A nail driven into a board increases in temperature. If we assume that 60% of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 m/s is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an 8.00-g aluminum nail after it is struck ten times?

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Welcome back everybody. We are taking a look at a hammer that is hitting an iron block. We're told that the mass of the hammer is 2.5 kg. Were told that it moves at a velocity of 10.2 m per second. We are told that the mass of this iron block that that is it is hitting is 500 g. Now we are tasked with finding what the transferred heat is. No, my apologies. The gain in temperature is So what this delta T. Is right here after the block is hit eight times. Well, I started as you see, I started writing out the formula we're going to need to use which is Q equals M C, delta T. And delta T. Is exactly what we want. But we don't have Q. We don't have em and we don't have see what we do have mm is going to be the mass of our block right here. But we still need the specific heat and the queue. Well, the heat transferred is just going to be the amount of heat transferred from the hammer to the block for every single hit. Since they're eight hits, this will be eight times the kinetic energy of the hammer. So this will be eight times while the formula for kinetic energy is one half times mass times velocity squared. Now this is going to be the mass of our hammer. So let's go ahead and plug in all of these values. We have that. Our heat is eight times one half times the mass of our hammer, which is 2.5 times our velocity of 10.2 which gives us 1040 jewels. Since we are dealing with an iron block as well, you can actually look up the specific heat and it says that it is 10400.450 joules per gram Celsius. Now that we have our heat, our specific heat and our mass we are go ahead. We are ready to find delta T. First. Let's isolate that delta T. Term. We have the Q. Is equal to M. C. Delta T. In order to get that by itself. I'm gonna divide both sides by M. C. As you can see on the right hand side, the EMC on top and bottom cancel out. And we get that delta T. Is equal to Q. Over M. C. So let's go go ahead and plug in our values here we have 1040 jewels of energy are transferred from the hammer to the block. Times the mass of our block which is 500 g times our specific heat of 0.450. Which when you plug all of this in your calculator, we get that. The temperature gain in the block is 4.62 degrees Celsius corresponding to answer choice. A thank you all so much for watching. Hope this video helped. We will see you all in the next one

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Key Concepts

Kinetic Energy

Energy Transformation

Specific Heat Capacity