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Trial [Trial 2Mass of waterMass of solid2 c0} c 22.$'€ 20.}'CInitial temperature2.0164 LLS' C Ia9"€Final temperatureATHeat change of...

Question

Trial [Trial 2Mass of waterMass of solid2 c0} c 22.$'€ 20.}'CInitial temperature2.0164 LLS' C Ia9"€Final temperatureATHeat change of solutionHeat change of calorimeterHeat change of the reactionAH;xn " per gram of' solidAverageShow calculations for trial I:

Trial [ Trial 2 Mass of water Mass of solid 2 c0} c 22.$'€ 20.}'C Initial temperature 2.0164 LLS' C Ia9"€ Final temperature AT Heat change of solution Heat change of calorimeter Heat change of the reaction AH;xn " per gram of' solid Average Show calculations for trial I:



Answers

Given these data, compute the heat of fusion of water. The specific heat capacity of water is $4.186 \mathrm{J} / \mathrm{cg} \cdot \mathrm{K}$ ). $$ \begin{array}{cl} \hline \text { Mass of calorimeter }= & \text { Specific heat of calorimeter }= \\ 3.00 \times 10^{2} \mathrm{g} & 0.380 \mathrm{J} /(\mathrm{g} \cdot \mathrm{K}) \\ \text { Mass of water }= & \text { Initial temperature of water and } \\ 2.00 \times 10^{2} \mathrm{g} & \text { calorimeter }=20.0^{\circ} \mathrm{C} \\ \text { Mass of ice }=30.0 \mathrm{g} & \text { Initial temperature of ice }=0^{\circ} \mathrm{C} \\ & \text { Final temperature of calorimeter }= \\ & 8.5^{\circ} \mathrm{C} \\ \hline \end{array} $$

In this example, they asked us to find the heat of fusion of water and they tell us that we have a kalorama ter water and ice and this enclosed system. So we're just going to assume that we're not losing any heat or energy to the environment or if we are that it's negligible. So first thing we have to do is we're gonna have to find the heat lost by the colora meter. We know that will be equal to the mass of the Colora meter times the, the capacity of the colora meter times to change in temperature. We're told that both the Colora meter and the water start added a an initial temperature of 20 degrees Celsius and have a final temperature of a 0.5 degrees Celsius. So changing temperature is just going to be our final minus our initial. That will give us negative 0.5 degrees Celsius, which is also equal to a change of 11.5 degrees or 11.5 kelvin mm. So, plugging in the values we have, we have the mass of our Colora meter as well as the the capacity of her colora meter. Plug that into our equation and we get a loss of heat of 1000 311 jules. Next we want to find how much he uh we lose from our water. So that's going to be the mass of the water times the heat capacity of the water times the change in temperature, which we already said the same as the change in temperature for our colora meter. When we plug all those values in, we have the mass firewater there and the heat capacity of our water there. So we can plug those values into our equation. I'm gonna get a loss of heat of 9000 627 point eight jewels. Then we want to find the he gained from our ice because it's experiencing a change in phase. It's going to be the mass of our ice times. T uh latent heat of fusion plus he the temperature of our water and our ice, right temperature gained from the water and our ice. So that would be the massive price times the heat capacity of the water times to change in temperature. And they tell us that you change in temperature of our ice. It started at zero degrees Celsius and had a final temperature of 8.5 degrees Celsius forgive us. A change in temperature of 8.5 degrees Celsius or 8.5 Calvin. When we plug other values. We we haven't oops. Another mass of our ice is 30 g. And we already have the capacity of our water. So you can plug those values in and we'll get her change. And he it's equal to 30 g times are latent heat of fusion plus 1000 67 point 43 jewels. We can then plug these three values into our original equation at the top here and we'll want to rearrange and saw for Al Sabah. If we're looking for the the latent heat of fusion for water, well, we do all that. We get that. The latent heat of fusion for water is equal to 329 0.1 jules program.

