Ch 21: Heat and TemperatureWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
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Ch 21: Heat and Temperature
Ch 22: Kinetic Theory of Ideal Gasses
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 25: Electric Force & Field; Gauss' Law
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Ch 39: Atomic Structure
Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics
Zeroth Law of Thermodynamics
Thermal Expansion
Introduction to Heat
Changes in Temperature & Specific Heat
Changes in Phase & Latent Heat
Temperature Change Across Phases
Phase Diagrams, Triple Points and Critical Points
Heat Transfer

Concept #1: Changes in Temperature Across Phases


Hey guys, we've talked before about how much heat is required to change the temperature of a substance in a single phase that was where we got our M CAT equation from. Recently we talked about phase changes and how much heat is required to change the phase that was when we talked about latent heat and the M L equation now we want to put it all together and talk about how to change the temperature of a substance across phase changes what if I have a sample of water at 50 degrees Celsius and I want to cool it down to -50 degrees Celsius well at 0 degrees Celsius there's a phase change between water and ice so I need to find the heat to cool it down to 0 the heat to freeze it and then the heat to cool it down to -50 as I ice. So let's talk about that now, remember that heat input while in a constant phase changes ops sorry wrong color, changes the substance temperature if you're in single phase and you're adding in heat you're changing the temperature. But adding heat during a phase transition or during a phase change does not. But you need it it is absolutely required to change the phase. During that phase change temperature is not changing but you need to keep adding in heat or keep removing heat to change that phase, what heat is required then to change the temperature of a substance when it is also changing phase that's the question we want to answer now that's just the sum of these points right here the heat to change the initial temperature to the temperature of the phase change so if I were to draw a temperature line right here. You're at some initial temperature, you need to add heat Q equals M C delta T to reach the temperature of the phase change then you need the latent heat required to change the phase how much of that latent heat do you need to change the phase depends upon the mass but you still need some heat to change phase so right here you're going to be throwing in some Q equals M L some latent heat just to change the phase at a constant temperature then you need to the heat to change from the phase change temperature sorry wrong color from the phase change temperature to the final temperature so now there's some final temperature your trying to go to and assuming there's no phase change in between these two you just go straight across again adding in heat with Q equals M C delta T. Now this just assumes one phase change if there are two phase changes then you're going to have another set of latent heats and another M CAT equation to take you from that second phase change temperature to your final phase change. As an example consider this graph which plots the temperature vs the heat input for ice starting at -15, why does is it, every time I end it , it ends with red its just bad luck or me not paying attention it starts at -15 degrees Celsius and it transitions from ice to water and ends at 30 degrees Celsius so starting here -15 degrees Celsius this is ice. This is in the solid phase, this is the transition of ice to water where no temperature change occurs and this is the water phase right if this were to keep on going eventually you would hit another phase change at what temperature? 100 degrees Celsius which is the boiling point of water where this would go from water to vapor which is what we call the gaseous phase of water and then the rest of this would be vapor this could continue the problem only says to go from -15 to 30 degrees but we could continue and then we hit another phase boundary at 100 degrees we'd have to put more latent heat in before we could be in a vapor and continue increasing the temperature. You can see that the total amount of heat then to go from -15 degrees Celsius to positive 30 is just the sum of these three heats the heat to raise the temperature from -15 degrees Celsius to the phase change temperature which is the boiling point, sorry ice to water is the melting point. That's what we call this phase change which for water is 0 degrees Celsius. That's