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# Specific heat and latent heat of fusion and vaporization

## Video transcript

let's talk about specific heat and the heat of fusion and vaporization let's say you had a container of some liquid and you wanted to increase the temperature you'd probably add Heat but how much heat should you add there's a formula for it let's try to figure out what should go in here it's going to depend on a few things for one it's going to depend on how much you want to increase the temperature so what is the amount by which you want to increase the temperature the more you want to increase the temperature the more heat you'll need to add it's also going to depend on how much of the material you have in other words the mass of this liquid in here the more you have the more heat you're going to have to add to change that temperature and it depends on one more thing the specific heat of that particular material so different materials will be harder to increase in temperature than other materials and if a material has a high specific heat it'll take more joules of heat in order to increase the temperature so let's get a little more specific let's say our liquid was water and it was at a temperature of 20 degrees Celsius let's say it was a big container had two kilograms of water in it now the specific heat you can look those up or sometimes they're given it turns out the specific heat of water is 4,186 joules per kilogram degrees Celsius and these units give you an idea of what the meaning of the specific heat is it's telling you that water takes 4,186 joules to heat up one kilogram by one degree Celsius so you can already tell this is going to take a lot of heat water has a very high specific heat it can store a lot of heat energy without raising its temperature by much and so let's say the question was we wanted to get this temperature up to 50 degrees Celsius how much heat energy are we going to add in order to get it to 50 degrees Celsius well we can come over to here the amount of heat that we're going need will be the mass is two kilograms so two kilograms times the specific heat which is 4186 times the change in temperature this really means T final minus T initial so what's T final going to be T final is 50 degrees Celsius that's where we want it to get minus the initial it started at 20 Celsius and now we can solve this for the amount of heat and if you multiply all this out you get 250 1160 joules that's a lot of heat energy just to get water to increase by 30 degrees Celsius that's why we often use water as a heat sink you can put a lot of heat in water and not change the temperature by all that much but that example was pretty simple let's look at a harder one let's say instead of heating the cube with a fire underneath we just take a hot piece of metal say we've got a 0.5 kilogram piece of copper and we drop it into the water we've heated up this copper and now we drop it in the same container of water and we want to know what equilibrium temperature will they reach the copper is going to cool down the water is going to heat up eventually they're going to reach equilibrium at some temperature what temperature will that be well we're going to need to know a few things I already told you the mass of the copper we're going to need to know its initial temperature so the initial temperature of the copper I said we made it really hot let's say it's ninety degrees Celsius we'll need to know the specific heat of copper and turns out the specific heat of copper is 387 and let's say the water has the same properties it had initially there are two kilograms of it the specific heat is always 4,186 for water and let's say it started at a temperature of 20 degrees Celsius so we know the equilibrium temperature the temperature at which these are going to meet is somewhere between 20 and 90 we have to figure out exactly where it's gonna be and the trick we're going to use is if you think about it the copper is gonna lose heat the water is gonna gain heat how do those heats compare they've got to be the same assuming no heat is being lost to the environment we're gonna assume no heats lost so you want this to happen in what's often called a calorimeter something insulated something that prevents any heat from getting out and if no he gets out then whatever heat the water gains has to be the same as the Heat lost by the copper so basically if you add up the heat from the copper plus the heat from the water you're going to get zero because one of these are going to be negative and one of these are going to be positive and they're going to be the same absolute value how do you find these we had a formula remember Q equals MC delta T which I like to remember because it looks like MCAT so MC this Delta looks like an a to me so it looks like Q equals Mk so I've got to use the mass of the copper there was 0.5 kilograms of copper times the specific heat which is 387 times the change in temperature I don't know the final temperature that's okay I'm going to name my ignorance and give it a variable here I'm going to call it T final - I do know the initial it started at 90 degrees Celsius so minus 90 degrees Celsius plus the heat gained by the water which we can use the same