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come to our third example video looking at specific heat in Kalorama Tree. In this video, we're going to consider the problem of dropping a block of 50 g of ice into water. The ice is at negative five degrees Celsius. We have one leader of water that's a 25 degrees Celsius. We'd like to know what the final temperature is going to be. Were given a few numbers here, including the density of water, which will allow us to calculate the mass notice when we convert from cubic meters two per leader that will have 1 kg per leader. That means we have 1 kg off water. Now we know that we need the total heat exchange to be equal to zero. And so that's going to be cue from the ice is negative, plus Q of the H +20 Now we have to be a little careful here because there are two steps in the amount of heat coming out of the ice because we have to heat it to five degrees Celsius and then we have to continue on do the latent heat of fusion, and then again, it has to continue to change temperature until it equilibrium. It's with the water here. So what we're going to have them is we're going to have first of all this five degrees Celsius, which is going to look like M ice times. Sea ice times will have five. Calvin will be the change in temperature. Plus we'll have the mass of the ice multiplied by the latent heat of fusion for it to melt. Plus, then we have. After that, the mass of the ice multiplied by the sea are multiplied by the sea of H 20 Because it's now become water is now liquid water. And we have ah t final minus zero Kelvin on the other side, we have remember Q equals M C Delta t, so they prefer the H 20 Then we will have the mass of the liquid water, which is 1 kg multiplied by the C H 20 that I've listed below multiplied by T f minus its initial temperature of 25 degrees Celsius, which is 298 Calvin. Okay, so this is a nice long equation for us to try and solve. We need to solve for t f, which means we want to get the terms with TF on a side by themselves, and then we'll be able to do this. So, uh, reorganizing, then what we should find is that we will have t f equal to okay. M ice sea ice times five. Calvin plus m ice Latent heat of fusion plus M water C H 20 times 298. Calvin. Okay, all divided by we have em water C H 20 minus M ice C h 20 If you want to go through and do the algebraic steps, you should come up with this exact equation here, and then you can go ahead and plug in all the numbers we have, remembering that massive water is 1 kg and that ice is given in terms of grams, so that will be 0.5 kg. And looking at this, then you should be able to come up with a solution for what the final temperature ought to be.

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