 We're, in this class, what we're going to do is we're going to work another example problem involving extra G destruction. And it's a rather long problem and consequently it'll take the entire lecture for us to solve the problem and I'll break it into different parts. But I'll begin with the problem statement. So there's the example problem, a rather long one with a lot of words in it. So what we're dealing with is we have a passive solar house. It's losing heat to the outdoors at 3 degrees C. At an average rate of 50,000 kilojoules per hour, the house itself is maintained at 22 degrees C during a winter night, we're specifically talking about one winter night for 10 hours. The way that we're doing this is that the house is heated by 50 glass containers each containing 20 liters of water heated to 80 degrees C during the day by absorbing solar energy. We have a thermostat within the house that is connected to a 15 kilowatt backup electric resistance here which turns on whenever necessary to keep the house at 22 degrees C. We're then asked to determine how long does that electric resistance heater need to be on during this long winter night. We're asked to evaluate the amount of extra G destroyed through this process and finally we're asked to determine the minimum work input required for that night in kilojoules. So like I said it's a long problem, we're going to take a couple of segments here to cover it but we'll begin like we begin all of the other problems by writing out what we know and then we'll go from there. So what is known is we're dealing with a passive solar home, so I'll sketch out the house and within the house we have a container and that has water in it. And we're told that during the day the water goes up to 80 degrees Celsius and the way that it's doing that is we have solar panels on the roof that are connected to our water tank. So we would circulate water into and out of that panel or the panels on the roof. We have radiation coming in from the sun and if it's sunny day it would be around a thousand watts per meter squared although it does depend upon where you are on the earth. The other thing going on we have the room temperature being maintained at 22 degrees C and the way that we're doing that is we have this backup electrical resistance heater which has a work I'm going to say minus 15 kilowatts because we're putting work into our system. And the other thing going on is we have a loss of heat during the long winter night in the amount of 50,000 kilojoules per hour. So what I'm going to do I'm going to draw a control boundary here and this will be the immediate control boundary for our problem. The other thing that we're told is way out here the surrounding temperature far away from the house is 3 degrees C which we can call T naught. So what are we after we're trying to maintain? T room equals 22 degrees C for 10 hours during the night and the water thermal mass consists of 50 20 liter containers and if you look at this that turns out to be one cubic meter of water that we have in those containers. So the things that we're trying to find is how long does the heating system come on, exergy destruction and minimum work input required which we saw outlined here at the bottom of the question statement. So with that what we're going to do we're going to go into our analysis for this problem and we're going to start with the first part. So let's first of all determine how much water we have. We know that we have one cubic meter but let's evaluate or compute the mass of water and the way that we'll do that is the density of water times the volume. We'll assume a thousand kilograms per meter cubed for the density of water which is pretty close and the volume one meter cubed we get a thousand kilograms for the mass of the water. So we want to figure out how long do we have to have the heating system on during the 10 hour night. We know the heat loss, we know the temperature of the water at the beginning, we know how much energy comes in through our electric resistance heater. This sounds like the first law. So let's apply the first law and this is going to be the first law applied to a fixed mass or closed system. Okay, what can we get rid of in this equation? That's what we always do when we go through these problems. Can we get rid of heat loss? No we can't. The house is losing heat to the surroundings. Can we get rid of work? No we can't because we have that electric resistance heater. Can we get rid of the internal energy? No we can't because the water itself is changing temperature. Can we get rid of the kinetic energy? Yes we can. The house is not moving. Can we get rid of the potential energy? Yes again. The house is not moving. Okay, so with that we whittled down the first law into something a little bit simpler and remember the sign convention for heat transfer into a system. If the heat flows in to the system it's positive and if the system does work it's positive. In this case we have heat leaving the system so it's going to be negative and we're doing work on the system so that's going to be negative. So what we get is a minus Q loss plus work electric. It's plus because it's a minus term and then a minus W that becomes a plus. And on the right hand side we have the change in internal energy of the water plus the change in internal energy of the air which is inside of our control volume. But wait a minute, we said the air was going to stay at 22 degrees C throughout this process. So that disappears and we can say no change. Okay, so that's our first law. Now what we're going to do, let's evaluate the amount of heat transfer coming out of the house. We said that this is going for 10 hours and the loss was 50,000 kilojoules per hour. So we can write 50,000 and then I'm going to write this in joules now times 10 to the 3 times 10 hours because we have this going for 10 hours plus the electric heat. We were told it's a 15 kilowatt heater. So 15 times 10 to the 3 multiplied by some unknown period of time. We don't know how many hours that will be or how many seconds I should say that will be. And then on the right hand side we can evaluate the change in internal energy of the water using MC delta T. So we can plug in our numbers and what we find is that delta T is equal to 17,170 seconds which equals 4.77 hours. So what that means is that we need to run the electric resistance heater for 4.77 hours out of the 10 hour evening. In order to maintain the temperature of the house at 22 degrees C. So the next thing we're going to look at in the next part of the problem will be the exergy destruction which will be in the next segment.