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A24.3 g sample of hot copper metal is added to a calorimeter containing 50.003 g of liquid bromine at 23.11 %C The temperature of liquid bromine increased to 44.59 ...

Question

A24.3 g sample of hot copper metal is added to a calorimeter containing 50.003 g of liquid bromine at 23.11 %C The temperature of liquid bromine increased to 44.59 %C. Calculate the initial temperature of the metal added to the calorimeter:Specific heat of liquid bromine is 0.226 Jg"C Specific heat of copper is 0.380 J/g"C,18.30 %€70.88 %C67.32 QC049.39 %50.58 QC

A24.3 g sample of hot copper metal is added to a calorimeter containing 50.003 g of liquid bromine at 23.11 %C The temperature of liquid bromine increased to 44.59 %C. Calculate the initial temperature of the metal added to the calorimeter: Specific heat of liquid bromine is 0.226 Jg"C Specific heat of copper is 0.380 J/g"C, 18.30 %€ 70.88 %C 67.32 QC 049.39 % 50.58 QC



Answers

Determine the original temperature of a $56 \overline{0}-\mathrm{g}$ piece of lead placed in a $165-\mathrm{g}$ brass calorimeter that contains $325 \mathrm{~g}$ of water. The initial temperature of the water and calorimeter was $18.0^{\circ} \mathrm{C}$. The final temperature of the lead, calorimeter, and water is $31.0^{\circ} \mathrm{C}$

Good day. The topic is about he transferred. We note that he flows from a hotter object to a colder object, electing he lost to the surroundings. The heat that is absorbed by the colder object is equal to the heat that is released by the hotter one. In one way of putting it, we say that the some of the heat that is transferred to or from the system is equal to zero. We note that the heat is associated with temperature change is solved as Q equals M. C. Delta T. Where Q is hit and miss mass C. Is the specific it, which value can be obtained knowing what type of substance there is and delta T. Is the change in temperature which is equal to the difference between the final temperature T. F. And the initial temperature. The Now let us suppose we are given with a corporate colour emitter that contains water in it. The mass of the corporate colour emitter is 55g and the mass of water that is in it is 250. Now the perimeter and water are thermal equilibrium at 18° issues when a lawyer is placed in it, The other is 75 g and the temperature of 100 people shoes, it is found out that after some time the temperature rises to 20.4°C shoes. Given that the specific heat capacity of corporate calorie meter is 0.0 93 that for what there is one we wish to find the specific heat of the alloy. So notice that we are labeling where we are letting corporate kilometer S. A water SB and sc so that we wish to find the C. four Cup for the other which is the specific it of the solution. We note that the heat Total heat must be equal to zero. And so we have a C A D f minus D A D A plus M V C V T F. Dynasty I B. And EMC C C TF. Dynasty I.C. is equal to zero. So we just have to substitute the event 55 times 0.093 Times 20.4 minus 18 plus 250. This is the mass of water. Yes it is times when Times 20.4 my nurse 18 plus the mass of the alloy times C. C. Which will look for this case Times 20th Point for my nurse. The initial temperature of the ally which is 100 Equals zero. So first let us simplify this. Doing that gives 12 point 276 plus 600 -5790 CC. Equals zero or disease. 612.276 -5 7 90 cc. Or this is equal to 612.276 Over 5 790. And this gives us 0.106 cal perry graham degrees Celsius. Or in two significant figures we can write it as 0.11 cal perry graham degrees and shoes. Thus the specific it of the alloy is point when, when cal program degree sell shoes, I hope that helps.

