that as a constant mass. So I'm going to say my first assumption is to treat my mass as being constant. Therefore, I'm calling this a closed system. Without assumption in place, that mass is the same it wanted to, so I'm calling it just a mass. The next information that we were given was that there was energy transfer by work from the system of 0.147 kilojoules per kilogram. So, is that a work-in or a work-out? You're right, it is a work-out because it's from the system. Next question, is it an uppercase W or a lowercase W? It's right, it's specific work. So this is specifically specific work-out and that is 0.147. And the reason that we know it's a specific work is because we were told an energy per unit mass. Then there is an elevation decrease of 50 meters, so I'll write that as delta Z is equal to negative 50 meters and an increase in velocity. Note that we can't just write that as delta V because the kinetic energy terms are squared and A squared minus B squared is not the same as A minus B quantity squared. So V1 and V2 are 15 and 30 meters per second respectively. Then next I know that the specific internal energy decreases by 5 kilojoules per kilogram, so little delta U because it's change in specific internal energy is negative 5 kilojoules per kilogram and gravity is 9.7 meters per second squared. So this example is supposed to combine a bunch of different quantities that all appear in the energy balance and make you solve for one other property. So we're bringing everything together, the entire kitchen sink into our energy balance and solving for one thing. So I will perform an energy balance on my system. I begin with delta E is equal to E in minus E out. Now we know that we have a transient process because we are told that the properties of this process are changing and these properties are describing the same mass, so they must be changing with respect to time. So I will include the left hand side of the equation delta U plus delta KE plus delta PE and E in because it's a closed system could be Q in and or work in and E out could be Q out and or work out. And then our next step is going to be to eliminate the terms that aren't relevant to my problem. First of all, can I eliminate delta U? I can't because I have a change in specific internal energy. Can I eliminate my change in kinetic energy? No, because I have velocities at the beginning and end of the process. Can I eliminate my change in potential energy? No, because I was told in elevation change. Can I eliminate heat transfer? No, because I'm solving for heat transfer, but I can eliminate one heat transfer and the logic here is basically I'm solving for one of the heat transfers and if I get a negative number, that means my direction is wrong. So I'm just going to choose arbitrarily a Q in or a Q out. You could think of that like calculating a net Q in or a net Q out. Whatever the case though, we are only solving for one heat transfer and I'm going to choose, let's say Q in. Do I get rid of my works? No, because I was told I have a work occurring. I have a work out, so I'm going to eliminate work in. There could be a situation where I have more than one opportunity for heat transfer and more than one opportunity for work, but if the situation was complex enough to warrant that, the problem would tell me about it. The fact that we weren't told about any other opportunities for work implies that there's only one work occurring and the fact that we only have one opportunity for heat transfer is a good indication that we're only going to include one heat transfer term. If we don't know enough to determine otherwise, we assume the simplest case. So I have delta U plus delta KE plus delta PE is equal to Q in minus work out. And then I am going to solve for heat transfer because that's what the problem actually wants. Therefore, Q in is equal to work out plus delta U plus delta PE plus delta KE. And then I can just start plugging stuff in. For work out, true or false, I'm just going to plug in 0.147 kilojoules per kilogram. False. That's a specific work out and this equation is total work out. So in order to write this in terms of things that I know, I have to recognize that specific work is defined as total work divided by mass because that's what we use for specific properties. Therefore, instead of writing total work out, I could write mass times specific work out. Then for delta U, am I going to plug in negative 5 kilojoules per kilogram? No, for the same reason. This is delta big U. This is delta little U. So in order to write a total internal energy in terms of what I know, I'm going to recognize that delta little U would be delta big U divided by mass as well. So I can write mass times delta little U specific internal energy change and that quantity I know. Then for delta P, I'm going to recognize that I'm taking the potential energy at the end of the process minus the potential energy at the beginning of the process and our total potential energy is represented as mass times gravity times height. Therefore, I would have mass 2 times gravity 2 times height 2 minus mass 1 times gravity 1 times height 1 and mass is the same, gravity is the same, so they factor out. So I'm left with mass times gravity times the change in height, which I'm going to call delta Z. And hey, wouldn't you know it? I know the delta Z. Then for kinetic energy, I'm going to have the kinetic energy at state 2 minus the kinetic energy at state 1. Total kinetic energy can be calculated by taking 1 half times mass times velocity squared. Therefore, delta big Ke is going to be 1 half times M2 times V2 squared minus 1 half times M1 times V1 squared and then I'm going to factor out the mass in the 1 half because 1 half is constant and mass doesn't change because I've assumed it's a closed system. Therefore, I have 1 half times mass times V2 squared minus V1 squared and again, A squared minus B squared