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Relrrenarlluerdtrtakes 32.4 Joules incrcase the temperalure of 1J.4 ENM; of solid gold from 20.5 t0 39.7 In the labanlary student finds Ial degrees CelsiusThe speci...

Question

Relrrenarlluerdtrtakes 32.4 Joules incrcase the temperalure of 1J.4 ENM; of solid gold from 20.5 t0 39.7 In the labanlary student finds Ial degrees CelsiusThe specitic heat gold calculated (rom het daluJK"CcnninenegRotty Enture OroupJUm @a Jurup autempts tcmainbno

Relrrenar lluerdtr takes 32.4 Joules incrcase the temperalure of 1J.4 ENM; of solid gold from 20.5 t0 39.7 In the labanlary student finds Ial degrees Celsius The specitic heat gold calculated (rom het dalu JK"C cnnineneg Rotty Enture Oroup JUm @a Jurup autempts tcmainbno



Answers

The specific heat capacity of silver is $0.24 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C} .$ Express this in terms of calories per gram per Celsius degree.

This is problem 91 from Chapter three of fifth Edition Introductory chemistry. This problem asks us. Two. Calculate the energy in heat that would need to be added to heat 25 g of gold from 20 Celsius to 75 Celsius. And it asks us to give our answers in jewels, killer jewels, calories and kilocalories. So this is a simple matter of calculating the heat needed to convert some material in this case, gold of some mass over some temperature increase. Now we are given all of these factors so the heat increase is equal to the specific heat. Capacity can be grabbed from Table 3.2 and is given a 0.13 Jules per gram degree Celsius multiplied by the Mass, which is given as 25.0 g multiplied by the change in temperature where final temperature is given as 75 minus our initial temperature, which is given as 20 and that is in degrees Celsius. Calculating this out, we are left with 178 0.7 five Jules, and given that the question asked us to express it in jewels, killer jewels, calories and kilocalories, we can then easily convert this 20.17 eight seven five killer jewels by simply dividing by a factor of 1000 moving that decimal 10000.3 places to the left. Now, to convert calories, we will need to multiply our 1 78.75 Jules by one calorie per 4.18 for jewels, which is a conversion factor that can be found in the book or online. Our jewels units will cancel and we will be left with 42 0.72 calories, which then, using the same conversion as above converting from calories kilocalories simply divide by 1000 moving the decimal 10000.3 to the left, they get 0.4 272 kilocalories.

In this problem, we need to calculate the amount of heat involved in four processes that are given, and we have to calculate the amount of heat either released or excerpt, being both used on calories. The first exercise is related to heating up a mask off 5.25 grams off water from 5.5 Sell shoes to 64.8 Celsius. Okay, again, this is a hit capacity problem under for the central formula that I'm going right here. He's amount of heat absorbed for release equals the mass of the substance times they hit the positive, the soft string, which is a constant that we finding that they will provide it times the different in the temperature or final temperature minus initial temperature. Okay, so this is the formula that we're gonna be referring through. The development of these exercise the amount of heat absorbed by water. In these process. He's in mice of water. My £0.25 times that he'd composite of war. E news, which is four point 184 yours. Program spur sells shoes times the difference in temperature just 68.8 sort of 64.8 by a temperature minus 5.5. Initial temperature on the on the delta is 59.3 Celsius, so it's just cancel grounds. Council on the answering Your species 1300 three 1000 300 on three use in calorie. The answer is even when you think the U they hit capacity constant the calories for water. This is one calorie her grounds. Purcell Shoes times does that the which is the same 59 points today. So is this Council Cramps Council, and in this case, the answer is 300 it's 111. God is you got in B exercise. We have to calculate the amount of water that is lost. The amount of heat that is lost when 75 grams of water goes down from 86.4. Sell shoes toe to point one sell, it says we do the same thing. Must off water times that hit capacitor border in mules. First times the difference in temperature, which is 2.1 final temperature minus 86.4 initial temperature. And the answer is that Delta bi East negative 84.3 salt is OK, so it's just gone solo Crimes Counselor for the answer. These negative 26,000 453 fuels cool to obtain the answer in calories. We do the same thing, but we use. They hit capacity constant in colonies for grants for sales is times that that the which is the same. We cancel their associates on the grounds we get. They got these 6000 323 galleries. Okay, Next we deal with hearing 10 grams of silver from 112 cells. Use to 275 Celsius. Good to solve these these case, we need to find the hit capacity of cylinder. The news Ah, first and then the calorie. So we multiply that 10 grams The mass times The hit capacity of silver in use, which is poin 235 fuels programs. Our cells use times the difference in temperature, which is to 75 final minus 1 12 You need chunk now that he's won 63 sell shoes. So it's just cancel crimes. Consul's and the answer ease. 383 fuels. Sorry. Yeah, 383 mules. To get these heat in calories, we used the heat capacity of silver in calories, which is 0.0 562 Carla is four grams per Celsius times the same built a T. So she's gone. So Grand council. And the answer is 91.6 galleries. Okay, Andi. Finally, we need to calculate the amount of heat that is lost when 18 grams of gold cools down. So it's amount of heat release by 18 grams of gold when gold cools down from 224 Celsius to 118. So let's use okay To calculate the amount of heat release by this massive gold, we need to find the hit capacity of gold first Mules, which is 0.1 to 9 fuels for grounds. Purcell issues times different in temperature. 1 18 minus 2 to 4 on delta is negative. 106 We can serve. The soldiers would cancel the grounds on the concept being used bees. Sorry, the answer in mules. He's negative. 246 yours to obtain this heat. This amount of heat release being calories. We use the incapacity off gold in calories, which is 0.3 08 Galleries, her grounds. Purcell Shoes times, Delta T. Which is the same They give 106 So it's just concert grounds. Council on the answer is negative. 58.77 Gallaghers. Okay, so these two are. The answer is related to the amount off heat released by gold this toe are the answers related to the amount toe heat absorbed by silver. These two answers are related to the amount of heat release by water, and these two are the answers related to the amount of heat out sort my war.

