Hello Today we'll be talking about Chapter 15 Question 78 which asks us to determine the specific heat oven alloy, um, that we don't know the composition of and we don't know anything really about it, except for a few given facts. So we know that 58.8 grams of this alloy M equals 58.8 grams and em alloy is dropped into a pot of water containing 125 grams of water and water, which all right M W equals 125 grams. The temperature of the alloy changes by about 106.1 degrees. Sir Delta T Howl oi equals 106.1 degrees Celsius. Similarly, the water T W goes up, and this is negative 106 because it by 10 0.5 degrees Celsius. And lastly, we just need to write out the equation that's going to describe the temperature of the water and the temperature of the metal of the alleyway. So cue the heat that is lost from the metal is equal to em alloy. It's alright m a times the specific heat of the alloy times Delta T outweigh. Now here's the important part. Let's draw a picture. We have this cup of water and here's our chunk of alloy and I'm gonna paint it red so that we know that it's hot. Initially, we have our chunk. We have our cup of water. The energy, which I will draw as red lines, is coming out of the metal of the alloy and going into the water surrounding it, which is why the temperature of the water is increasing. Thus, the energy that comes out of the metal Q is going to be the same magnitude and opposite sign negative Q as the energy going into the water. So M W, C W Delta T W. And so this is an important point in thermodynamics because we cannot create or destroy heat and energy. The energy leaving the metal must go in the exact same amount into the water. Now, if we look at the variables we have here, we can see that we know the two masses and we know the to temperature changes. And we also know we can look up the specific heat of water that leaves us with see a specific heat is the metal as what we're trying to calculate, and the only thing we don't know. Sure, we don't actually know either of the cues or the Q value, but we don't need to, because we know that these two numbers must be equal in opposite. We can actually rewrite the equation and drop que entirely So let's rewrite that down here. M a. C a. Delta T A. Is equal to negative M W, C W and Delta T W where again the only thing we're trying to solve for is C A. And so if we rearrange the equation by dividing both sides by the M ES and Delta T A, we can isolate the specific heat of the alloy, which is equal to negative. M W C W Delta T W divided by M A. Delta t A. And so we know the specific heat of water we can look that up so C P. C. W equals 4.184 jewels over grams degrees Celsius, and that gives us everything we need to solve this problem. So if we start writing this out again, we can just kind of move our numbers over here a little bit. Oops. We seem not to have taken our negative sign with us. Let's move everything over, and we'll just redraw the top of this tea, all right? And so now we can, right? Do we have negative 125 grams of water. Times 4.184 Jules per gram. Degrees Celsius times 10.5 degrees C. This gets divided by the massive the alloy, 58.8 grams of alloy times negative 106.1 degrees Celsius. And so what we get out is if we plug all these numbers in, we'll actually be calculating the specific heat. And so if we just check the units, we have grams above and below the fraction. And we have degrees Celsius above and below the fraction. And so the only thing we're left with is three units on the specific heat of water. The jewels per gram, degree Celsius. Yes, the others cancel out. And so if we multiply this out negative. 12 points a negative. 125 grams times 4.184 times 10.5 degrees Celsius. Divided by 58.8 and times negative 106.1. We can calculate that the specific heat of the alloy is what's right in black. 0.88 Jules her Graham degrees Celsius, 0.88 Jules over grams degree Celsius. And this means that if we plot, if we apply 0.88 jewels of energy to one gram of the metal, that gram of alloy would go up one degree Celsius. So hopefully this has helped you understand a little bit more about the conservation of energy in a system like this and how we can use this equation. Q E equals M c Delta T to calculate specific heats of unknown substances. Thank you.