An experiment measures the temperature of a 500 g substance while...

Question

An experiment measures the temperature of a 500 g substance while steadily supplying heat to it. FIGURE EX19.20 shows the results of the experiment. What are the (d) heats of fusion and vaporization?

Accepted Answer

Everyone in this problem. In a chemistry lab, a scientist continually provides heat to an 800 g sample and records the temperature changes. The final phase change is complete at Q equals 200 kilojoules. The data is presented in the graph as shown we are asked to determine two things. So the first is the heat of fusion and the second is the heat of vaporization or the substance based on the experimental results. We have our data in this graph. We're gonna come back to that in just a second. But first, let's take a look at the answer choices. We have four options, options A through D OK. And then each of them have a combination of the heat of fusion and the heat of Vapo vaporization. For the heat of fusion, we have either 30 kilojoules per kilogram or 50 kilojoules per kilogram. The heat of vaporization, either 120 kilojoules per kilogram or 100 kilojoules per kilogram. And then those answer choices A through D are just different combinations of those. So we'll look at those when we finish up this question. So let's go back to our graph here and what we can see is that we're graphing quantity of heat Q and kilojoules versus the temperature in degrees Celsius. OK. So the temperature is on that Y axis. And we can see that we have this kind of linear relationships, we have this increasing temperature. So as Q is increasing T is increasing upward and upward until we get to a point. And when we get to a certain point, this line becomes horizontal and that horizontal line means that we are undergoing a phase change. OK. So we think about the phase change and it starts at Q equals 40 kilojoules. It ends at Q equals 80 kilojoules. OK. Now, our temperatures are starting in the negative, they're becoming positive. OK? This is the first phase change. And so this is going to deal with the heat of fusion. OK? Because we're going from a solid to a liquid. And so let's do part one. And because again, because we're going from solid to liquid, we're looking at that first phase change, the one at the colder temperature. Now again, we begin the phase change when Q is equal to 40 kilojoules. And we end that phase change when Q is equal to 80 kilojoules, that's when that temperature starts to increase again. OK. The horizontal portion is our phase change. We wanna find the heat effusion. OK? We're given some information about the quantity of heat Q. So let's think about the relationship between the two of them. And during the phase change, we can say that the quantity of heat Q is equal to the mass multiplied by the latent heat effusion LF. And this is a positive value in this case because we are adding heat, they were going from solid to liquid, increasing the temperature. All right. Now, again, this equation deals with only this phase change. So we're not worried about going from negative 20 °C up to 20 °C, just the phase change itself in this horizontal portion. And when we look at that horizontal portion, OK, the quantity of heat Q is just gonna be the difference between the final heat and the initial heat. That's gonna be the amount of heat that occurred during that phase change. And so for us, Q is gonna be 80 kilojoules minus 40 kilojoules. We know that that's gonna be equal to the mass we're told is 800 g. We can convert this to kilograms by dividing by 1000. So 0.8 kg multiplied by our latent heat diffusion. LF, we have everything in this equation of except LF. So this is a simple equation we can solve. For LF, we get that LF is gonna be equal to 40 kilojoules divided by 0.8 kg. We've left our heat Q and kilojoules instead of converting to joules because the answer choices are in kilojoules per kilogram. OK. So we don't need to convert and then convert back we'll just leave it in that unit that we want. When we work this out, we get that the latent heat of fusion LF is gonna be equal to 50 kilojoules per kilogram. OK. So we're done with the first part of this question. We can actually take a look at our answer choices now and eliminate some of the options. So option A and D have that the latent heat of fusion is 30 kg per or sorry kilojoules per kilogram. That's not what we found. So we can eliminate A and DB and C have 50 kilojoules per kilogram just like we found. So either of those could be the correct answer. We're gonna move to part two to figure out which one it should be. Now, for part two, we're gonna do something very similar. But now we're looking at the next phase change. Now, we can see from our graph that, that phase change begins at 100 and 20 kilojoules. OK. That's when we have our next horizontal segment we're told in the question that it finishes at 200 kilojoules. And so even though in our graph, it doesn't show that this line is starting to go up again. What we're told in our question is that that phase change is complete at 200 kilojoules. So we can imagine that that line is gonna keep going upwards after that. Hm. All right. So just like before right now, we're dealing with the heat of vaporization because we're into that higher heat phase change. So we're going from liquid now to gas. All right. So for part two, we're going from liquid to gas, we're looking at the heat of vaporization, we begin our face change. When Q is equal to 120 kilojoules, we could see that from our graph and we end our phase change at Q equals 200 kilojoules. That's what we were told in the problem. The equation we're using is almost the same. We have Q is gonna be equal to M multiplied by L. But in this case, it's LV, the latent heat of vaporization. OK. So LV is what we're looking for. Hey, again, this is positive because we're adding heat, we're going from a liquid to a gas. The temperature is increasing. So the quantity of heat during this phase change. Well, it starts at 100 and 20 it ends at 200. So the difference between those two heats is gonna be the amount of heat throughout that phase change. OK? And again, we're looking at just the phase change. We're not looking at the heat when we rise from 20 °C to 60 °C as a liquid. OK? Just the phase change, just the horizontal portion, 200 kilojoules minus 120 kilojoules is going to be equal to 0.8 kg multiplied by the latent heat of vaporization. LV by simplifying and dividing LV. Is gonna be equal to 80 kilojoules divided by 0.8 kg, which gives us a latent heat of vaporization of 100 kilojoules per kilogram. So that is the second part of the problem and that is done. So we are now done with this question. You just need to pick an answer choice. So we had eliminated option A, we had eliminated. Option D option B has that the latent heat of vaporization is 100 kilojoules per kilogram. That's what we found. So option B looks good. Option C has 120 kilojoules per kilogram. Not what we found. OK. So B is the correct answer here. The latent heat of fusion is 50 kilojoules per kilogram. The latent heat of vaporization is 100 kilojoules per kilogram. Thanks everyone for watching. I hope this video helped you in the next one.

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An experiment measures the temperature of a 500 g substance while steadily supplying heat to it. FIGURE EX19.20 shows the results of the experiment. What are the (d) heats of fusion and vaporization?

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