Given mass of gold designated by em please. Five drops amount of energy. F. Dog. Yeah, designated by Q. Yes. One point the tree. Dude. And specific it of goddess. Yeah this is what we'll see is Did appoint 1-9 do for example degrees centigrade plugging the values we can create the change in the temperature as Q. Is because to em into sick into data. Yeah. Therefore that that that is changing temperature is equal to you upon M. C. value of queue that has been given at one T. T. Do upon. And this Deborah into expect the pacific here that you see. Get up 1129 You too program dig this integrated. There is a question that that be. Yes it was 2.06 degree centigrade. That's the changing temperature is 2.06°C..

In this problem, we're told that a sample of gold is dropped into a certain amount of water and that this raises the water temperature and that they come to an equilibrium final temperature of 27.5 degrees Celsius. So we know that the Q or the heat that goes in to the water must be equal to the heat that comes out of our sample of gold. And so, for equivalence ease, we can just do we know that Q is equal to mass time. Citizen Keat Times change in temperature and so weaken set the EMC delta t of the water equal to the emcee Delta t of our gold sample. And so let's do that really quickly. We know that the gold it's 100 Charlotte label this very quickly, so we don't get confused here. Cold, the top on one side and water on the other. We have 182 grams from gold. The specific heat given in the problem is 0.1 to eight ghouls program Kelvin and then our change in temperature. We don't know the change in temperature, but we do know the final temperature, so it's the final temperature given in the problem is 27.5. You know what? I'm taking up a little too much room here, so I will write these a vertical to each other 27.5, minus what? We're looking for our initial time. So that's That's the gold on top. I'll do water and blue to make this easier. The mass of the water 22.1 grams times this specific eat of water, which we know. It's 4.184 tools program Kelvin and then we have both. The initial and the final temperature of water, 27.5 is final and initial was 25. And so now the best way to solve this would be to do the math on both sides of the equation and isolate our final the variable. We're looking for initial temperature of gold, and so let's do that. The math on the right side of the equation for water. 22.1 grams times 4.184 times the temperature difference difference. Sorry, which is 2.5 and then let's divide over 182 and divide over our specific e of 1820.1 to 8, and we're We find that I'll go back to black for this 27 0.5 degree Celsius minus, the initial temperature is equal to nine 0.9. That's degrees Celsius. So the best way to find initial temperature then would be to add the initial temperature over to this side and subtract 9.9 from this side. So let's do that and in the field temperature to one side and subtract 9.9 from 27.5 and we find that the initial temperature, this 17 0.6 degrees Celsius and here is it our final answer.


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