 Let's start with this first problem. It says a coffee cup is filled with 100 grams of water at 25.0 degrees Celsius. A 5.00 gram piece of metal that has been heated to 100 degrees Celsius is added to the coffee cup and the lid is replaced and the water is stirred until the temperature of the water remains constant at 28.0 degrees Celsius. What is the specific heat capacity of the metal that was added to the coffee cup? So, how do we do this type of problem? Well, the thing you have to remember is that the heat that's flowing out of the system is flowing into the surrounding. So once you figure that out then it should be pretty straightforward. But let's just go ahead and write down all the variables that are all the stuff that they give us in the problem. So they give us the mass of water. They give us the initial temperature of that water at 25.0 degrees Celsius. They give us the mass of the metal at 5.00 grams. They give us the initial temperature of that metal at 100.0 degrees Celsius. And then they give us the final temperature of the water, which is 28.0 degrees Celsius. And consequently, of course, since it's equilibrated, that's the same temperature as the final temperature of the metal. That's about Tf of the metal 28.0 degrees Celsius. And this is, I should write, you really should label these as metal and water because it will get confusing in a second. Okay, so remember when we're doing the heat capacity equation, we need to find the difference in temperature. Now we go ahead and do that right now. So the change in temperature, of course, is going to be the temperature final minus the temperature initial. Write it out. So 28.0 degrees C minus 25.0 degrees C, which is 3.0 degrees C. And let's do the change in temperature for the metal. 28.0 degrees C, 72.0 degrees C. Okay, so we've got that. It's wondering what is the specific heat capacity of this metal. So that's what we're looking for. And given to you always is going to be the specific heat capacity of water, which is 4.184 joules per gram degree C. So you won't have to remember this. I just know this because I've done so many of these things. Okay, so essentially you're just going to set this side equal to the negative of this side and go. Okay, so what's happening? So the metal, if you want to think of it this way, is losing heat like that, right? So the water is gaining that heat. Okay, is everybody okay with that? So all we've got to do is set both sides equal to each other. So remember Q equals MC delta T, if you guys recall that formula. So M of the metal, C of the metal times delta T of the metal. So we're solving for that variable there. So C, is everybody okay with that sort of rearranging of the equation? If not, stay here. So let's just go ahead and do this. So we have all this stuff. So we have the mass of the water, the C of the water, delta T of the water, mass of the metal, and delta T of the metal. So we should be able to do this, just plug and show it. So 100.0 grams, 4.184 per 1 gram. And we've got Joules per gram degree C, right? Is that a good specific heat unit? Joules per gram degree C, yeah. So if you didn't know that, that's specific heat units. So negative times a negative down here, the whole thing is going to be positive now, right? So 100 times 4.184 times 3, where I got, or the specific heat capacity of this metal, it's got to be 2 sig figs because of this number here. It's going to be 3.5 Joules per gram. So is everybody okay with that? Everybody get that answer? Did you get that answer this time? So any questions on this one? Again, I think the real thing, there's a lot of these types of problems. You'll probably get one of these on your exam. You've got to remember this relationship, okay? And of course Q equals C and delta T. You know that already. Again, questions? Nope, okay.