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0WIoin0 { 1 ;oinc A76 copper calorimeter (c 385 J/kg "C) contains 271 & cf water 4184+ Mkg "Clat 11*€ dropped into the calorimctef; the resultin...

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0WIoin0 { 1 ;oinc A76 copper calorimeter (c 385 J/kg "C) contains 271 & cf water 4184+ Mkg "Clat 11*€ dropped into the calorimctef; the resulting temperature Is 27"€ What is the Wien 357 &of an alloy at 151*C I5 spccific hcat of the alloy? Answcr409.8(553.8)Hide Fcedback

0WIoin 0 { 1 ;oinc A76 copper calorimeter (c 385 J/kg "C) contains 271 & cf water 4184+ Mkg "Clat 11*€ dropped into the calorimctef; the resulting temperature Is 27"€ What is the Wien 357 &of an alloy at 151*C I5 spccific hcat of the alloy? Answcr 409.8 (553.8) Hide Fcedback



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A 55 -g copper calorimeter $\left(c=0.093 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)$ contains $250 \mathrm{~g}$ of water at $18.0{ }^{\circ} \mathrm{C}$. When $75 \mathrm{~g}$ of an alloy at $100^{\circ} \mathrm{C}$ is dropped into the calorimeter, the final resulting temperature is $20.4^{\circ} \mathrm{C}$. What is the specific heat of the alloy?

Everyone today will be solving a problem looking for the specific heat of in alloy. So what I've written here is just a summary of the information you've been given. So we're given the mass of the alloy, the mass of the water, the change in temperature of the alloy, which is negative, and the change in temperature of the water, which is positive. And what we're looking for is the specific heat of the alloy, which is C A. So, in order to solve this problem, there's one really important equation that will be using place. And that is the Q equals M C Delta T equation. So what the stands for is the heat energy of the system. Q is equal to the mass times this visit Keat times the change in temperature and we're going to be using this twice in order solve this problem. So the first time we use it, we're going to be solving for the heat energy of water. And so what that means is we're going to be using you mass of the water, the specific heat of the water and the change in temperature of the water to solve that. So we're going to plug in what we know. So Q W is equal to the mass of the water, which is 125 grams times the specific heat, which we know is 4.186 times the change in temperature of the water, which we know is positive 10.5. So we're just going to plug this in, and what we get is 5494 0.1 to 5. Is the he energy of water. So what we do with this? So next we need to know the relationship between the heat energy of the alley and heat energy of the water. And because these air in the same system, we know that Q. A. The heat energy of the alloy is equal to negative. Q W. So this gives us a lot of information. So if we then go down here and write the same equation we used above, Q equals and see Delta T. And if we do for the alloy than all of these are sub a. But then we can plug in negative. Q. W for this Q A. So that helps us a lot So we have negative Q w equals, um, a see a That's a t A. So now we only have one unknown and we're looking to solve for the specific heat of the alloy. So we're going to rearrange this equation in order to, um so for C A. So we see a hey equals negative, Q w no being divided by We're just gonna divide by m A and tell toe t a to get those on the other side of the equation. So now again, we're left. It's just putting in the things that we know. So see, a is equal to the negative 5494 0.1 to 5 divided by and A which is 58.8 times Delta t A. Which is negative 106.1. Okay, so again, we're just going to plug this in, and the two negatives cancel. So we get C A is equal to operate Cory a a one Jules her Graham. And that's it. That's our answer

Everyone today will be solving a problem looking for the specific heat of in alloy. So what I've written here is just a summary of the information you've been given. So we're given the mass of the alloy, the mass of the water, the change in temperature of the alloy, which is negative, and the change in temperature of the water, which is positive. And what we're looking for is the specific heat of the alloy, which is C A. So, in order to solve this problem, there's one really important equation that will be using place. And that is the Q equals M C Delta T equation. So what the stands for is the heat energy of the system. Q is equal to the mass times this visit Keat times the change in temperature and we're going to be using this twice in order solve this problem. So the first time we use it, we're going to be solving for the heat energy of water. And so what that means is we're going to be using you mass of the water, the specific heat of the water and the change in temperature of the water to solve that. So we're going to plug in what we know. So Q W is equal to the mass of the water, which is 125 grams times the specific heat, which we know is 4.186 times the change in temperature of the water, which we know is positive 10.5. So we're just going to plug this in, and what we get is 5494 0.1 to 5. Is the he energy of water. So what we do with this? So next we need to know the relationship between the heat energy of the alley and heat energy of the water. And because these air in the same system, we know that Q. A. The heat energy of the alloy is equal to negative. Q W. So this gives us a lot of information. So if we then go down here and write the same equation we used above, Q equals and see Delta T. And if we do for the alloy than all of these are sub a. But then we can plug in negative. Q. W for this Q A. So that helps us a lot So we have negative Q w equals, um, a see a That's a t A. So now we only have one unknown and we're looking to solve for the specific heat of the alloy. So we're going to rearrange this equation in order to, um so for C A. So we see a hey equals negative, Q w no being divided by We're just gonna divide by m A and tell toe t a to get those on the other side of the equation. So now again, we're left. It's just putting in the things that we know. So see, a is equal to the negative 5494 0.1 to 5 divided by and A which is 58.8 times Delta t A. Which is negative 106.1. Okay, so again, we're just going to plug this in, and the two negatives cancel. So we get C A is equal to operate Cory a a one Jules her Graham. And that's it. That's our answer

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.

