 We'll now perform the final part of the analysis. For example, problem with the solar house. So let's take a look back at the problem statement and the schematic. So there is a picture of the solar house that we're solving. We're maintaining the temperature at 22 degrees C and if we look at the problem statement itself, we were told for the third part to evaluate the minimum input required for that night in kilojoules. So that's what we're going to do now. Now minimum input or minimum work input is basically just reversible work in. So if our process required work to go in, the minimum work input that would be required would be the reversible work. So the minimum work equals the reversible work, which is reversible in. So what we're going to do, the equation for this is reversible in is equal to actual work in minus exergy destroyed. So that is the equation that we will use for this. Let's evaluate the actual work in. Remember we had the electric resistance heating device. That's where our work in is coming from. We get that for our actual work in. Now plugging this in and we get the reversible work in and we know the exergy destroyed because we've just evaluated that in part b of the problem. And what we find is the reversible work in is equal to minus 35372 kilojoules. So what does that mean? If we have a work input as being minus 35000 kilojoules, well think about processes. We said that if our system is doing work, the work would be positive and we also said that if we're doing work on the system, it would be negative. But given that this is a negative, this represents the amount of work that we could have extracted from the process, not done on the process. And so this is kind of significant because what it's showing us is had we operated in a reversible manner, doing things very efficiently as efficiently as we can thermodynamically speaking, we actually could have extracted work out of our heated water. And that's really quite surprising. And because that means that we could have done a much better job than what we originally did. And so what it's saying is that instead of consuming electricity, we actually could have generated electricity and we could have generated 35100 kilojoules of electricity. Now that is great, but how could we have done it? Well, if you look at the house, and let's go back here, what we had is a tank with water. And we were using this tank with water just to generate heat, which was keeping the house warm, which is nice and great and everything else. But had we actually been a little more clever, what we possibly could have done is we could have introduced a heat engine. And that heat engine, it's kind of messy, but I'll draw it there. That is our heat engine. It would have referenced some other state and that other state was our surrounding. So that would have been what it was rejecting the heat to. But through the process, we could have been doing work. And along the way, we could have been rejecting heat to the room, which eventually would go all the way to the surrounding. So that gives you an idea that we could have used a heat engine here instead of just dumping the heat into the room. And in the process, had we been able to operate reversibly, we would have been able to generate work, which is kind of an exciting idea. So if we had used a reversible heat engine, so let's go with this idea that we were actually able to generate work. So how much work are we getting here? Well, we said 35,372 kilojoules. And I'm going to say that that is for the 10 hours. That works out to approximately 982.6 watts. And if we look at a standard circuit and a house, and it operates with 110 volts, so I know that there's a sign conversion and things like that that may change this a little bit. But kind of from a hand-waving argument perspective, we can say that we potentially could have generated around 9 amps of power, which is roughly equivalent to one circuit in a house. Typically a house circuit would be 15 amps. So that's not bad. With our heated water, we could have done that. Now, what is the showstopper here? The showstopper with this problem is that when you have water at 80 degrees C, and you want to dump heat down to three degrees C, this is what we call low-quality heat. And it's difficult to come up with. This is low-quality heat. Given that the temperature is so low, it's very difficult to design a heat engine or a thermodynamic cycle that can take that low of a temperature and then convert it into a significant amount of work. And consequently, the data thermal of this process is going to be relatively low. And economics eventually come into play where it would make it difficult for us to do this. There are cycles such as organic Rankin cycles that are designed to be able to operate off of low temperature heat sources. And a Rankin cycle, an organic Rankin cycle, may actually be the thing to put here. But nonetheless, that's usually what causes us to have a little bit of difficulty in using this type of heat. So the purpose of the problem is to explore exergy destruction and look at other ways of doing things that we can use exergy for. However, I concluded by bringing in a little bit of reality saying that it's a nice idea. But usually we have certain restrictions with the way that we can do things. And that tends to be how do you deal with low-quality heat, which in this case 80 degrees C is kind of a low-quality heat. But with that, that concludes the example problem, giving you an idea as to what you can do with exergy destruction. And we again looked at two different approaches for solving an exergy destruction problem. So that concludes the lecture. Thank you very much.