for ice to water so that amount of heat is represented by this horizontal distance right here this is Q equals M C delta T that horizontal distance because this is Q that horizontal distance is just this heat that I'm indicating to go from negative 15 to the phase change then what you need is you need to add the latent heat of the ice to water transition that is just this width right here this is Q equals M L that's going to be the latent heat for melting and finally you need to add in the heat to raise the temperature from 0 degrees Celsius which is where it's at, at the phase change to the final temperature of 30 degrees Celsius right and that's this width right here which is another M CAT equation and this is an example of how to approach a problem to find out the total amount of heat when one phase change occurs if there was a second phase change then this third bullet point right here wouldn't say from 0 degrees to 30 degrees you keep going until you hit the second phase boundary which is the boiling point which is 100 degrees then there will be a fourth bullet point this is the latent heat for water to vapor which would be this width and then if you wanted to go to a final temperature of say a 150 degrees Celsius you would need another amount of heat to change the temperature from 100 degrees up to a 150 degrees Celsius so that's how to do it for two phase transitions, it's just the same repeated unit over and over but there is no more phase changes after two there's only two liquid to water, water to the gas. Let's do a quick example what heat is required to increase the temperature of 0.05 kilograms of ice from -15 degree Celsius to 30 Celsius exactly the graph that we just showed now we want to know exactly how much heat was represented by that graph that total heat by the way. Is just this entire horizontal distance that's Q total so if you knew like the indexing of this line if this line had dashes here lets just say that was 1 joule and that was 2 joule etc you could literally just measure out how long this was and that would tell you the total one but since we don't have that we have to do this the long way. So remember there are three steps first step we need to go from as ice from -15 degrees to 0 degrees this is an M CAT equation we know the mass is 0.05 what's the specific heat, we need the specific heat for ice not for water and we were told the specific heat of ice was this 2.05 kilojoules right per kilogram Kelvin and the change in temperature delta T is positive 15 degrees Celsius you're increasing it by 15 degrees Celsius and remember because this is a change it's exactly the same as 15 Kelvin so I can plug in 15 here all of this amounts to 1.54 kilojoules. Step two ice to water at 0 degrees this is a latent heat problem so we need M times the latent heat the mass is still 0.05 kilograms and we're told that the latent heat of fusion is 334 and once again this is kilojoules per kilogram so our answer is 16.7 kilojoules. Three now as water we're going from 0 degrees Celsius to 30 degrees Celsius this is another M CAT problem because there is a temperature change and were staying as a single phase the mass is 0.05 the specific heat now has to be for water we can't use specifically for ice we can't use a specific heat for water vapor in needs to be for water and the specific heat for water is 4.186 and remember this is kilojoules so our answer is going to be in kilojoules and this delta T is positive 30 degrees Celsius but that's equivalent to 30 Kelvin because it's a change in temperature so you just put 30 here multiplying all this together we get 6.3 kilojoules. So what is the total heat for everything for the entirety. For that whole process it's just going to be the sum of each individual step and remember if we had said what was the amount of heat required to convert ice at -15 degrees Celsius to water vapour at 150 degrees Celsius there'd be two more steps here. This step would not go from 0 to 30 it will go from 0 to the next phase change which is 100 the fourth step would be the latent heat for the phase chain from liquid to vapour and the fifth step would be the temperature change from 100 degrees Celsius to 150 degrees Celsius as gas. So this is just going to be 1.5 kilojoules I'm not carrying the 4 because everything only has one decimal place 16.7 kilojoules so I want some consistency and this whole thing winds up being 24.5 kilojoules so that's exactly how much heat is required to take ice at -15 degrees Celsius and convert it to water at 30 degrees Celsius. That wraps up our discussion on changes in temperature across the phase boundaries. Thanks for watching guys.

