Ch 19: Heat and TemperatureWorksheetSee all chapters
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Ch 02: 1D Motion / Kinematics
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Ch 19: Heat and Temperature
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Ch 39: Quantum Mechanics

Concept #1: Phase Changes and Latent Heat


Hey guys, in this video we want to talk about phase changes when we talked about adding heat to change the temperature we always consider the substance being in one phase even though I never explicitly said it turns out that the change phase you need to add in a bunch of heat called the latent heat or the latent heat of transformation so in this video we're talking about phase changes in latent heat let's get to it. Recall that substances commonly exist in one of three phases we have solids , we have liquids and we have gases now I said commonly because you also have plasmas, and Bose-Einstein condensates but we're not going to talk about those, those are way beyond the scope of any introductory physics course what the substance transforms from one phase to another this is known as a phase change right for obvious reasons sometimes the words aren't very complicated in physics sometimes they are luckily a lot of thermodynamics like phase change expansions, long expansions, linear expansions a lot of the words are very simple heat is required to change a substances phase just as it's required to change a substances temperature, now in the figure right next to me I graph the temperature versus the heat input into a substance. While during this first phase right heat is being input right and it's leading to a change in temperature. During the second plat the second phase this plateau right here he is being input but the temperature is not changing that's because all of the heat that's being input is going towards changing the substances phase temperature increases with heat input only so long the substance is not changing phase when the substance is changing phase heat input does not change the temperature so if you ever look at one of these T versus Q graphs whenever there's a plateau that means heat is being input but the temperature is not changing so you know you're in a phase change so all of the heat goes into the phase change for this third interval the heat is going into changing the temperature as well you can see that as you go horizontally the line rises so the temperatures going up for this fourth interval right here once again we have a plateau so this is a second phase change and in this last interval this goes on forever. Ideally, in reality, that's not true but ideally it goes on forever we have this all of this heat from now until any temperature only changing the temperature so these particular temperatures here are the phase change temperatures and they're important. First let's figure out what phases these are and then I will denote what temperatures they are at the lowest temperature what phase are you? At intermediate temperatures what phase are you? At high temperatures what phase are you? This question should be very easy to answer from any basic physical science class you had at low temperatures you're a solid, at intermediate temperatures you're liquid, and at high temperatures you're gas. So the first phase changes from solid to liquid which we call melting so this is the melting point or the melting temperature. The second phase change is from liquid to gas which we call evaporation so or boiling so this is the boiling temperature or the boiling point. The amount of heat required to change phase is known as the latent sorry wrong colour this is what we refer to as the latent heat. Sometimes it's called a latent heat of transformation. And the latent heat absolutely depends upon what phase change your undergoing if you have a T versus Q graph you can tell by the length of this what that latent heat is latent heat is given by L by the way so this is the latent heat for melting, just the width of that because this is on a Q plot so the width represented heat and the amount of heat is the latent heat so the width here is the latent heat for boiling now the latent heat gives us the amount of heat required per unit mass per unit mass to change phase now once again per unit mass is very important just like it was for the specific heat because per unit mass allows us to give this quality the latent heat to be inherent to a substance and not be dependent on the amount of substance obviously the more you have the more heat you need to put in to get to do anything but if you make the quality this measurement the latent heat per unit mass then it doesn't matter how much you have that quality is inherent to the material it only depends upon the material but this is going to be the Q for the phase change divided by M or you can write this as Q is just M times L, so you have your 2 equations for heat transfer you have your phase changing equation which I just gave and you have your temperature changing equation the M-CAT equation and the specific heat tells you how much heat is required to change the temperature the latent heat tells how much heat is required to change phase. So there are fundamentally different qualities but they're linked together if you want to analyse a substance throughout multiple phases as indicated