A blacksmith cools a 1.20-kg chunk of iron, initially at 650.0°C,...

Question

A blacksmith cools a 1.20-kg chunk of iron, initially at 650.0°C, by trickling 15.0°C water over it. All of the water boils away, and the iron ends up at 120.0°C. How much water did the blacksmith trickle over the iron?

Accepted Answer

Everyone in this problem, we have a 500 g block of aluminum heated to 400°C and dropped into water initially at 20°C were asked, what is the minimum mass of water required to cool down the block to 125°C. All right. So the first thing we wanna do here is look at the quantity of heat Q. Of the system. Okay. Because we have energy conserved. We have no heat, no net heat leaving or entering the system. This is going to be equal to zero. Okay, no quantity of heat Q. Of the system is gonna be made up of two things. Okay, We have Q related to the aluminum, we'll call it Q alum, we have the Q related to the water and we know that some of these things is going to be equal to zero. Alright, So how is this going to help us? We're looking for a mask a the mass of water. And recall that when we have Q. Okay, these equations are going to contain the mass. Okay, So let's go ahead and figure out what Q. Alum and Q. Water are. And use this equation to see if we can figure out the mass of water. M W. Okay, Alright, so, when we're looking at aluminum, yeah, We're starting at 400°C, and then the temperature is gonna drop to 125°C. Okay, and in that temperature range aluminum is not going to go through any phase changes. Okay, so all that's going to happen to aluminum is that it's going to go through a change in temperature. Now, when we're writing the heat Q. For a change in temperature, recall that it's going to be equal to m the mass. And in this case it's a mass of the aluminum. Okay. Time. See the specific heat capacity and again in this case the specific heat capacity of aluminum times delta. T. The change in temperature. All right, well, we know all of these things were told the mass of aluminum is 500 g. Okay, and remember to convert from g to kg, we divide by 1000. Okay, so this is gonna give us a mass of 0.5 kg. The specific heat capacity C for aluminum, this is something you can look up in a table in your textbook or that your professor provided and it's gonna be 910 jewels per kilogram degrees Celsius times delta T. The change in temperature while we're gonna go with the final temperature Of 125°C - the initial temperature of 400°C. And when we calculate this, do this multiplication. We find that we get negative 125, jewels. Okay, for the units we have kilogram times jewel per kilogram degrees Celsius times degrees Celsius. So the units of kilogram and degrees Celsius divide out and we're left with jules. Okay. Alright, So that's it for aluminum, we figured out that quantity of heat Q. Now let's move to water. Okay, what's going to happen to the water? The water is initially at 20°C. So it's going to be in its liquid form. The final temperature is gonna be 125°C. So what's gonna happen is that that water is gonna turn to steep? Okay. So we're first going to have a change in temperature. Okay. Getting that water up to 100°C And then we're gonna have a phase change of that water as it changes to steam. Okay. Once it's changed to steam, we're not containing that steam. Okay. So, it's going to be released and we're not gonna have any more change in temperature. Okay, so 100 is going to be the final temperature of that water or of that steam at that point. Okay. All right, So, what's this gonna look like when we have a change in temperature? We just did this with the aluminum. So we know that this is going to be the mass M of water in this case. See the specific heat capacity again, of water times Delta. T The change in temperature. And when we're looking at a face change, we have the mass M in this case M. W for the mass of the water times l the latent heat. And in this case we have the latent heat of vaporization because we're converting from a liquid to a gas. Okay, Alright. And we have a positive here because we are increasing our temperature, we're increasing the heat in the water when we change from water to seem okay, the temperature is increasing. So we have a positive there. All right, so let's give ourselves some more room to work and we're looking for the mass of water. Okay, So let's factor that out. We have M. W. Which is what we're looking for. We have seen the specific heat capacity of water which we can look up in the table in our textbook or that our professor provided jules per kilogram degrees Celsius times the change in temperature delta. T Then again, The final temperature is going to be 100°C. Okay. Before it goes through that phase change, We know that it's going to go from liquid to gas at 100°C. Okay, so that's the final temperature. Before our phase change. And our initial temperature is 20°C. We get -20°C. Okay. And then we're gonna add this latent heat of vaporization which is something you can also look up in the table in your textbook or that your professor provided 2256 times 10 to the exponent three in the unit here is jewel for kilogram. Okay, Alright. So if we work this out, we're going to have MW. The mass of the water times 2. 912 Times 10 to the six jewels per kilogram. Okay, in this second term our unit is joules per kilogram. In this first term we have jewels per kilogram degrees, Celsius times degrees Celsius. The degree Celsius will divide out and we're left with joules per kilogram. All right, so we have our heat Q. For the aluminum foil. Heat Q. Related to the water. Now we're gonna use that equation. We wrote at the beginning that the sum of these two quantities is going to be equal to zero. Okay and when we do that we're going to find that we have zero is equal to negative 125,125 jewels plus the mass of water. M W times 2.5912 times 10 to the six jewels per kilogram. Okay, the mass of the water is therefore going to be equal to 125,125 jewels divided by 2.5912 times 10 to the six joules per kilogram. Okay, when we divide jewel by jewel per kilogram, we're left with a unit of kilogram and we get 0. 483 kg approx. So if we go back up to our top to see our answer choices. Okay we round this to three decimal places. We see that we have answer b. Okay. The minimum mass of water required to cool down that block to 125°C is 0. kg. Thanks everyone for watching. I hope this video helped see you in the next one.

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A blacksmith cools a 1.20-kg chunk of iron, initially at 650.0°C, by trickling 15.0°C water over it. All of the water boils away, and the iron ends up at 120.0°C. How much water did the blacksmith trickle over the iron?

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