is not the same as A minus B quantity squared. That was a long sentence. So do I know all of these parameters now? Well, I know the mass. I know gravity. I know delta Z. I know delta U. I know specific work out. I know V1 and I know V2. So at this point, it's just a matter of plugging in some numbers. We want a quantity in kilojoules. So I'm going to have to have dimensional homogeneity among the things that I'm adding together. Therefore, I want everything to be in kilojoules as I add them together. So we begin with mass times specific workout and that mass is in kilograms and the specific workout is in kilograms. Therefore, taking 10 times 0.147 is going to yield kilojoules per kilogram directly. So I can just say 10 kilograms multiplied by 0.147 kilojoules per kilogram and that is a quantity in kilojoules. And just to keep my unit conversions kind of neat and orderly here, I'm going to actually break the addition across the lines of my paper so don't get confused. Then for mass, I have 10 kilograms again and then a little delta U. That's just kilograms, John. Not 10 kilograms. Try that again. 10 kilograms times delta low U, which is negative 5 kilojoules per kilogram. And then that kilograms cancels kilograms. I'm left with kilojoules again. And then I am going to take mass times gravity times delta Z. So 10 kilograms times 9.7 meters per second squared multiplied by negative 50 meters. And I want to get to kilojoules so I will start at kilojoules and work backwards. A kilojoule is a thousand joules and a joule is a newton meter and a newton is a kilogram meter per second squared. So joule cancels joules, newtons cancels newtons. Meters, meters, cancels meters, meters. Second squared cancels second squared. Kilograms cancels kilograms, leaving me with kilojoules. And then lastly, I have one half times mass times V2 squared minus V1 squared. So I will write that as one half times 10 kilograms times V2 squared, which I believe was 30, yep. 30 squared minus 15 squared meters squared per second squared. And again, I want to get kilojoules. So I will start at the kilojoule and I will work backwards. A kilojoules, a thousand joules. A joule is a newton times a meter. A newton is a kilogram meter per second squared. And a corrected N is a lot easier to read. Newton cancels Newton, joule cancels joule. Second squared cancels second squared. Meters and meters cancels meters squared. Kilograms cancels kilograms, leaving me with kilojoules. So I'm going to take 10 times 0.147 plus 10 times negative 5, plus 10 times negative 9.7 times negative 50 divided by a thousand plus one half times 10 times 30 squared minus 15 squared quantity divided by a thousand. And I will point out that I could have factored out mass, but I didn't want to do the unit conversions inside of a parenthetical statement for the purposes of this example problem. But if you were doing that and you totally could, then negative 5. I have to say that every time because the calculator emulator uses the minus on the numpad as the minus sign and the minus that is in the top row of QWERTY to actually be a negative sign. So if I type the wrong minus sign on my keyboard, it breaks. Then 10 times 9.7 times negative 50 divided by a thousand plus 0.5 times 10 times the quantity 30 carat to minus 15 carat to divided by a thousand. I bumped my iPad so it scrolled up and I get 10 times 1.47. It's 1.47 calculator now 1.4 all of calculator fall. Then 10 times negative 5 good and then 10 times 7.7 times negative 50 0.5 times 10 times 30 squared minus 15 squared. And those are both divided by a thousand cool. So I get negative 50.005 kilojoules. And that was my heat transfer in. So what does that imply about my heat transfer and negative heat transfer in implies that my heat transfer is actually out. So I would say Q out is 50 kilojoules. And that's the answer to my question. But before we move on, I want to take a minute to talk about the relative effect of each of these four quantities here. We were taking 10 times 0.147 kilojoules per kilogram plus 10 times negative 5 kilojoules per kilogram. And then we were adding potential and kinetic energy and both of those changes in kinetic and potential energy were relatively large. I mean the difference between 15 and 30 meters per second is like 50 miles an hour and 50 meters is a relatively large change in elevation and note that those changes in kinetic and potential energy are divided by a thousand. So their effect is drastically smaller than it appears to be. So even in this process where we have relatively large changes in velocity and elevation that effect on the final answer is almost negligible. So small in fact that we often neglect changes in kinetic and potential energy unless they're going to be huge that is large changes in velocity like hundreds of meters per second for most of the analyses that we do or they are the only thing that actually appears in the problem. Like if we're actually working towards an answer that is a velocity then we have to include kinetic energy. But if we have say an air duct where we have a slight change in kinetic energy unless it's significant we often neglect it. So we often neglect changes in kinetic and potential energy unless they are large or they are the thing that we're working towards or they are the source of energy for our problem and you'll get the hang of that assumption. But if you see us neglect changes in potential energy across a process that actually has a change in elevation like a foot or two probably not because there is no changes in potential energy. It's just that the effect is tiny. I mean if we had just considered the internal energy in this problem we would have gotten the same answer. We would have been negative 50 because that's five kilo joules per kilogram as opposed to the kinetic and potential energy which are divided by a thousand. That effect is in joules. Crazy right?