 Hello, welcome to another video of understanding thermodynamics. My name is Adrian, and today we are looking at the application, the first law of thermodynamics in closed systems. And thermodynamics, we study energy and the conversion of energy. And we have already encountered work, heat and internal energy as three forms of energy. And in this video, we will combine these three terms in the first law of thermodynamics. The first law states that energy cannot be created or destroyed. In this video, we will introduce the first law of thermodynamics for closed systems, what is also known as a control mass. And we will demonstrate its application. And lastly, we will also illustrate the role played by kinetic and potential energy in the first law. Let us consider a closed system undergoing a process from state one to state two, absorbing heat and doing work. The substance inside the setup is called the control mass. And heat is added to this control mass. And as the volume of the control mass increases, work is done by the control mass on the environment. As energy is conserved, this means that the energy contained in the control mass before the process starts, plus the energy added to the control mass must be equal to the energy contained in the control mass at the end of the process, plus the energy that was released or done by the control mass. We can formulate the first law using the notation we have already introduced in the previous video. The energy contained in the control mass in the beginning is equal to the mass multiplied by the internal energy plus the heat added to the control mass. The energy contained in the control mass at the end of the process is equal to the mass multiplied by the internal energy at the end. As this is a closed system, the mass remained the same, plus the energy that left the system in the form of work performed by the control mass. The units of mass, is kilogram, and the units of internal energy is kilojoules per kilogram. Therefore, the unit in this equation is kilojoules. Now let's do an example. The example states that 0.1 kilogram of air in a piston-cylinder arrangement at 25 degrees Celsius is heated at a constant pressure of 200 kilo Pascal to a temperature of 500 Kelvin. Now we need to determine the heat transfer. We can write down the equation we need to solve, which was the equation we saw in the previous slide. Now we know the mass is 0.1 kilograms. We know the initial temperature and can therefore read the value of internal energy at state one from the tables. Q in is the quantity that we need to determine. And similarly, U2 or the internal energy at state two can also be read from the tables. The work performed is a constant pressure process, and this is easily determined. We know the pressure as it's a constant pressure of 200 kilo Pascal, and then we need to multiply that by the change in volume. So if we substitute all the values in, the only two unknowns here is the volume at state two and state one, which we can use the ideal gas law to get these values, and we get an answer for the work done as 5.79 kilojoules. Now that we know the work done, we can go and solve for the only unknown in the equation, which is the heat transfer. And we get an answer of 20.47 kilojoules. Now in general, we need to consider potential and kinetic energy of the control mass as well. Usually they are small in comparison with the other terms in the first law, and we will now illustrate this by doing a few examples. Now let's start off with potential energy first. The center of gravity of the control mass is indicated here as a red dot. In the expansion process, the volume increases, and therefore the center of gravity in this configuration also rises, and you can see the change with the red dot shown. If this process takes place in a gravitational field, the potential energy of the control mass would have increased. Now if we consider the previous problem, let us assume that the center of gravity was raised by one meter. We can calculate the change of potential energy, and we get an answer of 0.00981 kilojoules. The energy necessary to raise the center of gravity will have to be supplied by heat added to the system. You will note that its value of 0.00981 kilojoules is much smaller than the work done by the system, which was 5.79 kilojoules, as well as the change in internal energy, which was 14.7 kilojoules. It may be safe to check, but in closed systems, the change in potential energy is usually negligible compared to the other terms and not taken into consideration. Next, let's look at kinetic energy. Consider a lead bullet traveling at 200 meters per second. It comes to a complete standstill against a target plate. Now calculate the temperature rise before any heat exchange with the target plate takes place. Now we can assume the volume of the bullet remains constant, therefore no boundary work is done. The specific heat of lead is 0.13 kilojoules per kilogram Kelvin. The energy before the collision with the target plate is the internal energy plus the kinetic energy of the bullet. The energy after the collision before any heat transfer takes place is only the internal energy of the bullet. All the kinetic energy is converted into internal energy, and we assume no noise is generated by the collision. The first law of thermodynamics therefore becomes m mass times internal energy at state one plus the kinetic energy equals mass times internal energy at state two. The change in internal energy can be calculated using the specific heat of lead multiplied by the change in temperature equals the kinetic energy, and the temperature rise can then be calculated as 235 degrees Celsius. When deciding when to take kinetic energy into account, consider the change in velocity of the center of gravity of the control mass. It should be evident from the problem where the kinetic energy plays a role and should be taken into consideration. Usually though, in the processes we consider, the velocity of the control mass is very low, if not zero, and kinetic energy can thus be ignored. So in summary, the first law states that energy cannot be created or destroyed, only converted from one form to another. Energy in the beginning plus the energy entering the system is equal to the energy at the end plus the energy that has left the system. We are often interested in the conversion of heat into work such as internal combustion engine, and the potential energy and kinetic energy of the control mass does not change much and is usually not taken into consideration. And that's it. The course notes which these slides are based on is available on my website odreonsblog.com. I am also available on Twitter, my Twitter handle is at asv90. If you've got any questions, do not hesitate to connect with me and ask them and I'll be happy to answer them. Thank you very much for watching and I will see you in the next video. Bye.