 An ideal gas initially at 1 MPa and 120°C receives shaft work at a rate of 1.2 kW, while 900W of heat is being rejected. If changes in kinetic and potential energy are assumed to be negligible, how much does the internal energy of the system change over the course of an hour? First question we have to ask ourselves, is this an open or a closed system? We were given no indication as to an entering or exiting mass, therefore we are treating this as a closed system. Then the next question is, do we treat this as a steady process or a transient process? We're going to treat it as a transient process because time matters. The key here is we are analyzing the process over the course of an hour. So time is significant, we can't make the simplification of treating this as steady state despite the fact that we were given energy transfer rates and not magnitudes. So I will start with the assumption of closed system. I will define a system somewhat arbitrarily and recognize that we have entering shaft work a rate of heat rejection. With my system defined I can perform an energy balance. I will start by saying that the change in energy of our system is going to be equal to the energy entering minus the energy exiting. Because it's a transient process, the left hand side matters. I have delta u, delta ke and delta pe as possibilities. On the right hand side we have heat transfer and work as the two opportunities for energy to cross the boundary of a closed system. Then I will neglect terms that aren't relevant to my analysis. I have work in the inward direction so I'm going to get rid of work out. I was described only one heat transfer and that's in the outward direction and it's unlikely that there's going to be a complicated enough setup for there to be heat transfer in and out at the same time. If it was complex enough to warrant that then I would presumably have that given in the problem statement. You could also think of that as a network and net heat transfer if that helps. Then I was explicitly told to neglect changes in kinetic and potential energy. I don't need to list that as an assumption because I wasn't making that assumption on my own. That was built into the problem statement. And I'm leaving delta u because that's what the problem actually wants, which is a pretty good indication that I should leave it in my analysis. Then I recognize that I don't have a magnitude of work in, I have a rate of work in. So I'll recognize that that rate of work in is going to be the work in occurring divided by duration because we are analyzing across the entire process. I can write work in as a very sloppily written quantity and also I can write that as rate of work in times duration. Same goes for heat transfer, that rate of heat transfer rejected is the magnitude of heat transfer divided by duration, therefore q out could be written as q dot out times duration. When I make those substitutions I have delta u is equal to w dot in times duration plus minus q dot out times duration. I'm going to factor out duration leaving me with delta t times power input minus rate of heat transfer rejected. My duration was one hour so I will write one hour and then I am subtracting the power input minus the rate of heat rejected so I have 1.2 kilowatts minus 900 watts. In order to subtract them they need to be in the same unit so I'm going to write that as 1200 watts minus 900 watts. Note that I could also do 1.2 kilowatts minus 0.9 kilowatts, that would be fine as well. Potato potato. And then I recognize that the problem wants an answer in megajoules so I will begin to break apart my units here, a megajoule is 1000 kilojoules, a kilojoule is 1000 joules, a watt is a joule per second and I'm out of space so I will scooch this over. And then one hour contains 60 minutes and most minutes contain 60 seconds. So then hour cancels hour, joules cancels joules, kilojoules cancels kilojoules, watts cancels watts, minutes cancels minutes, seconds cancels seconds leaving me with megajoules so with a little help from my calculator we can compute an answer. It's going to be 1 times the quantity 1200 minus 900 times 60 times 60 divided by the quantity 1000 times 1000 times 1. And I get 2725th, thank you calculator, or 1.08 megajoules. So as a result of this process the internal energy of this system is going to increase by 1.08 megajoules over the course of an hour.