Calorimetry with Temperature and Phase Changes - Online Tutor, Practice Problems & Exam Prep

Calorimetry problems can involve both temperature changes and phase changes. To solve these, use heat and temperature diagrams to visualize the processes. For example, when mixing water at 15°C with ice at -20°C, aim for half the ice to melt at 0°C. The calorimetry equation is $q=-q\_b$, expanded to account for multiple processes. Use $q=mc\Delta T$ for temperature changes and $q=mL$ for phase changes.

Calorimetry Problems with Temperature AND Phase Changes

Hey, guys. So up until now, we've seen calorimetry problems that have only involved temperature changes. But you're going to need to know how to solve calorimetry problems where materials are not only just changing temperature but also phase. What I'm going to show you in this video is that we're going to combine what we know about calorimetry. We are going to use a lot of the same sequence of steps with just a couple of new ones, and we're going to combine it with what we know about latent heats and phase changes. We are going to be using these heat and temperature diagrams to visualize what's going on inside your calorimetry problems. Let's go ahead and check this out here. We're going to jump straight into our example. So, I have an insulated cup that combines 2 different materials. I've got some mass of water that's at 15 degrees Celsius, and then I'm going to put some ice in it to cool it down, and the ice is at negative 20 degrees Celsius. I want to combine these things, so two things happen. I want exactly half of the ice to melt, and I also want the final temperature of that mixture to be 0 degrees Celsius. Alright?

So we know we are going to have to use calorimetry equations, this QA is negative QB, but the very first thing I like to do, what I like to call step 0, is to actually draw the temperature versus q diagram and figure out what the initial temperatures are, so that I can sort of work backwards and find out what that final temperature is going to be. So let's go ahead and check this out. The water is at 15 degrees Celsius, which is going to be here in the water part of the diagram. So this is 15 degrees Celsius. The ice, on the other hand, is somewhere over here below the freezing points. So this is going to be negative 20 degrees Celsius. Now both of these things, right, are going to exchange heats, and they're going to meet somewhere in the middle. What we want is exactly half of the ice to melt, and we also want the temperature to be 0 degrees Celsius. So what that means is that the temperature is 0 degrees Celsius over here on this sort of dotted line but is going to be anywhere along this horizontal line here. Remember, the temperature is the same throughout this whole process. It's just that the phase is changing from ice to water. So what's going on here is if you kind of visualize as you input more heat, at this point, this is the freezing point, melting starts. So we have melting starts here. And then what happens is you input more heat, more ice gets converted to water. And then, basically, what happens over here is here is where you have complete melting. So this is complete melting over here. What we want is for this final mixture to exactly be half ice-melted. So what that means is that we're actually going to be looking here at this halfway point. We know the temperature that's final here is going to be 0 degrees Celsius because we're along this line, and we want exactly half the ice to melt. So we're going to go exactly halfway across this horizontal line. So that's really it. So let's go ahead and get to our first step, which is going to be writing out the calorimetry equation. So, we've got that q_A is equal to negative q_B. What we have here is we're going to have 1 q per change. What's going on here is if you think about it, we actually have 2 processes that are going on for the ice. The ice, in order for it to get from negative 20 degrees Celsius all the way to melted, has to do 2 things. The first thing ice has to do is actually get up to the freezing point of 0 degrees. I'm going to call this QA1. And the next part is it actually has to melt somewhat or possibly so that half the ice melts, so that's going to be QA2. The water on the other hand only actually has <<- HTML Error: Page break marked incorrectly in text processing: Fixed >>.