Practice: The temperature needed to keep CO2 frozen is -78.5Β°C. What heat is required to sublimate 5kg of dry ice and then raise the gaseous CO2 to a room temperature of 27Β°C? The πΏπ‘ π‘’π‘π‘™π‘–π‘šπ‘Žπ‘‘π‘–π‘œπ‘› for CO2 is 571 kJ/kg, and 𝐢𝐢𝑂2,𝑔 = 0.79 kJ/kg K.

Example #1: Heat Removed to Freeze Water Vapor


Hey guys, let's do an example. You have 50 grams of water vapor initially at 120 degrees Celsius. If you want to cool this vapor down to a temperature of -30 degrees Celsius, how much heat do you have to remove from the vapor? I'm not going to rattle off all the constants. Everything else I have that's a constant, you guys can just look at that. So we're going now from ice sorry from vapor to ice. If you want to get that water down to -30 degrees Celsius, this is definitely ice. So we're going from vapor to water to ice. This is going to take 5 steps. Notice that the phase change temperatures, this one's 100 Celsius, this one is 0 degrees Celsius. Now let's just do this. It's going to be a tedious calculation but the only way to solve it is to do it. Step one, as vapor, we're going from our initial temperature of 120 degrees Celsius to the first phase change which is at 100 degrees Celsius. We're using our M CAT equation because this is a change in temperature. We were talking about 50 grams of ice which is 0.05 kilograms, the specific heat of ice is 2.1 kilojoules per kilogram Kelvin so our answer is going to be kilojoules and our temperature change here is clearly 20 degrees Celsius. Which is the same as 20 Kelvin. Always keep that in mind, change in temperature same in Celsius as it is in Kelvin. So this is 20 and if we plug that in our answer is oh sorry I made one just tiny little mistake here this delta T is -20 and this is -20 Kelvin and it should be negative not only because you're dropping the temperature so that delta T should absolutely be negative but when you plug in a negative number here your Q was going to be negative. So far every problem that we've done has involved a positive Q. What does a negative Q mean again? It means heat is leaving your substance and if we want to take water vapor and cool it down to ice, we absolutely need to remove heat from it. So this is -2 kilojoules. Step two, this is vapor to water at 100 degrees Celsius. This is a latent heat. The mass was 0.05 right and the latent heat of vaporization we are told is big, 2260 kilojoules. So this isn't in joules this is still in kilojoules per kilogram. So this is a whopping -113 kilojoules to convert that water vapor into water. Step three is as water, we're going from 100 degrees Celsius to 0 degrees Celsius. 0 degrees is the next phase boundary. This delta T is -100 degrees Celsius. By the way I messed up once again on the sign just because I haven't done any problems like this yet. Q2 is negative, this is negative. You put a negative sign there because we're removing heat. The magnitude ML is the same whether you're going from vapor to water or water to vapor, but if you're going from vapor to water you need to remove heat so you need to put that negative sign there. I already had the negative sign in the answer but I forgot to put it in the equation. Now as water dropping 100 degrees Celsius we're using M CAT equation again so still 0.05 degrees. Now the specific heat of water is 4.186 kilojoules per kilogram Kelvin and this is -100 Kelvin, this whole thing becomes -21 kilojoules. So so far predominately the biggest energy was the amount of energy to condense all that water vapor into water. That took 10 times basically the amount of heat that it took to drop it from 100 degrees Celsius to 0 and it took 100 times the amount of heat to go from 120 one twenty to 100 degrees. So now four is the next boundary which is water to ice at zero degrees Celsius. This is a latent heat so we're using ML, negative ML, I keep forgetting, -0.05 and the latent heat of fusion for water is 334 and it's also in kilojoules right kilojoules per kilogram so our answer is -17 kilojoules. So you see for the same mass of water, it's a lot less energy to convert it between ice and water as it is to convert it between water and vapor. This is all for the same mass. And the last step is as ice, so we're going from 0 degrees Celsius to -30 degrees Celsius. This is an M CAT equation, mass is 0.05 as it was the specific heat of ice is 2.1, this is still in kilojoules and the change in temperature is -30 degrees Celsius which is the same as -30 Kelvin so this comes out to be -3 kilojoules. A really long and tedious problem but not difficult. You're only applying two equations. The total heat then removed is going to be Q1 plus Q2... etc. Those are all negatives. So this is -2 minus 113 minus 21 minus 17 minus 3 and that all works out to be -156 kilojoules. So you have to remove 156 kilojoules of heat from that sample of water to get it from water vapor at 120 degrees Celsius down to ice at -30 degrees Celsius. Alright guys, that wraps up this problem. Thanks for watching.