A 2.85-g lead weight, initially at 10.3 °C, is submerged in 7.55 g of water at 52.3 °C in an insulated container. What is the final temperature of both substances at thermal equilibrium?

Video transcript

Welcome everybody. Let's take a look at our next problem. It says a 2.85 g lead weight initially at 10.3 °C is submerged in 7.55 g of water at 52.3 °C in an insulated container. What is the final temperature of both substances at thermal equilibrium? A 51.8 °C B 42.0 °C, C 52.8 °C or d 49.1 °C. So let's think about what is happening here. We have this lead which is cold being submerged in a certain amount of hot water. We can just sort of intuit that what's going to happen is the lead will warm up until the lead and the water at the same temperature. That's thermal equilibrium. They're no longer exchanging heat, they have exchanged heat until they reach the same temperature. What is that temperature? So the first thing to note is this final temperature will be the same for both substances. So equal for both. And what is going on particularly as this happens, our lead, which is cold. So our cold lead absorbs she and our hot water. Well, that heat that the lead is absorbing has to come from somewhere releases heat or loses heat. More specifically, the heat lost by the water is absorbed by the lead. So the heat lost equals the heat absorbed. That's key here because that allows us to set two equations equal to each other to solve for what we want to know. We want to know our t final. But our other unknown here is the amount of heat. But since we know these two amounts of heat must be the same, we can use that. So we're going to start with the equation for Q or heat which equals M C delta T telling us that when we have a temperature change of a specific amount of a substance that helps us solve for the heat that's lost or gained by that substance. So in this case, we know that the heat absorbed by the lead. So heat absorbed by PB just abbreviation for lead, which is positive Q subscript PB has to be equal to the heat released by the water which will be negative Q h2o. So in that case, we can have two equations here since our QS are equivalent but negative, just one is the negative of the other. That would say that M so the mass of lead. So M PB by CC being the specific heat capacity, let's recall. So that will be specific to our material So C of PB times delta T of PB equals negative. Now, I'm going to put parentheses around this whole term because the negative next to an equal sign is really easy to lose track of. So parentheses, negative mass of water times C of water times delta T of water. Now, of course, since delta T equals T final minus T initial, we are looking for the final temperature of both substances. So final temperature and of course, T final of water will equal the tea final of lead. So at first, we think we might have two unknowns, but we just have one unknown, the T final for both. So let's write out this equation, we could solve the whole equation for T final before we even start. But it's a little easier to plug in the numbers and get rid of some terms that way. Since with that delta T RT final is going to be part of a subtraction problem. So let's expand that out with some values here. We do need to look up the specific heat capacity for water or rather that we should really know we use it so often in chemistry good to have memorized, but we need to look up the specific heat capacity for lead. So I've just written those off in a little box to the side, our specific heat capacity of water, which again, we should know 4.18 joules per gram degrees Celsius. And I've looked at the specific heat capacity of lead, 0.128 joules per gram degrees Celsius. Note often in thermodynamics problems, we need to convert our temperatures to Kelvin. But in this case, our specific heat capacity has the temperature as degrees Celsius. So we can leave it that way. That's kind of nice. So let's start plugging in the mass of lead is equal to it says we have 2.85 g. So we have 2.85 g multiplied by C. So the specific heat capacity is 0.128 joules per gram degrees Celsius. Now times delta T, so that would be our TF I'm going to write just TF since again, just got one T final for everybody minus our initial temperature for lead. So our initial temperature was 10.3 °C and that's going to equal. Now we have all the terms for water. See why we put the parentheses in here. Don't lose the negative. So I'm going to put brackets here since, well, parentheses for other terms negative uh times the mass of water. So 7.55 g multiplied by that specific heat capacity. C so 4.18 g per or excuse me, not grams, joules per gram degrees Celsius multiplied by T final minus our initial temperature of water. 52.3 °C, close the parentheses, close the brackets running out of space a little bit here. So you can see how it's easier with the temperatures having a subtracted term to plug in my numbers first and simplify this a little bit. So let's simplify our um our multiplication in front of our temperature subtraction there. So 2.85 multiplied by 0.128. So that term is going to simplify to 0.3648. Let's take a look at what happens to our units. In this case, our grams cancel out. So the units that are left here are joules per degree Celsius. And that's going to be multiplied by T final minus 10.3 °C. And that's going to equal, let's do that multiplication there and we will get in parentheses again, we've got that negative carrying through negative 31.559. Again, our grams cancel out here. So we're left with joules per degree Celsius multiplied by T final minus 52.3 °C. So now essentially I distribute my terms so that I can start to get my T final by itself. So I have 0.3648 joules per degree Celsius multiplied by T final and then subtract and then I multiply that 0.3648 by my temperature. So I get minus 3.7574. And now how, what happened to my units here? The degrees Celsius canceled out in that case. So I have jewels as my units there. Note I rounded just a little bit. I'm going to end up if I look back at my numbers that are given to me, I'm going to end up with three sig figs, three significant figures in my answer. So I did round a little bit, even though it's an intermediate step, I've still got some extra significant figures. So as not to introduce any significant rounding error, but just makes this long equation just a little bit more manageable. So now distribute my terms here and I have negative 31.559 joules per degree Celsius multiplied by T final. Now notice I have a negative multiplied by a negative here. So this will turn into a plus 1650.5 35 Jules. So now we'll bring all our t final terms over to the left. So we'll end up with 0.3648 joules per degree Celsius. And note this is a negative um number here in front of this, the other T final. So plus 31.559 joules per degree Celsius. That addition term multiplied by T final equals. And then I can subtract 3.7574 Jews from both sides to combine those terms. And that will give me 1654 0.293 Jules. And now I just have to divide both sides by that coefficient of TF once I do that addition. So I'm just going to scroll up a little bit. So I've got room, I have 31.9238 joules per degree Kel degrees Celsius. If you wanted to say Kelvin multiplied by TF equals 1654.293 joules divide both sides to isolate TF. So then I have TF equals notice that my jewels will cancel out once I do that division my degrees Celsius for inverse, when multiplied by TF, so I will end up with my T final in the proper units of degrees Celsius, rounding to three significant figures. I will end up with T final equaling 51.8 degrees Celsius. So there we go, I have my temperature. It's in the right units. So again, just a good double check. Does this seem reasonable? That's always our good check. Did we mess up the math? We have the lead at about 10 degrees, the water at 52 degrees. And we see that the temperature has come down a little bit but not very much as we would expect sort of this realistic scenario. So gut check level, this is good. So just we have to match it with a multiple choice and choice. A is indeed 51.8 °C. So that is our answer. And again, the key, the main key here was recognizing to use that Q equals MC delta T and then to set those two heats equal to each other that the heat absorbed by the lead was equal to the negative of the heat released by the water. See you in the next video.