So in this example, they want us to find the heat of vaporization of water and a water, steam kalorama. Our system, we're just going to assume that we're not going to lose any energy or heat to our environment. And our system is just going to contain these three um substances or objects. So we know that the heat transfer or the heat required to produce a certain temperature right, has to be conserved. So the uh heat transfer of the water plus the heat transfer of the colora meter, plus the heat transfer. The steam. I'll have to equal zero. So if one is losing heat, another one has to be gaining that heat. What? And then we know that the heat required for a certain temperature is Q. R. The equation for it is Q equals the mass times the specific heat capacity times to change in temperature. Right? And if that substance or object is changing states it's going to be Q. Is equal to the mass times the latent heat. So this would be, for example if it was going from a gas or solid or solid to a gas, um solid to liquid whatever. So mass times the latent heat plus the mass times the he capacity. The specific heat capacity times to eat change in temperature. So what they want to know is this the latent heat, the heat of vaporization. So in order to find that though, we need to first saw for Q sub W use, etc. And Q says so Q sub W is going to be the mass of the water times the specific heat capacity of the water times the change in temperature of the water. We're told that the water and Colora meter start at a temperature of 15 degrees Celsius and then reach a final temperature of 62 degrees Celsius. So the changes temperature is going to be the final temperature minus t initial temperature. And when we plug in, the values that we have, we get the he is 39,000 348 0.4 jewels because it's positive. We know that it's gaining this heat right? The water is getting hotter, which makes sense. Mhm. Next the he transfer of the capacity are Colora meter, which will be the mass, the Colora meter times the specific key capacity of the Colora meter times to change in temperature, which will be the same for the water. And again, we have all these values over here, the specific heat capacity for the water and the Colora meter, the masses for both of them as well. So let me plug in all those values. We get that the he uh specific heat is 5000 358 chul's. And again, because it's positive, we know that it's gaining this this energy. Lastly, we need to find the heat required to produce the known temperature for the steam. And because the steam we're told starts at a temperature of 100 degrees Celsius and goes to a temperature of 62 degrees Celsius. Um It's going to be experiencing a change in states. So it's getting cooler. So that would be the mass of esteem times the um heat of vaporization plus the mass of esteem times the specific heat capacity of the steam times to change in temperature. And we know that it started at 100 degrees. I went to 62 degrees Celsius will be 60 to minus 100. When we saw for that we get negative 18 point five g times t heat of vaporization. What? Plus sorry, in this case would be minus 2000 942 6 to 7 to jules. Right? So we can plug this these three values into our original equation. That stated the he transfer of the Colora meter lets you transfer the water plus he transferred the steam. Oh has to equal zero. So we can plug those in and we'll get 39. Let's do it in order 5358 tools plus 39,348 0.4 jewels minus 18.5 g times Elsa v minus 2942.72 jewels. I'll have the equal zero. They want to know what the heat of vaporization of the water is. So we're going to rearrange this to solve for Elsa V. And we'll get negative 18.5 L. C. V equals negative 39,000 300 48.4 jewels minus 5300 58 jules plus 2000, 942.76 tools divide by the negative 18.5 and we get the latent heat of vaporization to be 2000, 257 0.5 jules program.

Simple. They want us to find the the heat of vaporization of water in our system. And in order to do that, we're going to find the either loss or a gain of heat from each of our substances. So in this case it's going to be other Colora meter, water and steam. We know that the change of our colora meter plus the change of our water plus the change of our steam has to equal zero. Right? The energy is conserved, assuming that we're not losing any to our environment. Where if we are losing some, it's negligible. So first thing is we want to find how much heat our colora meter is gaining. So we know that's going to be the mass of our colora meter times the heat capacity of our Colora meter times to change and temperature, we know the change in temperature is just going to be the final temperature minus the initial temperature. We're told that the Colora meter and the water have the same final and initial temperature. So it's going to be, yeah, 62 degrees Celsius minus the initial temperature of 15 degrees Celsius, which will give us a change in temperature of 40 seven degrees Celsius, which can also be written as a change change of 47 kelvin. So, plugging that information, our equation, our temperature, the mask for Colora meter as well as the the capacity of our Colora meter. We get a gain of 5000 358 chores doing the same for our water. Then we know the gain of heat transfer of he is going to be equal to our mass for water times the the capacity of water times he change in temperature. The change in temperature is gonna be the same as it was before that. We're just going to take our mass for water and our capacity for our water plug those values into our equation and we'll get a gain of 39,000 348.4 chul's. Lastly, we want to calculate the heat transfer of our steam and we know that's going to be the heat released by our steam as it goes through its change of phase plus the change of temperature of the water from the steam. Right? So it's going to be the mass of our steam times the heat capacity of the water from our steam times the change in temperature of our steam. We know the change in temperature in this case is still going to be the final temperature minus the initial temperature. We're told that our steam is originally 100 degrees Celsius with the final temperature of 62 degrees Celsius. So our final change in temperature for our steam is going to be negative 38 degrees Celsius, Which can also be written as a change of negative 38 Calvin plugging that as well as our the capacity for water and are massive. Our steam we get are accused Abbas to be equal to negative 18.5 grams times are Elsa be which is just are um he of evaporation plus I'm sorry c minus minus 2000 942 0.76 tools now that we got all three of these values. We can plug these three and to our original equation and rearrange too soft for Elsa be right because we're looking for the heat of evaporation for water and when we do all that and software LCV we get that. It is equal to 2200 57.5 jules program

