There are a number of different types of problems that we run into when we're talking about specific heat. Let's take a look at a few variations and see how some of those work. So remember a specific heat problem or specific heat is really a relationship between the amount of heat that's being transferred, the change in the temperature and the amount of material that we're working with. So the specific heat is energy over amount and change in temperature. That means that if we take that specific heat and we cancel out the amount, cancel out the change in temperature, we're left with energy. Let's start with this problem. How much energy is involved? So how much energy, how much heat must be removed from a 62 gram block of aluminum to change its temperature from 46 to 9 degrees Celsius? Visually, I've got my block of aluminum. It's 62.354. At the beginning and the end, I've got an initial temperature and a final temperature. And here I've got the specific heat of aluminum, 0.897 joules per gram degree Celsius. The only thing that's missing is the amount of heat, the amount of energy that we're working with. Here's my generic problem. Plug in the specific heat for aluminum, the amount of aluminum I'm working with and the change in temperature that I'm observing. I can solve that out and get an energy. In this case, it's negative 2,090 joules. This is another one of those potentially ambiguous, sig fig numbers. Is this zero significant or not? If we look back, 0.897 is only three sig figs. And I've got three or fewer sig figs in all my terms, or three or more sig figs in all my terms. So this is a three sig fig number. That zero is not significant, so I don't have a decimal point to after it. It might be easier. It might be less ambiguous if I report that as kilo joules instead of joules. We can also look at how much material, how much stuff we're dealing with. So another problem. When 28 kilojoules of energy is added to a block of glass, the temperature increases from 12.5 to 27.5 degrees C. What is the mass of the block of glass? So again, setting up, I don't know the amount. I don't know the mass. I do know the beginning and the final temperature, so the temperature change. And now I know the amount of energy involved. And I looked up the specific heat of glass, and it was 0.8392 joules per gram kelvin. Couple things there. First of all, I managed to find a four sig fig specific heap. That was nice. But this one is also in units of kelvin. So I've got to keep that in mind when I look at the temperature. So again, plugging into my generic formula. I've got 0.8392 joules per gram kelvin. Grams is what I'm looking for. That's the amount. And I've got my temperatures in kelvin. That's the unit I had to be careful about. And we just convert those by adding 273.15 to the Celsius temperature to get to kelvin. It's equal to the amount of energy I moved around. And solving that for X, I get that the amount was 2243 grams. And in this case, we don't have that pesky zero causing us sig fig ambiguity. Another nice thing about this particular problem is I'm dealing with a material that isn't a pure material. Isn't pure element. Isn't necessarily even a pure substance. Glass can be a mixture. It can be a variable mixture. So having a heat capacity for that mixture, heat capacity or specific heat still works as long as I know what I'm working with. We can also start to explore the temperature part of these problems. So how cold does it get when 1.277 kilojoules of energy is removed from a 327.629 gram block of tungsten? Initially at 73.91 degrees, see what's the final temperature of the tungsten block. So 327.629 grams is my mass. I'm removing. So removing energy means that's a minus 1.277 kilojoules of energy. I'm starting at 73.91 and I'm going to some other temperature. Specific heat that I found for this one is listed here 24.79. Now this is, again, watch the units. Jewels per mole kelvin. So now I've got an amount in moles and my temperature is in kelvin. One of the things to always make sure you pay attention to here in this problem, I am removing heat from a substance. So as a really quick check, if it starts at 74 degrees Celsius, the final temperature better be less than 74 degrees Celsius. We know what direction this has to go. Just make sure that our answer goes in the correct direction. Plugging in, I've got my specific heat, 24.79 joules per mole kelvin. Now I've got moles for my amount. So I've got to take that into account and convert my grams of tungsten to moles of tungsten. And my temperatures are in kelvin. So I need to include that 273.15 factor in on there. And my energy is in joules and it was given in kilojoules in the problem. So I need to convert that over. And solving out for X, I get the initial temperature with 318.16 kelvin or subtracting 273.15 from that 45.01 Celsius. Now as that double check, I started at 74, I removed a bunch of energy and I went down to 45. So the direction on that temperature change makes sense. What about the other temperature? When 18 kilojoules of energy is added to 173.617 grams of steel, the final temperature is 126.54 degrees C. What was the initial temperature? Again, we've got steel. That's another mixture. So that's fine as long as we know the heat capacity or the specific heat of what we're working with, we can plug in values for the same thing. 173.617 is my mass. That's how much energy is moving. I don't know what temperature it started at, but I know what temperature it ended at. And I've got a specific heat. This time, the specific heat I found was in kilojoules per kilogram times degree C. Let's plug it in. 0.466 kilojoules per kilogram. I just converted grams to kilograms without an explicit conversion factor. Degree C, I can plug in. There's my X. That's what I'm solving for. And I've got kilojoules. Doing that, I've got negative 101 degrees Celsius. So that piece of metal started out pretty darn cold, but heats up quite a bit when you add that much energy to it. And finally, what else is missing? Now, here we've got a situation where I have a 38 gram block of some metal. And when I add 174 joules of energy to it, it increases in temperature from 17.62 to 52.92 degrees C. Is that metal copper, which has a heat specific heat of 0.385 joules per gram Kelvin? Or is it gold that has a specific heat of 0.129 joules per gram Kelvin? Same approach. Let's just plug it in. Now, this is my unknown. I know the mass. I know the amount of material I'm working with. I know the temperature change. Remember, we've got Kelvin. So we better keep that in Kelvin. And I know how much energy was involved, solving that down for X. I get a specific heat of 0.129 joules per gram Kelvin. And my block is gold. I wouldn't mind having a 38 gram block of gold in my pocket right now. So we can solve for any of the terms that are in this problem. That depends on what we're looking for and what information we know or can find out. Specific heat problems are pretty straightforward once you get used to them, but you've got to get used to them. So go out there and do a bunch, and they'll start to make a little bit more sense, and you'll get a little bit more comfortable with them. Good luck.