formula for MCAT so the mass is 2 kilograms the specific heat is 4,186 T final I still don't know T final but I do know the T initial the initial temperature is 20 degrees Celsius and I've got to set this all equal to zero ran out of room there sorry about that this looks a little intimidating now you've got this big equation you've got your unknown hidden in here is this solvable yeah it's solvable look you've only got one unknown the unknown is T final these are both the same variable this is the temperature at which the water and copper are going to meet this whole term here this orange term is going to come out to be some negative number because the copper is going to lose heat energy and the water term is going to come out positive because it's going to gain heat energy the two will cancel out they'll give you 0 that's the condition that we're requiring we just have to solve for T final which means we multiply all this out combined the T final terms and then solve for T final first I'm just going to multiply everything out I'm going to combine the T final terms and the two terms that don't have T final then I'm going to move everything over to the other side I solve for T final and I get 21.5 eight degrees Celsius and when you look at this you might think hey we must have screwed up 21.5 eight the water started at 20 it barely increased its temperature at all and yeah that's what we were saying is that this specific heat for water is so high you can add a lot of heat and it doesn't change its temperature much note that we could have added in this container this container might absorb some of that heat so we could have had another term over here Q of the container and taken that into account or we could have dropped another cube of something in here we could have added that over here if you add up all the Q's from everything involved that might gain heat or lose heat you can set that equal to zero because if no heats getting out that heat just has to be interchanged within here no heats going to be created or destroyed it just gets transferred amongst the materials that are interacting so this is the key problem-solving idea when you're doing these specific heat problems you set it up with this and then you solve for the unknown in this case it was T final sometimes thing you won't know would be the mass of one of them or the specific heat of one of them regardless you solve for the thing you want to find let me ask you another question let's say we took the same amount of water two kilograms at a temperature of 20 degrees Celsius but this time I want to know how much heat do I have to add in order to boil all of this water into steam well the first thing we need to do is get this to the boiling temperature and the boiling temperature of water is 100 degrees Celsius so Q I first have to use M cap MC delta T the mass is to the specific heat of water is 4,186 the change in temperature will the boiling point of water is a hundred so I have to get this water from twenty to a hundred that means T final is a hundred t initial was twenty and I get that the heat that needs to be added in order to get this to the boiling temperature is 669 thousand seven hundred and sixty I'm being sloppy with significant figures but this is the number you get from that calculation but that's not enough to boil it that's just the heat required to get the water up to 100 degrees Celsius that's not enough if you get water to 100 degrees Celsius and let it sit there it'll just sit there it won't actually boil you've got to keep adding Heat how much more heat are we going to have to add once this water gets to a hundred in order to boil all of this water into steam for that you need to know about the heat of fusion and vaporization in this case since it's boiling the heat of vaporization because we're turning liquid into vapor if we were turning liquid into a solid it'd be the heat of fusion the formula for the heat of fusion and vaporization looks like this Q the amount of heat you need to add in order to change the phase this is what happens when you change phase heat of fusion and vaporization specific heat is what happens when you change temperature so this calculation showed us how much heat we needed to change the temperature by eighty degrees Celsius this calculation is going to tell us once we're at a hundred how much heat do we need to add to change the phase of all of this water into vapor and the formula for the heat of fusion and vaporization looks like this Q equals ml M is the mass so the more mass you have and the more heat you have to add L is the latent heat of fusion or vaporization this is a number similar to the specific heat but instead of telling you how much heat you have to add in order to change the temperature this is telling you how much heat you have to add in order to change the phase and it turns out the latent heat of vaporization for water is huge two million two hundred and sixty thousand joules per kilogram this means it takes two million two hundred sixty thousand joules to turn one kilogram of water which is at the boiling point into one kilogram of vapor so if I'm trying to get this water from 20 degrees turned into vapor first I need to MC delta T get it to 100 degrees Celsius