A steel tank is completely filled with 1.90 m3 of ethanol when bo...

Question

A steel tank is completely filled with 1.90 m3 of ethanol when both the tank and the ethanol are at 32.0°C. When the tank and its contents have cooled to 18.0°C, what additional volume of ethanol can be put into the tank?

Accepted Answer

Hey everyone in this problem. We have a student who's completely filled a quartz Erling Myhre flask a volume 250 ml with glycerin liquid. The flask and the glycerin are in thermal equilibrium at 25°C. The flash was placed in a refrigerator at 10°C and were asked at 10°C, how much more glycerin can the student pour into the flask. Alright, so we're looking for how much more glycerin the student can pour into the flask, we're looking for some change in volume and that change in volume is due to a change in temperature. So let's recall that we can relate the change in volume to the change in temperature through the following equation delta V, the change in volume is equal to beta K. The coefficient of volume expansion. V not the initial volume times delta T. The change in temperature. Alright, so we're looking for this delta V. Let's write out the information we know and see where we, that gets us case if we have enough information to figure this out. So we know that v not the initial volume is ml. Okay, This flask is completely filled. It completely filled 250 ml. So we know that we have 250 ml of glycerin. And because we have this flask of 250 ml, we know that the courts is enclosing 250 ml as well. Okay, so our Vina is 250 ml. Our delta T. Well, that's going to be the final temperature minus the initial temperature. Okay, the final temperature is 10 degrees Celsius, initial temperature is 25 degrees Celsius. So we get 10 degrees Celsius minus 25 degrees Celsius gives us a delta T of negative 15 degrees Celsius. And what about these beta values these coefficients of volume expansion. These you can get from a table in your textbook or that your professor provided. Okay, we need to consider too, we need to consider beta for the courts and we need to consider beta for the glycerin. So beta for the courts is going to be equal to 0.12 times 10 to the negative five. And the unit here is per degree Celsius. And beta for the Glycerin is going to be times 10 to the -5. And again the same unit per degree C. So we know vina, we know delta T and we know our betas. So let's go ahead and find delta V. And again we have two cases we need to consider, we need to consider it for the courts and for the glycerin. So let's start with the courts. Delta V is going to be equal to beta. And in this case Beta Q for the court, 0.12 times 10 to the negative five per degree Celsius. The not 250 mL times delta T, negative 15 degrees Celsius. Okay, the unit of degrees Celsius will divide out, which leaves us just with the unit of milliliters for the change in volume, which is a unit that we would be looking for. And we multiply to get negative 0.0045 ml. Okay, so this is a change in volume of the courts. Let's do the same for glycerin. Give ourselves a bit more room here and we're gonna have delta V. And let's call this one delta V Q for the courts. And we'll call this delta V G. For the glycerin. We get beta. Okay, that coefficient of volume expansion for glycerin 49 times 10 to the negative five. Pretty grease sauce. Yes, 250 mL. The initial volume and the change in temperature negative 15 degrees Celsius. Again, the degrees Celsius will divide out. We're gonna be left with delta V G is equal to negative 1. mL. Okay, so we found the change in volume for both the courts and the glycerin. Now we need to figure out how does that relate or how does that contribute to how much more we can pour into the flask. Okay, all right, so let's think about this. The courts is the flask on the outside. Okay, so if we have a delta V, that is negative. Okay, the courts is going to be smaller. The containers smaller. So that means we have less room to add glycerin. Okay, we're talking about delta V. Of the glycerin. The delta V is negative here. That means we've lost some volume in our glycerin, which means we have that much more space to add glycerin after the temperature change. Okay, so our delta v total is going to be equal to delta v. G minus delta v Q. Okay, because losing volume in the glycerin means that we have more room to add glycerin, but losing volume in the courts means that our container is a little bit smaller and we have less room to add in that liquid. Okay, so we're going to subtract that one. Alright, so we get negative 1. mL minus negative 0.45 mL. Okay, and watch the signs here, we have minus and negative. This is going to give us plus. Okay, we're gonna end up with negative 1. ml. Hey, that is our delta v. That is a change in volume. So we can approximate what that means. Is that the change in volume is going to be negative 1.833 mL case or volume is going to be that much less which gives us that much more room to add glycerin. So we can add 1.833 mL of glycerin after that temperature change. And if we approximate we're going to have answers. See we can add 1.83 mL of glycerin. Thanks everyone for watching. I hope this video helped see you in the next one.

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A steel tank is completely filled with 1.90 m3 of ethanol when both the tank and the ethanol are at 32.0°C. When the tank and its contents have cooled to 18.0°C, what additional volume of ethanol can be put into the tank?

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