A lava flow is threatening to engulf a small town. A 400-m-wide, ...

Question

A lava flow is threatening to engulf a small town. A 400-m-wide, 35-cm-thick tongue of 1200°C lava is advancing at the rate of 1.0 m per minute. The mayor devises a plan to stop the lava in its tracks by flying in large quantities of 20°C water and dousing it. The lava has density 2500 kg/m^3, specific heat 1100 J/kg K, melting temperature 800°C, and heat of fusion 4.0×10^5 J/kg. How many liters of water per minute, at a minimum, will be needed to save the town?

Accepted Answer

Hey, everyone. So this problem is dealing with heat transfer and phase changes. Let's see what it's asking us. We have a molten steel spill that is m wide and 18 centimeters thick. The stream is advancing at a rate of 0.7 m per minute at a temperature of 1610 degrees Celsius. We are cooling and solidifying the metal, the steel with water that is at 22 degrees Celsius. It gives us the steel's properties which we will use to solve the problem. And we're asked to calculate the flow rate in liters per minute required to stop this um flow of molten steel and to solidify the steel. So our multiple choice answers here are in units of leaders for a minute. We have a 2.28 times 10 to the five B, 1.47 times 10 to the three C, 6.33 times 10 to the five or D 3. times 10 to the three. Ok. So we can start this problem with our heat balance where we know that the amount of heat for the steel, the amount of heat energy and the steel plus the amount of heat energy with the water is equal to zero because the water and the steel together are essentially a closed system. And so that means that negative, the heat energy of the steel is equal to positive the heat energy for the water. So when we break this term down, recall that we are dealing with phase changes. So first, we have liquid steel and that's solidifying a stop. So, solid steel, so our Q S term is going to be two parts. First, we're gonna have our MC delta T. So we have the mass of the steel multiplied by the specific heat capacity of the steel multiplied by delta T the change in temperature of the steel. And then we also have the term which deals with the phase change. So we can recall that that's going to be the mass of the steel multiplied by the latent heat of fusion of the steel. And so we are dealing with fusion for the steel because we are going from a liquid to a solid. Similarly with water, we'll have the mass of the water, the specific heat capacity of the water and the delta T of the water. And we're going to add the mass of the water multiplied by the latent heat of vaporization of the water because the water is going from liquid to steam, latent heat e vaporization of the water. OK. And so now when we look at each of these terms. We aren't given the mass of the steel and the problem. But we are given the density kg per meter cubed. And we're given the um volume of steel of this molten stream. So we can recall that mass is equal to density multiplied by volume. And in turn, volume is going to be our length times are wi times our um height or depth at this point, the the um of the of the street. So those three parameters of the street and so that is going to equal 7850 kg per meter cubed multiplied by the width of the stream which is 15 m, the thickness which is 18 centimeters or 180.18 m. And then the length of the stream, we're gonna use that 0.7 m per minute because we're solving this in a per minute um basis. So we will just use that 0.7 m. When we plug that in, we get a mass of 1.484 times 10 to the four kg as our massive steel. So now we have that term, our specific heat capacity was given to us. Um And the problem as 500 Jews per kilogram, Kelvin, our delta t not super clear, but it does tell us that the solidification temperature of steel is 1500 degrees Celsius. So the delta T of steel is going to be the initial temperature. It's going to be what it takes to get from the solidification temperature from the initial temperature. So our solidification temperature is the final state of our steel that's 500 or sorry, 1500 degrees Celsius minus our initial temperature which is 16, 10 degrees Celsius. And so that equals negative 110 degrees Celsius. And then our laten heat of fusion is given in the problem as well as two times 10 to the fifth jewels per kilogram. OK. So we now have all of the terms on the left hand side of the equation, the right hand side, we are solving for the the flow rate in meters per minute. And that's going to come out of the mass of water. So this is gonna be our target a term in this problem. We can recall the specific heat capacity of water. It wasn't given to us in this problem, but it's a common constant we can recall is 100 and sorry, 4186 jewels per kilogram. Kelvin. And then our delta t of water is going to be similarly the final temperature of water, which is the boiling point when it vaporizes to steam. So that's 100 degrees Celsius minus the initial temperature of the stream of water, which was given in the problem is 22 degrees Celsius. And so that's 78 degrees Celsius. And then we can also recall that the latent heat of vaporization of water is a constant 2.25 times 10 to the six jewels per kilogram. And so now looking at the right hand side of the equation, we have all the terms we need as well. So we can take this to a plug-in shag to so for the mass of the water. So the first thing we're going to do is pull out that term. So the mass of the water is equal to 4186 jewels per kilogram. Kel Kelvin. In this case, I'm just gonna write Celsius. We can recall that a degree Kelvin is the same magnitude as a degree Celsius. And that's multiplied by 78 degrees Celsius plus that latent heat of vaporization 2.25 times 10 to the sixth jewels per kilogram. And that side of the equation equals our mask, the steel. So that first term is gonna be negative 1.484 times 10, the fourth kilograms multiplied by 500 Jews per kilogram. Celsius multiplied by negative Celsius minus 1.484 times 10 to the fourth kilograms again. And then this is just multiplied by, sorry. It's a little bit neater. Our um heat laten heat of fusion or steel, which was given in the problem as two times 10 to the fifth jewels per OK. And now the latent heat of fusion is a negative term when we are going from a liquid to a solid. So we are going to make sure that that term we're writing as a negative. And when we plug that into our calculator, the mass of the water comes out to be 1.47 times 10 to the third kilograms. Now, we can recall that one liter of water equals one kg. And so to stop the or to solidify the molten steel, we would need 1.47 times 10 to the third liters per minute. And so that is the final answer to this problem. And when we go back and look at our multiple choice answers, it aligns with answer choice B that's all we have for this one. We'll see you in the next video.

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