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Change of state example

Specific heat capacity and enthalpy of vaporization example: calculating how much energy it takes to vaporize 1.00 kg of ethanol starting from 20 degrees C.  Created by Sal Khan.

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Video transcript

I have this problem here from chapter five of the Kotz, Treichel, and Townsend Chemistry and Chemical Reactivity book, and I'm doing this with their permission. So they tell us that ethanol, C2H5OH, boils at-- let me do this in orange-- it boils at 78.29 degrees Celsius. How much energy, in joules, is required to raise the temperature of 1 kilogram of ethanol from 20 degrees Celsius to the boiling point and then change the liquid to vapor at that temperature? So there's really two parts of this problem. How much energy, in joules, to take the ethanol from 20 degrees to 78.29 degrees Celsius? That's the first part. And then once we're there, we're going to have 78.29 degrees Celsius liquid ethanol. But then we also need the energy to turn it into vapor. So those are going to be the two parts. So let's just think about just raising the liquid temperature. Let's figure out how we're going to do that. Just the liquid temperature. So the first thing I looked at is well how many degrees are we raising the temperature? Well we're going from 20 degrees Celsius-- let me write Celsius there just so it's clear-- 20 degrees Celsius to 78.29 degrees Celsius. So how much did we raise it? Well, 78.29 minus 20 is 58.29. So our change in temperature is equal to 58.29 degrees Celsius or this could even be 58.29 kelvin. And the reason why we can do that is because differences on the Celsius scale and the kelvin scale are the same thing. The kelvin scale is just a shifted version of the Celsius scale. If you added 273 to each of these numbers you would have the kelvin temperature, but then if you take the difference, it's going to be the exact same difference. Either way you do it, 78.29 minus 20. So that's how much we have to raise the temperature. So let's figure out how much energy is required to raise that temperature. So we want a delta T. We want to raise the temperature 58.29. I'll stay in Celsius. Actually let me just change it to kelvin because that looks like what our units are given in terms of specific heat. So let me write that down. 58.29 kelvin is our change in temperature. I could have converted either of these to kelvin first, then found the difference, and gotten the exact same number. Because the Celsius scale and the kelvin scale, the increments are the same amount. Now, that's our change in temperature. Now how much ethanol are we trying to boil? Well, it tells us right over here. It tells us that we're dealing with 1 kilogram of ethanol. And everything else they give us is in grams. So let me just write that 1 kilogram, that's the same thing as 1,000 grams. We could just write it here. 1.00 kilogram is equal to-- or let me write it this way-- times 1,000 grams per 1 kilogram. These cancel out. This is the same thing as 1,000 grams. Although the reality is we only have three significant digits-- this makes it look like we have four. So we have 1,000 grams times 1,000 grams. And then we just multiply this times the specific heat of ethanol. The specific heat capacity of ethanol right here, 2.44 joules per gram kelvin. So times 2.44 joules. Let me write it this way, 2.44 joules per gram kelvin. You see that the units work out. This kelvin is going to cancel out with that kelvin in the denominator. This gram in the numerator will cancel out with that grams. And it makes sense. Specific heat is the amount of energy per mass per degree that is required to push it that 1 degree. So here we're doing 58 degrees, 1,000 grams, you just multiply it. The units cancel out. So you have kelvin canceling out with kelvin. You have grams canceling out with grams. And we are left with-- take out the calculator, put it on the side here. So we have 58.29 times 1,000-- times one, two, three-- times 2.44 is equal to-- and we only have three significant digits here. So this is going to be 142-- we'll just round down-- 142,000 kelvin. So this is 142,000. Sorry 142,000 joules. Joules is our units. We want energy. So this right here is the amount of energy to take our ethanol, our 1 kilogram of ethanol, from 20 degrees Celsius to 78.29 degrees Celsius. Or you could view this as from 293 kelvin to whatever this number is plus 273, that temperature in kelvin. Either way, we've raised its temperature by 58.29 kelvin. Now, the next step is, it's just a lot warmer ethanol, liquid ethanol. We now have to vaporize it. It has to become vapor at that temperature. So now we have to add the heat of vaporization. So that's right here. We should call it the enthalpy of vaporization. The enthalpy of vaporization, they tell us, is 855 joules per gram. And this is how much energy you have to do to vaporize a certain amount per gram of ethanol. Assuming that it's already at the temperature of vaporization, assuming that it's already at its boiling point, how much extra energy per gram do you have to add to actually make it vaporize? So we have this much. And we know we have 1,000 grams of enthanol. The grams cancel out. 855 times 1,000 is 855,000 joules. So it actually took a lot less energy to make the ethanol go from 20 degrees Celsius to 78.29 degrees Celsius than it took it to stay at 78.29, but go from the liquid form to the vapor form. This took the bulk of the energy. But if we want to know the total amount of energy, let's see if we can add this up in our heads. 855,000 plus 842,000. 800 plus 100 is 900. That's 900,000. 50 plus 40 is 90. 5 plus 2 is 7. So it's 997,000 joules or 997 kilojoules. Or we could say it's almost 1 megajoule, if we wanted to speak in those terms. But that's what it will take for us to vaporize that 1 kilogram of ethanol.