in this graph ops sorry I think I hit page down.Bring myself back a little technical difficulty there is a difference specific heat sorry yeah like I said there is a difference specific heat for each phase water has three common phases vapor water ice each has a different specific heat and has a different latent heat for each phase that I said the three phase changes are gas to liquid, liquid to solid and gas to solid ignoring those weird phases like plasma and condensates there are three common phases so there are three possible phase changes.We call these different things depending on what the person who was originally talking about it way back when felt like calling it and the direction, gas to liquid is called condensation specifically it's also called fusion that's gas to liquid.Liquid to gas is called evaporation or vaporization the two on the right fusion and vaporization are very common, they're much much more common than condensation and evaporation by the way liquid to gas also called boiling right the temperature at which liquid converts to gas is typically called the boiling point not the evaporation point or the vaporization point it's typically called the boiling point so that's another word. Liquid going to solid is typically called freezing and solid going to liquid is typically called melting and a gas going into a solid is typically called a deposition and a solid going to a gas is called sublimation later on we'll talk more about this weird phase change gas directly to solid skipping liquid it does happen in one very very common example is dry ice, dry ice is carbon dioxide in a solid form and at normal pressures at atmosphere pressures it skips the liquid phase it goes straight from a solid to a gas so it sublimes from dry ice to gases carbon dioxide. The latent heat is actually the same for the phase change regardless of which direction you went so I gave you all these different names for the different directions of phase changes the latent heat is the same regardless of which way you go it's just one direction requires an input of heat into the substance one direction requires a removal of heat from the substance so in one case you're going to have +Q and in one case you're going to have -Q but the L is the same. Let's do an example. You have a styrofoam container of ice while you're at the beach if heat enters the container at 93 joules per second how long will you have before the 5 kilograms of ice inside the cooler melts the latent heat of fusion sorry the latent heat of fusion for ice is 334 kilojoules per kilogram, so we have our little styrofoam container filled with our ice of which we have 5 kilograms so the mass of the ice is 5 kilograms and we want to assume that this ice is hanging around at 0 degrees which is the temperature the ice typically exists at that's also the melting temperature of ice so how much heat is required to completely melt the ice that's just however much heat is required to convert 5 kilograms of ice from solid to liquid then if we know how quickly the heat is coming in and we know how much heat is needed we can find the amount of time. So first we know that Q is M.L. Now really quickly we need to decide right away is this going to be positive or is this going to be negative or are we going to add heat to the ice or remove heat to the ice we already sort of gave this away because we said heat enters but remember that ice exists at a lower temperature than water does so you have to add heat in order to get from ice to water, so this is going to be positive in other instances if we were trying to freeze water it would be negative so this is going to be 5 kilograms.This is going to be 5 kilograms times the latent heat which is 334 if you multiply this together we get let me write this down here because I need to save for a lot of space 1.67 times 10 to the 6 joules now really quickly I want to rewrite this in kilojoules just because oh sorry what am I even doing here I misread my notes. The latent heat remember is in kilojoules this answer is 1670 kilojoules but our rate at which heat is entering the container is in joules per second so we want to convert this from kilojoules which is what our answer is going to be since our latent heat is in kilojoules we want to convert that to joules so that we can use the rate at which heat enters Sorry about that guys I just looked at the wrong part of my notes 1.67 times 10 to the 6 joules now we are ready to find out how much time. So 93 joules per second is how much heat is entering per unit time so if you want to find the time we have to multiply the time up and divide 93 over right so delta T is Q over 93 which is 1.67 times 10 to the 6 divided by 93 this is why it was important that our heat be written in joules because this is 93 joules per second and you need the joules to cancel and so this is going to be 17957 seconds you can leave the answer like this but this is not a good answer to this is not a good form to leave the answer its not a good unit if you divide by 60 you see that this is 299 minutes also not great but getting a little better divide it by 60 one more time we get 5 hours there we go.I'm going to choose to leave my answer in a nice easy number like 5 hours. Alright guys that wraps up our discussion on phase changes and latent heat. Thanks for watching.