Good day. The topic is about heat transfer. Heat is a form of energy that flows from a hotter body. Talk holder body. Now, if we think that hit close to the surroundings, we can assume that the heat that is, I'm sorry, by a colder body must be equal to the heat that is that is released. Made a hotter one or another rates of another way of putting it. The sum of the total heat that is transferred to or from the system must be equal to zero. Now, when a body accepts or absorbs or releases heat, its temperature may change or its face male. Um also change. If he that is associated with temperature changes evident, then we solve for that heat as Q equals M. C. Delta T. Where Q. Is the heat and mr mass cease to. Is the specific it. Which value can be obtained if we know the identity of the substance delta T. Is the change in temperature. Which can be solved as the difference between the final temperature T. F. And the initial temperature T. A. Whereas when he causes the face change to the substance, then we solve that heat. Ask you because sometimes l where is the light and heat which depends on the type of phase change the substance undergoes. So for example, in the south and cause undergoes vaporization, then that heat is latin heat of vaporization. Now let us suppose we want to find the heat of fusion of ice. Give me the information on the mass of the whole perimeter the mass of water is added to it and the mass of ice that is added to the water And that the initial temperature of water that is contained in a kilometer is 38°. After some time that was found at data water, ice and perimeter called the term in equilibrium at 5°C. And that we assume that the temperature of ice is initially and 0°C. So if this assure me in this case, that will hit escaped in the system and that is no heat escapes in the surrounding. So we can say that the cuticle when the heat so that is transferred to or from the perimeter. As the heat water lets the heat ice is equal to zero where calorie meter definitely uh exhibit change in temperature. So we saw for the heat associated with that S. M. C. T. F minus D. A. Where M is the mass of the perimeter, which is equal to 60 times its specific it. Which is given to be equal to point when times the finance temperature minus initial temperature, Final temperature being five in the initial temperature is 38. Not that since the parameter contains water. So it must also take or or it must also have the same initial temperature instead of water. So this gives us -198. Can next for water we have the same water undergoes Temperature change from 38 down to five. So now for the mess we can get the mass by the difference in the mass of the kilometer with the water and the mass of the kilometer only. So that will be 4 69 60. So that would be 400 times the specific water which is one cal program particularly sell shoes And the finance temperature is five. The initial East 30. So solving this one gives us negative 1000 or 30,200. Okay. And for ice. So not that is in this case when first undergo melting that his face changed. So for face change we have sometimes L and this end here is specifically future and are melting. So this is this L. F. Here that we wish to find in this problem and after melting the melted ice or that is now in the form of water will increase its temperature from zero which is the melting temperature of water to the final temperature of the mixture which is five. So for that we will use M C delta, T or T F. Dynasty I. Now the massive ice can be obtained by taking the difference between taking difference of the massive calorie meter plus water plus ice and the mass of kilometer with water. So that difference will give us the mass of ice. So that will be 618 -460 And that's equal to 150. So this is 158 times LF plus 1 58 times the specific it now of water. Since this is no longer ice, this is already more water since ISIS melted. So that will be one and the T. F. Is five minus zero. This simplifies into 1 58 L. F. Plus 7 90. Right? So like the weekend now combine them in of single equation that we've been 198 less negative 13,200 plus 1 58. L. F. Plus 7 90 is equal to zero. And we can simplify this. Doing that gives us -12608 Plus 158 n. F. Equals zero. Or this will be L. F. Equals Went to 608. Oh very uh 158 and this equals 79 points eight cal Okay. Okay. Or we can also have it as simply 80 80 count quicker. Thus the heat of fusion of Water is 80. Camper. Okay, so I hope this helps


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Lel Aanaranuardh #oFndaFatu € ,t nhichia @ 21 entry unauanrat cuaaeaaudxen madrdeezTn erorerdtn udcn 0ler mt , nnant crnretathta mtu l te &bid Ol yeur aniaer # you nced to ttat Over Ihon Gick Cn tho #ash canAC ad [email protected]
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