and then I need to add to that another amount of heat an amount of heat M times L the mass is two kilograms of this water L for water is two million two hundred and sixty thousand so I get that it will take six hundred and sixty nine thousand seven hundred and sixty joules to get the water up to the boiling point and it will take another four million five hundred and twenty thousand joules to turn that water into vapor which in total gives five million 189 thousand seven hundred and sixty joules in order to get this from twenty all the way up to two kilograms of vapor now there's one more thing that I want to show you let's get rid of this let's say instead of starting with water you started with some ice let's say you started with a big piece of ice three kilograms worth of ice and it was very cold this wasn't just a zero let's say this started at an initial temperature of negative forty degrees Celsius the question I want to know how much heat would we have to add in order to turn this three kilogram block of ice into three kilograms of water vapor but not just get it to the point of being vapor I want to get it to hotter than 100 degrees Celsius I want to get this to a T final of a hundred and sixty degrees Celsius how much heat would I have to add in order to do that and to track this to visualize this a little better I'm going to track what the temperature is on this vertical axis as a function of how much heat we've added into this system let me quickly show you how not to do this a naive approach would say okay Q equals MC delta T my M is three kilograms my C is well whatever my C is we'll talk about that in a minute delta T all right delta T my final temperature is 160 degrees Celsius my initial temperature is negative forty you can't forget the negative sign and I just plug in my specific heat and I find my value this is wrong you cannot do it this way for one what specific heat are you going to put in the specific heat of water the specific heat of ice the specific heat of vapor they all have different specific heats and for two there's going to be phase changes first the ice turns into water and then some later point the water turns into vapor you can't just neglect and gloss over those phase changes so this is not how you do it here's what you should do we're going to start at negative 40 degrees Celsius I know this is above the axis over here but just pretend like this isn't the zero point for this vertical axis and we're going to add Heat and that's going to bring this temperature up to zero degrees Celsius and we have to stop there we have to pause at zero degrees Celsius because we have to pause every time there's going to be a phase change so how much heat does this correspond to right here we can use MC delta T and there's three kilograms the specific heat of ice is about two thousand and ninety and the final temperature is zero our initial temperature so T final minus our initial is negative forty don't forget that negative sign and if you calculate that you get that that amount of heat is 250 thousand eight hundred joules but that's just getting the ice up to the melting point now we need to melt it what's this graph going to look like while it melts well the temperature of the ice is going to stay constant while this thing melts as you're melting the ice cube the temperature doesn't change all that energy is going into breaking those bonds and turning this ice into water so how much heat does this correspond to it's a phase change so you got to use Q equals ml the M is three kilograms the latent heat we can't use the latent heat of vaporization this is a solid turning into a liquid that's latent heat of fusion that we need in the latent heat of fusion for water is about three hundred and thirty three thousand joules per kilogram which gives you nine hundred and ninety-nine thousand joules of heat in order to turn this ice at zero degrees Celsius into water at zero degrees Celsius now you see how this works we've got to take this water at zero up to what not up to one hundred and sixty up to one hundred degrees Celsius because that's where it's going to turn into steam and whenever there's a phase change you've got a pause because your specific heats going to change its value so this Q that is required right here we can do MC delta T specific heat of water is 4,186 delta T is final it's 100 initial is zero we get 1 million two hundred and fifty five thousand eight hundred joules now we've got to turn that water into steam how much heat is that going to take that's a phase change so we've got to use ml the masses three kilograms the latent heat of vaporization is two million two hundred and sixty thousand which means we need six million seven hundred and eighty thousand joules in order to turn this water into vapor and we've got one more step to go we've got to turn this vapor at a hundred into vapor at a hundred and sixty we've got to do one more MC delta T the mass of the steam is three kilograms the specific heat of steam is about two thousand and ten the final temperature is 160 the initial temperature was 100 which gives us three hundred and sixty one thousand eight hundred joules that's how much heat it would take to get this ice cube at negative forty into vapor at 160