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Current time:0:00Total duration:10:31

AP.Chem:

ENE‑2 (EU)

, ENE‑2.E (LO)

, ENE‑2.E.1 (EK)

, ENE‑2.E.2 (EK)

let's look at the heating curve for water a heating curve has temperature on the y-axis in this case we have it in degrees celsius and heat added on the x-axis let's say it's in kilojoules let's say we have 18.0 grams of ice and our goal is to calculate the total heat necessary to convert that 18 grams of ice at negative 25 degrees celsius to steam at 125 degrees celsius so we're starting with ice at negative 25 degrees celsius and first we need to heat up the ice to zero degrees celsius which we know is the melting point so on our heating curve we're going from point a to point b to calculate the heat necessary we need to use the equation q is equal to m c delta t where q is the heat added m is the mass of the ice c is the specific heat of ice and delta t is the change in temperature which is the final temperature minus the initial temperature so we're trying to calculate q and we know the mass of our ice is 18.0 grams the specific heat of ice is 2.03 joules per gram degree celsius and for the change in temperature it's final minus initial so the final temperature would be zero degrees celsius initial is negative 25 so 0 minus negative 25 gives us positive 25 degrees celsius so grams will cancel out degrees celsius cancels out and this gives us q is equal to 9.1 times 10 to the second joules to two significant figures or we could also write 0.91 kilojoules now that the ice is at zero degrees celsius we know it's going to melt so we're going to go from point b on the heating curve to point c and to calculate how much heat is necessary to melt the ice we need to know the heat of fusion of ice which is equal to 6.01 kilojoules per mole so we need to figure out how many moles of ice we have well we're starting with 18.0 grams we divide by the molar mass of h2o which is 18.0 grams per mole and the grams will cancel and give us one mole so we have 1.00 moles of ice and we multiply that by 6.01 kilojoules per mole and the moles cancel out and give us 6.01 kilojoules now that all the ice is melted we have liquid water and so on our heating curve we're going to heat that liquid water from 0 degrees celsius to 100 degrees celsius which is the boiling point of water so we're going from point c to point d on the heating curve to calculate the heat added we use the q is equal to m c delta t equation again so we're solving for q the mass is still 18.0 grams but the specific heat now since we have liquid water we need to use the specific heat of liquid water which is 4.18 joules per gram degrees celsius and for the change in temperature the final temperature is 100 so 100 minus 0 gives us positive 100 degrees celsius so grams cancel out degrees celsius cancels out and we find that q is equal to 7.52 times 10 to the third joules just correct that three there 7.52 times 10 to the third joules which is equal to 7.52 kilojoules once we reach point d in the heating curve we're at the boiling point of water so the heat that we add now is going to go into turning the liquid water into gaseous water so going from point d to point e we're doing a phase change we need to know the heat of vaporization of water and that's equal to 40.7 kilojoules per mole and we already know we have one mole of h2o so one mole times 40.7 moles the moles cancel and it takes 40 40.7 kilojoules of energy to convert the liquid water into gaseous water or steam next we're going to heat the gaseous water from 100 degrees celsius to 125 degrees celsius so we're going from point e to point f on the heating curve and to figure out how much heat we need to add we use the q is equal to m c delta t equation one more time so we're solving for q and we still have 18.0 grams this time we need to use the specific heat of steam which is 1.84 joules per gram degree celsius the change in temperature would be 125 minus 100 or positive 25 degrees celsius so grams cancel units cancel out and we get q is equal to 8.3 times 10 to the second joules to two significant figures which is equal to 0.83 kilojoules finally we need to add everything up so going from point a to point b on the heating curve so think about just the x-axis this time all right so going from point a to point b we calculated that to be equal to 0.91 kilojoules and then from point b to point c we calculate that to be 6.01 kilojoules from c to d so this distance here was 7.5 to from d to e this was the uh this was the big one here this was equal to 40.7 40.7 kilojoules and finally from e to f we calculated this was equal to point eight three point eight three kilojoules and when you add everything up this is equal to fifty six point zero kilojoules so that's how much energy it takes to convert 18.0 grams of ice at negative 25 degrees celsius to gaseous water at 125 degrees celsius next let's think about the slopes of the different lines on our heating curve so let's look at the line going from b to c and also the line going from point d to point e both of these lines represent phase changes going from point b to point c was going from a solid to a liquid and going from point d to e was going from a liquid to a gas and since the slope of both of these lines is zero that means as you add as you add heat on the x-axis the temperature doesn't change so during a phase change all the energy goes into disrupting the intermolecular forces that are present and they don't go into increasing the temperature so there's no increase in temperature during the phase change think about going from point d to point e this was converting our liquid water into gaseous water so as as the heat is being added all that energy goes into breaking the intermolecular forces between water molecules and pulling apart those liquid water molecules and turning them into gaseous water molecules so it's only after all of the liquid water molecules are converted into gaseous water molecules that's when we see the temperature increase again so talking about from point e to point f everything is now in the gaseous state and then we see the increase in temperature finally let's compare the slope of the line from a to b to the slope of the line from c to d if we look at it the slope of the line from a to b is a little bit steeper than the slope of the line from c to d the reason for the different slopes has to do with the different specific heats from a to b we use the specific heat for ice which is 2.03 joules per gram degree celsius from c to d in our calculation we use the specific heat for water which is 4.18 joules per gram degree celsius the higher the value for the specific heat the more energy it takes to raise the temperature of a substance by a certain amount so if we think about comparing these two let's say we try to raise the temperature of ice by 25 degrees celsius so think about this distance here on the y-axis we would have to put in only a small amount of energy to get ice to increase its temperature by 25 degrees celsius we think about that same temperature change on liquid water so if we tried to increase the temperature of liquid water by that same amount 25 degrees we would have to put in more energy so on the x-axis we have to put in more energy to accomplish the same change in temperature and that's because liquid water has a higher specific heat since it might be a little bit hard to see on that diagram let's think about let's think about putting some heat into a substance here so let me just draw a horizontal line and then we're trying to accomplish a certain temperature change so i'll draw a vertical line those two give me those two give me a line with a slope so let's say we're trying to accomplish the same change in temperature so i'll draw this y distance the same as before but we have a higher specific heat so it takes more takes more energy therefore this x distance is going to increase and when we increase the x distance we see that the slope decreases so the greater the value for the specific heat the lower the slope on the heating curve

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