 Okay, so let's try this problem. This one is a more involved problem. You actually have a lot of calculations to do in this one. So this says calculate the heat required to convert 100 grams of ice at negative 40 degrees Celsius to steam at 125 degrees Celsius, okay? So it's at negative 40 degrees Celsius currently. What state of matter is it in if it's water? Liquid? No solid. It's a solid, right? Because why is it a solid? Because negative 40 is below what temperature? Which is what temperature? Zero degrees, right? Okay, so right now we're in the solid, but we're going all the way up to 125. What, what a state of matter is the water in at that point in time? Gas. Gas. So when starting from a solid, what do we have to do before we get to the gas? Liquid, right? So we got to go from solid to liquid to gas. So we've got to do a lot of different calculations here, okay? So first things first. I have all of this other information written here. This information was not given to us in the problem. Remember, this is from that table that was given to us in the book, okay? So I don't want you to memorize this information. If I want you to do a problem like this, I give you this information, okay? So I just wrote it there because we're going to have to use all of these little bits of information, okay? So when I look at this, C of ice, C is the specific heat, okay? And then heat of fusion, that's how much energy it takes to melt something. Heat of vaporization, remember that's how much energy it takes to take the liquid water and make it into steam where gas is water, okay? So when we're looking here, well, the first thing we have to do is figure out, well, what temperature does ice convert to water? So what is that, the melting point? What's the melting point of water? Zero degrees Celsius, right? So zero degrees Celsius. So why don't we write down, well, what's the boiling point of water as well? A hundred degrees Celsius. Yeah, a hundred degrees Celsius. So why don't we write that down too? Because like we said, we've got to go through both of those points, right? Okay. So remember the equation for finding the amount of heat, okay? So I'm going to write it down, okay? So remember, it's Q, which is heat equals MC mass times specific heat times what? Delta T, very good, okay? So Delta T means what, Veronica? The change in temperature, okay? Should know that, like that, okay? So change in temperature, we're going from, well, we've got to get from negative 40 to what temperature? Over there, what temperature? Zero, right? Okay, so we've got to do that first. That's the change in temperature. So the change in temperature is going to be zero degrees Celsius minus, minus 40 degrees Celsius. And this will be, we'll make this zero. For our significant figures. For us, that's going to be 40.0 degrees Celsius change, okay? That makes sense, right? Going from negative 40 to zero, that's 40 degrees Celsius change, right? Is everybody okay with that? Okay, wonderful. So now we're just going to say, well, what's the heat given off by that first ramping going from negative 40 to zero? Okay, so the mass of ice is 100 grams over here. The specific heat of ice is given to us here, 0.50 to 0.50. And watch how I write this. I put the calc on top and then divided by one gram degrees Celsius. I do that so we can cancel out our units. So that makes sense, right? Okay, wonderful. And then the delta T, well, we found that in the step before. So we're just going to multiply that by 40.0 degrees Celsius. Okay, so are we cool with that step? Okay, wonderful. So now we can cancel, cancel, cancel, cancel. Notice we're left with calories, okay? That's an energy unit, okay? We're good there. So what do we got here? 100 times 40, well, divided by 2 or times 0.5 is 2000. So the two significant figures will say 2.0 times 10 to the third, calories. So try to calculate this on your own, that's what, just to make sure you're able to do this. Okay, so that only went from negative 40 to 0 degrees, right, ladies? Look here, we only went from negative 40 to 0, right? But we want to get all the way to 125. So this is not the right amount of heat, that's just a part of it, okay? So we're going to write that down somewhere and then go on to the next step, okay? So we're going to say that Q1 is going to be 2.0 times 10 to the third calories. So I'm going to erase this portion and we're going to start the next part. I'll post this video so you can go through it step by step a little slower. Okay, so the next portion is the actual melting of the ice, okay? So you've got to melt the ice before it becomes liquid water, okay? So that's where this heat of fusion comes in, okay? So we're done