Considering aluminum Kalorama Ger, originally thermal equilibrium with water. So the mass of deal in minimum is 100 grams mass of the water is 250 grams, and the initial formal equilibrium temperature that they're both that is 10.0 degrees Celsius. They were going to introduce two blocks of metal. The first block of metal is made of copper. Its mass times of C is 50.0 grams and its initial temperature temperatures of copper is 80.0 degrees Celsius and then the second block is the maid of a mystery metal. We don't know what it is. I'm calling M sub m for mystery for its mess, and it's mass is 71 0 grams and its initial temperature cheese of mystery is 100 degrees Celsius is now. We've got free sick pigs carried on all of this and then finally were given that the system finally settles out to a final equilibrium temperature of 20 once year of degrees Celsius. So the first thing we might ask is what is the specific heat of our mystery metal and we could go to table 20 0.1 in the textbook, which will give us reference for the other specific heats. We have a specific he'd aluminum as 900 jewels per kilogram degree Celsius. The specific heat off water is 41 86 jewels per kilogram degree Celsius and the specific heat of copper is given to us as 387 Jules per kilogram Degree Celsius. So what's the next step Next? We probably want to look at the conservation of energy we know and right are Calera Metric equation. We now that Kisum cold is equal to the negative queues of hot. And what I mean by that is just that all of the energy used to raise the temperature of our colder substances here The water in the aluminum is equal to the and negative energies lost from the harder substances the metal cubes going from hot colt so we can expand. That would just are standard U equals M c Delta team structure. Where am AST? Mass sees the specific he and L t is the final temperature minus two initial and we could go ahead and expand this so massively aluminum I'm suspensive heat of the aluminum times daredevils t is going from 20 to 1 or start 20 to 10. So that's just 10.0, degrees Celsius and thats plus the Q or water. So that's ems of water times to specific heat of water again that they're both at 10 degrees. To start, that's again. They're guilty is just 10.5 degrees Celsius, and this is equal to negative of the master copper times, this specific heat of the copper and again there. Delta T is going from the final temp of 20 degrees to the initial town, minus the initial town off 80 degrees. So this doesn t generates a negative 60. That's why we have this negative sign because of the direction. Um, and then we're just gonna add the mystery few so massive, mystery, specific heat of mystery. And again, this one's gone from 120. So it's negative. 80 Greece, Elsie's And you might realize we actually have all of these values except for the specific heat of our mystery, uh, variable. So we can rearrange this whole equation real quick into just the specific heat of our mystery Middle, which is this is kind of a cool fraction we got going here. 10 times to quality. Sometimes a quantity of mass of the aluminum specific heat of the aluminum plus mass of water specific eat of the water minus this guy because it would be plus name taking over they ever saw 60 times just massive copper specifically to the copper. All of that fired by it's coefficients 80 times the mass of Mr Metal on. Let me rewrite this a little more clearly as we substitute in our actual values. Onda We can see that c m equals 10 times 0.100 k g eyes 900 Jules over kg degree Celsius plus zero 0.250 kilograms. That's the mass of the water. I'm so specific. Heat of the water 41 86 jewels over kilogram degrees Celsius minus 60 times 0.50 that's a mess of the Copper Times to specific Ito's copper, which we looked up to be 3 87 jewels per kilogram degree Celsius. All of that over just 80 times the mass of the mystery metal, which is 0.70 kilograms. That's, uh, 80 degrees Celsius. Oh, to conjure up a final specific heat of 18 22.1 Jules per kilogram, degrees Celsius. And that is the specific heat of our mystery metal. So the second question you might ask is Okay, so can we figure out what it is on? We can go back and look at Table 20 0.1 again, and we see it's not actually on there. There's no values on this table that are exactly what our value is. So no, we cannot confident way we cannot identify the substance because the specific, um, it's just that the heat isn't on that table. However, stables, not exhaustive, doesn't have every substance known to man. And you might notice that these specific heat for beryllium is 18 30 which is actually pretty close. All things considered on this table and therefore it might be a logical hypothesis to think that, perhaps, are substances, maybe an impure sample on Alloy of brilliant. But again, we can't just confidently idea because we don't have the exact specific heat on our table.


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