 Alright, so let's take a look at some more algebraic problems solved using a tape diagram. So I have an object with a temperature of 160 degrees, and I'm dropping it into water to cool it off, and the water has a temperature of 40 degrees. The object cools while the water warms until they're both at the same temperature. That's actually a fundamental law of physics here. And it turns out that if the object's temperature falls five times as much as the water's temperature rises, then we want to determine the final temperature of the water and the object. And we could set this up as an algebraic equation, but let's see what we can do here. So again, our fundamental idea behind a tape diagram is every tape represents a quantity, and our equation is going to correspond to the equality of the lengths of two tapes. So it seems I have two quantities here. The object, the hot object, at a temperature of 160 degrees, so maybe that temperature is going to be 160. I also have the water at 40 degrees. That's our cold object. And they have two tapes, and because their magnitudes are different, their tape lengths are going to be different. All right, so let's take into account what the description is. We know the object's temperature falls five times as much as the water's temperature rose. So we might represent this by indicating the water's temperature has increased by some amount. Here's our new magnitude. And at the same time, our object's temperature has dropped by five times as much. So five of these blocks, and here's the important thing, it's dropped to the same temperature. And so there's our equality. New water temperature is the same as object's temperature minus five of these blocks. And so I'll draw that. So here's my object's temperature. I've taken five of these blocks out, and then here's my new temperature. And those are the things that are supposed to be equal. Now, we could try and solve it this way, but it's actually going to be a little bit easier if I can actually see the two tapes of equal length. I can see that this is equal to this bit of a tape, but it's a lot easier if I actually have two tapes that I can actually see that have equal length. And so what I might do is I might take this block here, and if I extend it by one, two, three, four, five of these blue blocks. So if I take this water temperature and extend it by five of the blue blocks, I'm going to have two tapes of equal length. Well, now, wait a minute. This tape here was the same as this tape here. Again, what this was, is this was the original object temperature dropped by five blocks. In other words, the original tape is 160. And so now I have two tapes, still of equal length. And now I can proceed to solve this. Now let's get rid of that 40. And so now I have six blocks, is 120, which tells me that each individual block must be of size 20. Now if I go back to my original figures representing the new temperatures of the waters in the object, well, I know that now that these blocks are of size 20. So that says my water temperature, 40 and 20, is now 60. And the object temperature, 160 minus five of those blocks, is going to be also 60 degrees, which will be the same temperature, which is what we wanted, because we wanted to make sure that the two ended up at the same temperature.