with this one, now we go to the next one, heat of fusion. Remember, fusion means the melting of the ice to the liquid water, okay? So the units kind of give you a clue, okay? Whenever we're going, we're ramping up from one temperature to another, then we're going to use this formula here, okay? But notice here we're not changing the degree Celsius so that degree Celsius isn't in there anymore, okay? So the heat given off in this step is just going to be the mass times the heat of fusion here, okay? So the mass again is 100.0 grams times the heat of fusion here and there, okay? So notice cancel, cancel, and we get, we'll call this Q2. And eventually what we'll do is add up all of the Qs, okay? So what do we got? 100 times 0.8 calories, okay? Oh, it's 80, probably on the video everybody was like, already figured it out, okay? Obviously it's 8,000, so 8.0 times 10 to the third calories, like that, that's Q2, okay? So let's put that over here. So what's the next step? So now we're liquid, what do we want to do? Gas. Yeah, we got to get it, well, we got to heat it up first. Before it can become a gas, we've got to heat it up to what temperature? I'll let you guys think about it while I'm erasing this. To 100 degrees Celsius? To 100 degrees Celsius because that's when the liquid converts to the gas, right? Does that make sense? Yes. Okay, wonderful. So it's kind of a logical step, you're getting it, right? Yeah. A logical step-wise process. Okay, so in this case we're ramping, so what do we use? Q equals MC delta T, but we got to figure out what delta T is. So what temperature are we at right now? Do you remember what temperature are we at right now? Zero degrees Celsius. Zero degrees Celsius and we want to get to what? 100 degrees Celsius. 100 degrees Celsius. So to do delta T we say 100 degrees C minus zero degrees C, like that. Okay, and of course that's 100 degrees C, like that. And we use the specific heat of water using that, so this is going to be Q3, if you will. So mass is 100 grams times 1.001 degrees C times 100 degrees C. Cancel, cancel, cancel, cancel, like that. Okay, so 100 times 100 times 1, that's 10,000, okay? So I'm just going to do it in the same times 10 to the 3, okay? Because if we do it times 10 to the 3 then we can add them all up, okay? So we do that, it's going to be 10.0 times 10 to the 3. Okay, so that's Q3. So what we've done so far. Okay, so now we're at 100 degrees Celsius but we want to get to 125. Can we go directly from 100 to 125 or do we have to do something else first? What do we have to do before? When it went from solid to a liquid. The heat separation? Yeah, the heat of vaporization. So that's all right. So what did we do? We just multiplied the mass times the heat of vaporization because it's staying at the same temperature, remember? So those intermediate steps you've got to watch out for. So this one is Q, what do we have? 4. So the mass still is 100 degrees C, the heat of vaporization, 540 cal per 1 gram, sorry about that. Cancel, cancel like that. And then 100 times 540, that's going to be 540,000, okay? So we'll say 54.0 times 10 to the 3. Calories like that. So again, I just want to keep it in that 10 to the 30 so we can add it up real easily. And then the last step, right, so we're done with that step. The last step is the steam part. So what do we have to figure out first? Total heat energy? Well, that's what we're going to figure out eventually but first we have to figure out the change in temperature, okay? So the change in temperature, this is the final change. We're going up to 125, so it's going to be 125 degrees C minus 100 degrees C. So that's going to be 25.0 degrees C like that. So now Q5 is going to be the mass 100.0 grams times C. In this case is 0.48 cal per 1 gram degrees C like that. And then multiply that by 25.0 degrees C. Cancel, cancel, cancel, cancel. 1200 but let's just make sure it's 100. Okay, so 1.2 times 10 to the third say Q5 is 1.2 times 10 to the third calories. And now what we can do is just add all those numbers up over there, okay? So it's going to be 2 plus 8 plus 10 plus 54 plus 1.2. So the total, so Q total if you will, going to be Q1 plus Q2 plus Q3 plus Q4 plus Q5 like that, okay? So those are all right there. And I get for this 75.2 times 10 to the third calories like that. But let's make this into a better number. Let's do it into kilo-calories, okay? So for every 1 k-cal there's 1,000 calories or 10 to the third. So notice what we can do. We can cancel, cancel, cancel, cancel. And what we get is 75.2 k-cals of energy, okay? So that's how you do that problem. It's a long problem but very logical in how you do it, okay? Any questions on it? Okay, wonderful.