Boiling Water at High Pressure. When water is boiled at a pressur...

Question

Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20 * 10^6 J/kg and the boiling point is 120°C. At this pressure, 1.00 kg of water has a volume of 1.00 * 10^-3 m^3 , and 1.00 kg of steam has a volume of 0.824 m^3. (b) Compute the increase in internal energy of the water.

Accepted Answer

Everyone in this problem. A car radiator is filled with pure water. The radiator cap keeps the pressure at 3.5 atmospheres and pushes the boiling point of water to 140°C at 3.5 atmospheres. The heat of vaporization is 2.11 times 10 to the six joules per kilogram and were asked what would be the increase in the internal energy delta U. Of water. If 250 g of steam were formed from the water at this temperature and pressure. Okay. And we're told that at 3.5 atmospheres in 100 and 40 degrees Celsius, the density of water and steam are 928.5 kg per meter cubed and 1.93 kg per meter cubed respectively. Alright so we're giving a lot of information in this problem. Let's just go through and write out all the information we're given. Okay? And then go from there. So the first piece of information we're given is this pressure p. Okay. And we're given that this is 3.5 atmospheres. We want to convert this to Pascal's our standard unit. So to do that we take 3.5 we're gonna multiply by 1.13 to 5 times 10 to the exponents five pascal's. Okay. Which is gonna give us about 3. times 10 to the five pascal's. Okay. And in terms of your significant digits here and how many you should leave an intermediate calculations. Always check with your professor on what convention they're using or what convention your textbooks using. Okay, Alright. So, we have that pressure. We have this temperature of 140 degrees Celsius. We're told that the heat of vaporization is 2.11 times 10 to the six joules per kilogram. So that's given by L V. K. The latent heat of vaporization 2. times 10 to the six jewels per kilogram. All right, what else are we told? We were told the mass and the mass of the water? The steam is 250g. Okay, and again, converting this into our standard unit of kg. Okay, this is going to be okay. We can multiply by one kg per 1000 g. Okay, The unit of gram cancels. So to go from grams to kilograms, we divide by 1000 and we get 0. kilograms. Alright. The other information we're given is these densities. So we're given the density of water. Were given the density of steam. So let's just write them on the right hand side here, we have the density of water Which is 928.5 kilograms per meter. Cute. In the density of steam. 1.93 times whoops. Not times anything. Just 1.93 kilograms per meter. Cute. Okay. Alright, so that's all the information we were given. Okay, We've converted to our standard units. So everything's ready to be used. Now, what are we trying to find? We're trying to find the increase in internal energy. Okay, so, we're trying to find delta U. Now, let's recall Delta U. Ok. Is equal to the heat Q minus the work. W All right, well, what do we know about? Well, in this case we're talking about vaporization. Okay, we're going from water to steam. So we're talking about the quantity of heat. Q. Okay. This is gonna be M times L V. Okay, because we're talking about the heat related to a phase change. So, mass times the latent heat. And in this case it's the latent heat of vaporization because we're going from water to see. Alright, now this quantity M L V is positive. It's positive because we're adding heat. Okay, we're going from water to steam. The temperature is increasing. We're adding heat to the system. Okay, So M L V positive. Okay, what about work? We're told that the radiator cap keeps the pressure constant. Okay, we have that constant 3.5 atmospheres. If the pressure is constant, then the work is equal to the pressure. P times of change in volume V two minus V one. Okay, so we have a way to write that work. W so now to calculate delta U. We need M L V. P and this change in volume. We know em we know LV and we know p those were given in the problem. Those are values that we wrote down. We don't know the volumes. V one and V two. Well, let's think about this. Were given information about density were given information about mass. We know that we can relate the density the mass and the volume. So let's go ahead and do that. Okay, recall that the density row just equal to the mass per volume. Okay, Mass divided by volume. So if we're thinking about the water, okay, the density of the water is 928.5 kg for me, they're cute. The mass is 0.25 kg and the volume we're gonna call this V one. Okay, Because the initial volume we're starting with water, so that's going to be one and then our steam will be V two because that's the final state or afterwards. Okay, Alright. So if we do the same thing for steam, the density 1.93 kg per meter cubed is equal to that same mask and we have the same mass of water transforming to steam And the Volume V two. All right. Now, if we saw for these, we multiply up the V one, we're gonna have 0.25 kg divided by 928.5 kg per meter cubed. The unit of kilogram will divide out and we're gonna be left with the unit of meters cubed, which is what we want for volume. So our units check out there and we get that V one is equal to 2.69 25 times to the -4 meters cubed. And similarly for V two we have 0.25 kg divided by 1.93 kg per meter cubed. And we're left with 0.12 m. Cute. Alright, so now we have V one and V two. We know p we know LV. We know em we can get back to our equation for Delta, you that quantity that change in internal energy that we're looking for. So now Delta U. Is going to be equal to M the mass 0.25 kg times L. V. The latent heat of vaporization 2.11 times To the six jewels per kilogram minus the pressure. 3. 375 times 10 to the five Pascal's times change in volume. V two minus V one. So we have 0.1295 m cubed minus 2. times 10 to the negative four m. Cute. Alright, so we have this big long equation. And if we work this out On our calculators, we're going to get approximately 480. 0.93. Ok. In the unit here we have kilogram times jewel per kilogram. Okay, the unit of kilogram divides. So in this first part we have a unit of jewel in this second part we have pascal times meters cube which is also a jewel. So our unit is going to be jewel here. Which is what we want when we're talking about energy. Now if we write this in scientific notation with two significant digits like the answer choices have we have 4.8 times to the five jewels. Okay. And so that is going to be our change in internal energy Delta. You can even notice that this is a positive quantity which relates to an increase in internal energy. Which is what the problem asked us to do is to find the increase so that positive indicates that it was indeed an increase. So if we go back to our answer choices, okay. We see that we found the change or the increase in internal energy to be 4.8 times 10 to the five jewels which is going to be answer C. Thanks everyone for watching. I hope this video helped see you in the next one.

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