Practice: You have an unknown substance, X, which you want to measure the latent heat for. You find that it requires 1.7 x 106 J of heat to completely evaporate 5 moles of substance X. If substance X has a molecular mass of 225 g/mol, what is 𝐿𝑋,𝑒𝑣𝑎𝑝?

Example #1: Finding Information About Heat Changes in Data


Hey guys let's do an example 0.34 kilograms of an unknown substance Z is in a liquid state at 0 degrees Celsius you can add heat to it at a rate of 0.01 kilojoules per second and record the temperature, plotting your results in the following figure during your experiment the substance changes phases from liquid to a gas. Find A the specific heat of Z in the liquid phase B the temperature of the phase change from the liquid to gas C the latent heat of vaporisation for a Z and D the specific heat of Z in the gaseous phase. So first of all let's try to figure out what's going on here we saw a lot about how to work temperature verses energy sorry versus heat graphs this is a temperature versus time graph but we are adding heat at a constant rate so we can relate time to heat. Heat is just going to be 0.01 kilojoules time sorry this is Kilojoules per second times delta T. So in this time interval right here this was 15 seconds 0.15 kilojoules if heat was added in this time interval right here which is 5 seconds 0.05 kilojoules of heat was added in this time interval which was 30 seconds 0.3 kilojoules of heat was added. So we can absolutely convert this to a T versus Q graph if we wanted to but I'm just going to write the equivalence right here this is equivalent to 0.15 kilojoules this is equivalent to 0.05 kilojoules this is equivalent to 0.3 kilojoules now they ask us about a bunch of stuff they ask us about the liquid phase the gas phase the temperature of the phase change the latent heat etc. So first want to figure out where is the liquid phase where is gas phase remember that the liquid phase always occurs at the lower temperature the gas phase always occurs at the higher temperature so this is the liquid phase and this is the gas phase the problem also explicitly stated that it was in a liquid state at 0 degrees Celsius this right here is the point at 0 degrees Celsius so you know it has to be a liquid there, so either way you want to figure it out that's fine that means that this transition right here that is the change from liquid to gas so we had four problems to answer A was the specific heat as a liquid, B was the temperature of liquid to gas, C was latent heat of vaporization and D was the specific heat of the gas. Part B we can answer right away very easily just by looking at this graph we know that this is the phase change from liquid to gas what temperature does that occur at 65 degrees Celsius. Boom that one's done, it's also very easy to find the latent heat of vaporisation this right here is that phase change I just told you that the phase change takes 0.05 kilojoules to occur so if we know how much heat the phase change takes how do we find how much the latent heat is? Well for Part B the latent heat we know is always going to be the Q of the phase change divided by the mass and we're told that we have 0.35 kilograms of this mass so what we have is we have a heat of 0.05 kilojoules divided 0.35 kilograms. Just checking my notes where the heck is latent heat right here 0.143 kilojoules per kilogram. I'm sorry this was part C by the way not part B those two are immediately apparent. What's the latent heat? What's the temperature of the phase change? Now to find out the specific heat that's going to be a little bit more difficult to find a specific heat all that we can rely on is our heat transfer equation the M CAT equation. So what we're going to need to do is we're going to need to find changes in temperature that correspond to changes in heat sorry heat exchange in one phase and then we can find the specific heat right the specific heat is going to be how much heat is required to change a temperature delta T for some mass M. Well look in a liquid we already know to go from 0 degrees to 65 degrees Celsius it takes 0.15 kilojoules so we can solve for part A right part A says that C is a liquid. Which is Q over M delta T, how much heat is input to increase its temperature 65 degrees as a liquid 0.15 kilojoules. How much mass do we have right that still 0.35 that doesn't change what was the temperature change during this heat input 65 degrees Celsius which is the same as 65 Kelvin. So plugging that all in we get 6.59 Sorry we get 0.00659 kilojoules per kilogram Kelvin this is one of those instances where the heat is so small that we don't want to leave it in kilojoules so this is 6.59 times sorry not times anything just joules per kilogram Kelvin we just want to move this decimal place over three points to decrease it by a thousand to get rid of the kilo in kilojoules, so we've already answered three of the questions to approach the fourth question it's the same thing.Sorry I need a little more space right here the last question C for the gas is going to be the amount of heat input while it's a gas divided by the mass times how much temperature changed while it was a gas, well we know that 0.3 kilojoules were input during those 30 seconds it was a gas and during that time the temperature changed by 32 degrees Celsius. So Q is 0.3 kilojoules the mass is still 0.35 kilograms and the temperature change is 32 degrees Celsius which is the same as 32 Kelvin. Plugging all this in we get 0.0268 kilojoules per kilogram Kelvin and if we move once again this is over one to three to get rid of the kilo and kilojoules.We arrive at 26.8 joules per kilogram Kelvin for the specific heat in the gaseous phase. So that's all four answers even though our graph was temperature versus time not temperature versus heat which is what we're used to because there was a constant rate at which heat was being added It was easy to convert from time to heat and then it was almost like we could change this to a graph of temperature versus Q which we absolutely know how to use. That wraps up this problem. Thanks for watching guys.