 Okay, so this problem says how much heat in joules is required to heat a 43 gram sample of aluminum with a specific heat capacity, 0.903 joules per gram times the degree Celsius from 72 degrees Fahrenheit to 145 degrees Fahrenheit. So we've already gleaned all the information that the problem is giving us that we're going to need for this problem, okay? The one thing, so I guess going back to, again, y'all's answers in general, a lot of us understand how to do this problem. It's not that. The thing we're having trouble with is reminding ourselves that the degree Celsius and the degree Fahrenheit have, they're measuring different amounts of temperature change, okay? So you can't just subtract your 145 from 72 and get that answer and go with that or do the problem that way. You've got to actually convert each one of those Fahrenheit temperatures to degree Celsius temperature first, okay? Because, again, the magnitude of the degrees Fahrenheit is different than the magnitude of the degrees Celsius, okay? So you just got to remember that. So how do we do this? Well, okay, so we just use that and we're going to use that for each one of these things, okay? So in the first case in this one, we're going to say 22-32 times 5 divided by, so degrees Celsius. Is everybody okay with what I've done? Leave those trailing digits if you like to. I'm just rounding into the significant things, okay? Same thing here. If you prefer dividing by 1.8, that's fine too, okay? This is just the way that I was talking about, way back when. 113 times 5 divided by 9 gives us 63. So when we take 63 and subtract 22, we actually get, so what I'm doing now is delta T, right? And I'm subtracting T2 from T1 here, or T1 from T2, I mean, and I get that, I get 41 degrees Celsius, okay? If you convert that Fahrenheit degree, you get a different answer, and that's where a lot of people messed up, okay? What was that answer you got? 22, so, or 23 degrees, right? As you can see, 23 degrees and 41 degrees are quite significantly different, okay? And this hopefully instills the magnitude difference inside of your head between the Fahrenheit and the Celsius degrees. Is everybody clear on what I'm saying? Okay. And this was really the big problem for everybody, okay? Everybody knew Q equals MC delta T, and once you get all of this stuff, it's just a regular old problem. The next thing that people had a hard time, well, of course, significant figures always, okay? But the real issue was when you don't put your units into this equation and try to cancel them out, you oftentimes forget what units you're supposed to get to, okay? So I encourage you once again, you know, I think this is a daily encouragement, okay? That you put your units into your calculations, because we're solving for heat. Heat is units of joules, okay? It even asks you joules. You can put kilojoules or whatever, but since this problem says joules, that's what you want to put. Some people put joules degrees Celsius, joules, grams, you know, that's not joules, okay? So just keep that in mind. The whole thing is, is every one of these algebra problems will cancel out to give you the correct units. And if you get, like, for energy units, joules, you know you've done the problem correctly, okay? That's what I'm really trying to say, but just help you out. So 43 grams, that's the mass here, right? C, 0.903 joules divided by 1 gram degree C. Okay, I like to write it like that to remind myself that I can cancel out because the grams in the numerator, grams in the denominator. And then the delta T, 41 degrees C. Again, I do it that way so I can remind myself to cancel out. And look what comes naturally from just doing that, joules, right? That's the units that I want, okay? So no big deal if you like to just do your calculations without your units, okay? When I look at your work, that's not what I'm looking for necessarily, okay? But if I see you're putting the wrong units on there, I'm going to make a comment that says all your units will make your life a lot easier, okay? Okay, so now it's just multiply, multiply, multiply. Okay, so 41 times 43 times 0.903. And I get 1591.89, or 989, okay, joules. But I don't write all of those numbers. Why? Because of significant digits. There's only 2, 3, 2. So what am I going to use? 2, right? So 1.6 times 10 to the 1, 2, 3, joules. Or like what Harlan was saying, you can just put 1,600 joules without that decimal point there. That says 2 significant figures as well. Either one of these answers is correct. Although it might be easier for you guys right now to think about it this way, okay? Is everybody okay with what I'm saying? I think you guys are doing great, you know? It has nothing to do. I think it's just these little misconceptions, you know? And once you get over the fact that anything could be a convergent factor is very brutal. Or, oh yeah, I need to plug in my units to make sure I get the right units. Or, oh yeah, the degrees Celsius and degrees Fahrenheit are different magnitudes. The two temperature scales are different magnitudes. Once you get those little bits of information nudged into your brain, then I think you'll be set with these types of problems. Any questions on this one? Okay, wonderful.