Okay for this problem, It's not too tough. It just requires a little bit of keeping track. Of things were given the following scenario. We have an unknown metal. We call it my junk metal. We have a sample of copper metal and we also have a constant pressure. Keller immature with water in it. And I know the mass for each of these and the mass for my unknown metal is 20.3 grams. The massive my compra mental is 28.5 grams and the mass of my water is exactly 100. Gramps, make that a little bit easier to read. Okay, I also know my initial temperature for each one of these in the initial temperature for each metal is 80.6 degrees c and the initial temper for temperature for the water is 11.2 degrees C. And I also know final temperature for all three. And that's the same for all three 13.7 degrees Celsius, 13.7 degrees Celsius and 13.7 degrees Celsius. Okay, Now I could look up the specific heat for copper, and that is 0.3 85 J over GC degrees and water specific heat is 4.184 j over GC degrees. That should be everything I need. Let's go back and make a plan, and I want to make sure my cameras on and it is. So let's switch colors and make a plan for a plan. Step one. We're gonna find q of the water, and we're going to use Q equals M C Delta T Step two. We're gonna find que for the copper using the same formula and Step three. We know that Q of the water equals que of my unknown plus you of the copper. And if you'll recall, we just calculated this in this so we will be able to figure out queue for the unknown. And then q of the unknown equals the mass of the unknown, which we know see, which is are unknown plus the delta T which we know. So we're going to solve Um, let's we're going to solve this equation for Let me get my We're going to solve this equation for C, and then we'll be done. Let me get my pin back here and we will start on the next page. Okay, Step one Q for water Mass was 100 grams. C is 4.184 j over GC degrees, and my delta T is ah for water. It's gonna end up being. I'm gonna jot this down here 13.7 degrees C, minus 11.2 degrees C, and that equals 2.5 degrees Celsius. So solving for Q for water punching all those numbers in the calculator, we get 1000 46 jewels. There's our first Q. Now let's dio Q four cop perk you Where. Copper Same equation. We just have different numbers. We're gonna have 28.5 grams, gonna have 0.3 85 j over GC degrees, and our delta T will be 80.6 degrees C, minus 13.7 degrees C, and that is 66.9 degrees c. Do the math for my copper and we get 700 34 0.1. Jules. Okay, our third order of business. We know that Q H two is equal to queue of are unknown plus Q of our copper, and we know this value in this value, so cue over unknown is going to equal que over age 20 minus que of the copper And we just found those values. So it'll be 10 for six Jules minus 734.1 jewels, and that will eke well, 311. Let's make it 312 Jules. And that equals Q of the unknown. Okay, Last and final step for the unknown Sea of the unknown equals que over mass and delta t And let's all sub in Seoul. So we said 312 and we're dividing that by 20.3 grams and 66 0.9 degrees C. And I'm gonna double check because I think when I did this the first time, I used to money Sig Figs for my I used 3 11.9 So let me punch these. Make sure rounds the same. Yes, it does. So when I round to the correct numbers Sig Figs here I get zero point 230 j over G C. Degrees and there it is. Problem solved.

So the specific heat is the heat that re required to increase the temperature of one g of a substance by 1°C. And he absorbed by the water comes from the metal which increases the temperature of the water. Where the heat absorbed by the water is cube. And calculated by the following equation is equal to the mass, multiplied by the specific e capacity, multiplied by the change in temperature. And so when we substitute our values into this equation, What we get is 26.7 g. The maths Multiplied by 4.18 Jules per gram per degrees C multiplied by 5.61°C Where Q. is equal 6- 6.1 jules. And so that is the heat absorbed by the water. And so next we can look at the heat provided by the metal where we're looking to determine the specific heat capacity of the metal. And so what we need to do is rearrange the equation that we just used. And so what we get is the specific heat capacity is equal to the Q. The heat divided by the mass multiplied by the change in temperature. And so we plug our values into here. So we've got cute which we just calculated negative 6-6.11 jules where the sign is negative because the heat is evolved from the metal And then we divide that by 19.6 g multiplied by negative 31.06°C. And what we get is 1.028 jewels program per degree Celsius

Okay, so we're putting a metal in water so he lost equals minus heat gained. So we can use M. C. Delta T. Is negative M C delta T. I'm gonna start by putting our metal on the left here. So its mass is 19.6 g and its specific heat capacity is unknown. And it's changing temperature is final minus initial. So 30 -6167. So that's going to equal negative the mass of our water 26.7. Its specific heat capacity is 4.184 And the change in temperature is 30 minus 25. So if we go ahead and start to simplify all that, we get negative 620.73 times the specific heat capacity -558 5 6. So specific key capacity of our metal At this .900 fuels program degrees Celsius.


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