Cooling a Chunk of Iron

Hey, everybody. So welcome back. I've got a great problem that's going to pull together a lot of great stuff from calorimetry. So I want you to imagine that you're a metal worker. You've got a really hot 1.6 kilogram chunk of iron that's at 600 Celsius. It's piping hot. Now, to cool this thing down very quickly, you're going to drip some water over it at 20 degrees Celsius. A lot of metal workers do this because water has a really high specific heat, so it can absorb a lot of energy from that really hot iron and cool down very quickly. So, you do this and the iron cools down to 130, and then all of the water that you dripped boils away once it makes contact with the iron. So, basically, what we want to figure out in this problem is how much water did we drip over the iron. So, that's going to be m_w over here. Right? It's the mass of water. So, the first thing I like to do in these problems is set up the diagrams for both of these substances. That's what I want to do here. So I've got the water and the iron right here. So, remember what this looks like for water. We have the ice portion, the phase change, the water portion, the phase change, and then the steam portion. Right? Then what happens is for iron, iron has a really high melting point of 1500, way hotter than anything we're going to work with in this problem. So, it really just has only one sort of part here. The temperature change, it does eventually melt but just way after anything we'll deal with in this problem. So, let's get started. We have to figure out the initial and final temperatures. So, in the iron graph, I'm just going to plot 600. It's going to be somewhere over here. It doesn't have to be to scale. This is going to be my initial point for the iron. The iron cools down to 130, so it goes down the graph and ends up somewhere over here. So the iron really only has one change. It's just doing a temperature change, moving down the graph. The water, on the other hand, is going to start off at 20 degrees Celsius. It's going to start off somewhere here in the water part of the diagram. Remember this is steam and this is ice. Now what happens is the water absorbs all the heat from the iron and it starts to warm up. So, we're going to start off here, this is my initial, and we're going to go all the way up until we reach the boiling point because we're told that all of the water starts to boil away, which means that once you get up to the phase change, you're going to move all the way across because all of the water boils away, and you end up over here on the diagram. Alright? So what happens here is that the water has 2 steps. This first diagonal piece is going to be q_w1. The second one is going to be q_w2. It's 2 different steps, so we're going to need 2 different equations. Alright. So now that we've done this, it's going to make the rest of the problem much simpler because the next step here is we're going to write our calorimetry equation Q_A equals negative Q_B and we're going to use 1 Q per change. So what happens is on the left side the thing that gains the heat is going to be the water. So this is going to be q_w and then on the right side, we're going to have negative q for the iron. Now the w here, remember, has two changes. We have q_w1+q_w2. So I'm going to do q_w1+q_w2= negative q for the iron. Now the iron only has one change because it's just only doing one, temperature change like this. Right? So there's only one term. The next step is we just have to replace our q's with either m_cats or Δm_Ls. Remember the rule is m_cat for the diagonal parts, Δm_L for the flat parts. So this q_w1 here is a diagonal So this is going to be m_cΔt, but this is going to be for water. So this is going to be m_wc_w, and then Δt for the w. Then for the flat part over here, we're going to have Δm_L, and this is the boiling point. So we're going to use the vapor, a latent heat of vaporization. So this is going to be Δm_Lv. We use Δ when we have whenever we have like a partial melting or boiling or something like that. Now on the right side, we're only going to have a m_cΔt for this piece. This is going to be negative m, this is going to be m_Ic_IΔt_I like this. So really what happens is we're going to have a negative m_Ic_IΔt_I. Alright. So we want to do really is figure out what's this water this mass of the water over here. Now I should also mention that remember you use the Δm's whenever you go only sort of partway across the phase change. If you do the entire thing like we have in this problem here, then what happens is this m and this Δm are the same. So all the water heats up and then all the water boils away. So that means that Δm is equal to m. So what this means here is we can actually sort of combine these two terms. So we have m_w, then we have c_wΔt_w + m_wL_v. So this is going to make our equation a little bit simpler like this. Alright. So now what I can do is these are my target variables. So what I can do here is just use an algebra trick. Right? I've got these 2 are the sort of greatest common factors of these terms, so I want to combine them. If I add them together, I can do something like this. I can do m_w and this is going to be c_wΔt_w + L_v. Right? If you distribute back, you should get back to this expression over here. So this is going to be equal to negative m_Ic_IΔt_I. Now all I have to do, the last step here, is just to move this to the other side, and I can just start plugging in some numbers, but we're basically almost done. The last thing we have to do is just solve for our target variable. Okay? So that means our m_w is going to be I've got a negative. Now what's the mass of the iron? So this is going to be 1.6. Now the specific heat for the iron is actually given right here as 470. So this is 470. Now what's the temperature change? Well, basically, it's just final minus initial. This is my final and this is my initial. Right? The 130 and the 600. So basically, what I can do is say 130 minus 600 divided by, and then I just have the specific heat for water, which is 4186 times the Δt for the water. Now what happens is the water goes from 20 and then it finally ends up at 100. So this is kind of one of those rare calorimetry problems where the substances don't necessarily have to end at the same temperature. So, really, what this is going to be, this is going to be Δt_w is just going to be 100 minus 20. Then what we have to do is add the L_v. So this is the latent heat of vaporization, 2.256 × 10⁶. Now just be really careful when you plug this out in your calculator, but this is basically just all the numbers. Just you could just do the first part and then, the top part and then the second part. But what you'll end up with here is you'll end up with 0.136 kilograms of water. So it's actually not a whole lot of water and that's kind of what we said. Right? Water has really high specific heat, so it actually doesn't take a whole lot of it to cool something down very rapidly. Anyway, so that's it for this one, guys. Let me know if you have any questions.

