 So, this one says a 63.0 gram piece of aluminum with specific heat of 0.25 calories per gram degree Celsius. So, notice that calories is a little different. It's a different energy unit than you're used to. At 25.0 degrees Celsius is warmed by the addition of 325 calories of energy. And the final temperature of the aluminum. So, it gives you calories of energy so you don't have to convert between calories and joules. But, it does kind of throw you a loop because it doesn't say what's the change of temperature. It says what's the final temperature. So, you have to recall, and this is a mass equation because it gives you the mass, not the number of moles. Okay, so you have to recall the heat equation q equals mc delta t. Okay, is everybody okay with that? And then it gives you some of these variables. It gives you m, which is the mass of the aluminum piece, 63.0 grams. It gives you the specific heat, which is c, 0.215 cal per gram degree c. And I'm going to change that to cal over so we can cancel our unit eventually. And then it also gives you, remember, the other thing we're looking for is the change of t, right? But if you recall, that's going to be tf minus ti. So, there's a little equation there, right? So, do we have these things? Well, tf is really what we're looking for. So, we need to know delta t before we can do tf. And we know ti is 25.0 degrees Celsius. Okay, so the cool thing about these ones is a lot of time you won't have to change. Normally, you're changing your temperatures to Kelvin or whatever. These ones are in degree Celsius. And if you look at the specific heat, that's also in degree Celsius. Okay? So, let's go ahead and do this. So, oh, and the energy we need, too, is 325 calories. I looked at it, I was like, oh, man, there's one missing 30. Okay, so, first thing we want to do, of course, is isolate the delta t variable. Is everybody okay with that? So, delta t equals q divided by mc. So, q is 325 calories times 1 over m, 1 divided by 63.0 grams, 1 over c. So, in that case, it's going to be 1 gram degrees Celsius, or 0.215 calories. Cancel, cancel. So, we're left with degrees Celsius, which is a good temperature value, right? So, we should be confident in our calculation. 325 divided by 63 divided by 0.215. And so, I get 23, so 24.0 degrees C is change of t. And remember, change of t equals tf minus ti. And we're looking for tf. So, tf equals change of t plus ti. Like that. So, we're going to move this to this side. So, change of t is 24.0 degrees C plus ti is 25.0 degrees C. So, that's going to be 49.0 degrees C. So, the final temperature of that is going to be 49 degrees Celsius. Any questions about this one? Pretty straightforward. So, this is one of the harder ones that you can do. I could imagine, like, if they had given us this energy, the heat, and kilojoules or something like that, we would have had to convert it to calories, because the specific heat is given to us in calories. Just remember, cancel your unit 10. Would you really have to remember that? Which one is key? The 4.1, 4.1. So, it's good to keep it in your head. Usually, I can't say what they'll give you on the phone, you know? So, usually they expect you to do SI to SI units. So, probably they'll give you that. Any other questions about that?