An asteroid with a diameter of 10 km and a mass of 2.60*10^15 kg ...

Question

An asteroid with a diameter of 10 km and a mass of 2.60*10^15 kg impacts the earth at a speed of 32.0 km/s, landing in the Pacific Ocean. If 1.00% of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of 10.0°C), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about 2*10^15 kg.)

Accepted Answer

Hey everyone in this problem. We have a highly energetic object with a massive two kg that passes through a horizontal water pipeline containing water at 30°C. The objects initial speed is three km a second and it loses all its kinetic energy through viscous drag in the water. We are asked if 25% of the lost kinetic energy goes to boiling the pipeline water. Well, the rest is lost to turbulence, how much water will evaporate due to viscous drag. Alright, so we want to know how much water is going to evaporate if 25% of the lost kinetic energy goes to boiling that water. Okay, so let's start by finding that kinetic energy. Okay, what is that kinetic energy? Well, recall that the kinetic energy K is equal to one half M V squared. So this is gonna be equal to one half. Okay, and our problem, we're told that the mass is two kg, so one half times two kg times the speed. Three kilometers per second. Okay, Now, in order to convert our three kilometers per second to meters per second and we're gonna multiply by 1000 we get 1000 m per kilometer in the unit of kilometer, will cancel. And we're going to be left with meters per second. Okay, so that's just a little aside on our conversion from kilometers per second to meters per second here, we want to use meters per second are standard unit. So when we have VR speed squared, this is gonna be three times 10 to the three m per second. All squared. Okay. And if we work this out one half times two is just one. So we just get three times 10 to the three squared, and we end up with nine times 10 to the six. Okay. And our unit is gonna be kilogram meters squared per second squared. And recall that that's just equivalent to a jewel. Okay. So we have nine times 10 to the six jules is our kinetic energy K. Alright. Now, 25% of that. Lost Kinetic energy goes to boiling the pipeline water. Okay, So let's find 25% of K. Well, that's gonna be 0.25 K. Which is equal to 0.25 times nine times 10 to the six jewels. It's going to be equal to 2.25 Times to the six. Alright, so this is how much energy? 2.25 times 10 to the six jules is gonna go into boiling our pipeline water. Okay. Alright. And when we're thinking about the energy that we need to bring water to a boil, okay, starting at 30°C. What do we need? Okay, well, we need a change in temperature. Okay, because we're at 30°C. And in order for water to boil it needs to be at 100°C. Truth. And then we need to actually boil it. So we need a phase change when we're talking about boiling water. We're talking about vaporization. All right, so let's write this out in terms of an equation. Okay, So the heat required, or the energy required the quantity of heat Q. To bring this water to a boil. It's going to be a change in the temperature. And we have a change in the temperature, recalled that the equation is given by M. And in this case w massive water, K. C. Specific heat capacity. And again, of water times delta T. The change in temperature. And then when we're talking about a phase change in this case, we're talking about vaporization and recall that when we talk about a phase change, our equation is going to be M. And again the mass of the water here times L the latent heat. And in this case it's L. V. The latent heat of vaporization. Because we're vaporizing and we have a plus here. Okay. And this is plus because in our phase change we have heat entering into our material. Okay, When water is boiled into steam, it's increasing temperature. It's gaining heat. So, we have a positive. Okay. All right. So, we have this equation. Let's just look at it and think about what we know what we're trying to find. Okay, well, Q. Who was it amount of energy that we have to boil the water, which we found. Okay, so we know Q 2.25 times 10 to the six jewels M. W. The mass of water. Well, that's what we're looking for. Okay, so we're looking for this quantity. M W. Okay. See the specific heat capacity of water we know or we can look up in our table um the change in temperature we know okay, in the latent heat of vaporization again we can look that up in our textbook. So let's fill in the information we know 2. times 10 to the six jewels. Okay, is our value for Q. And that's going to equal to M. W. Let's factor that out. Okay. C. W. Specific heat capacity for water. Again look it up in the table in your textbook or that your professor provided? 4190 jewels per kilogram Calvin. Okay, times the change in temperature. So that's going to be the final temperature which is 100°C. Okay we're gonna add 273.15 to convert this to Calvin. And then we're gonna subtract The initial temperature 30°C. and again add 273. to convert this to Calvin. And then we're gonna add lv the latent heat of vaporization and this you can also look up in the table in your textbook or that your professor provided 2256 times 10 to the exponent three. The unit here is jules per kilogram. That should be a big square bracket to close that out. Alright so now we have one variable that we're looking for. Mw Everything else is number. So it's just a matter of simplifying and solving for M. W. At this point. Okay, So we're gonna leave the left hand side alone. 2.25 times 10 to the six jules is equal to M W. K. Simplifying. What's in the brackets? We get 2. 9, 3 times 10 to the six jewels per kilogram. Okay, and when we divide we get the mass of the water. M. W. Is equal to we have jewel divided by jewel per kilogram, which is gonna give us a unit of kilogram. We get 0. kilograms. And if we look at our answer choices, we can round and see that we have answers. See okay, the amount of water that is going to evaporate due to the viscous drag is going to be 0.883 kg. Thanks everyone for watching. I hope this video helped see you in the next one

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