Here’s what students ask on this topic:

What is calorimetry and how is it used to measure heat transfer during temperature and phase changes?

Calorimetry is the science of measuring the amount of heat transferred to or from a substance during physical or chemical changes. It involves using a calorimeter, an insulated device that measures temperature changes. For temperature changes, the heat transfer is calculated using the equation $q=mc\mathrm{\Delta T}$, where $q$ is the heat, $m$ is the mass, $c$ is the specific heat capacity, and $\mathrm{\Delta T}$ is the temperature change. For phase changes, the equation $q=mL$ is used, where $L$ is the latent heat. By combining these equations, calorimetry can analyze complex processes involving both temperature and phase changes.

How do you solve a calorimetry problem involving both temperature and phase changes?

To solve a calorimetry problem involving both temperature and phase changes, follow these steps: 1) Draw a temperature vs. heat diagram to visualize the process. 2) Identify the initial and final states of the substances. 3) Write the calorimetry equation $q=-q{b}_{}$. 4) Expand the equation to account for multiple processes, such as $q{a}_1+q{a}_2=-q{b}_{}$. 5) Replace $q$ with $\mathrm{mc\Delta T}$ for temperature changes and $\mathrm{mL}$ for phase changes. 6) Plug in known values and solve for the unknowns. This method ensures accurate calculation of heat transfer during complex processes.

What is the difference between specific heat capacity and latent heat in calorimetry?

Specific heat capacity ($c$) is the amount of heat required to raise the temperature of 1 kg of a substance by 1°C without changing its phase. It is used in the equation $q=\mathrm{mc\Delta T}$. Latent heat ($L$) is the amount of heat required to change the phase of 1 kg of a substance without changing its temperature. It is used in the equation $q=\mathrm{mL}$. Specific heat capacity applies to temperature changes, while latent heat applies to phase changes.

How do you calculate the amount of ice needed to achieve a specific temperature in a calorimetry problem?

To calculate the amount of ice needed to achieve a specific temperature in a calorimetry problem, follow these steps: 1) Identify the initial temperatures and masses of the substances involved. 2) Write the calorimetry equation $q=-q{b}_{}$. 3) Expand the equation to account for multiple processes, such as $q{a}_1+q{a}_2=-q{b}_{}$. 4) Replace $q$ with $\mathrm{mc\Delta T}$ for temperature changes and $\mathrm{mL}$ for phase changes. 5) Plug in known values and solve for the mass of ice needed. This method ensures accurate calculation of the required amount of ice.

Why is it important to consider both temperature and phase changes in calorimetry problems?

It is important to consider both temperature and phase changes in calorimetry problems because real-world scenarios often involve both processes. Ignoring either aspect can lead to inaccurate calculations of heat transfer. Temperature changes involve specific heat capacity, while phase changes involve latent heat. By considering both, you can accurately determine the total heat transfer, ensuring precise results in applications such as material science, engineering, and environmental studies.