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Raainig Tane hour;21 minutex 31 setondsQuestion Completion statusMoving to the next question prevents changes to this answer:Question 5The following data were obtai...

Question

Raainig Tane hour;21 minutex 31 setondsQuestion Completion statusMoving to the next question prevents changes to this answer:Question 5The following data were obtained in part of E302:.Initial lemperature of melal specimen = 10O"CInitial temperature of water = 27,5'CSpecific heat of aluminum = 0 2174 callg coFinal temperature of the syster = 30'CSpecific heat of water 00 calg €oWhat is the 'experimental value of the specific heat capacity of aluminum metal speciren?

Raainig Tane hour;21 minutex 31 setonds Question Completion status Moving to the next question prevents changes to this answer: Question 5 The following data were obtained in part of E302:. Initial lemperature of melal specimen = 10O"C Initial temperature of water = 27,5'C Specific heat of aluminum = 0 2174 callg co Final temperature of the syster = 30'C Specific heat of water 00 calg €o What is the 'experimental value of the specific heat capacity of aluminum metal speciren?



Answers

An aluminum calorimeter with a mass of $100 \mathrm{g} \mathrm{con}-tains 250 $\mathrm{g}$ of water. The calorimeter and water are in thermal equilibrium at $10.0^{\circ} \mathrm{C}$ . Two metallic blocks are placed into the water. One is a $50.0-\mathrm{g}$ piece of copper at $80.0^{\circ} \mathrm{C}$ . The other has a mass of 70.0 $\mathrm{g}$ and is originally at a temperature of $100^{\circ} \mathrm{C}$ . The entire system stabilizes at a final temperature of $20.0^{\circ} \mathrm{C}$ (a) Determine the specific heat of the unknown sample. (b) Using the data in Table $20.1,$ can you make a positive identification of the unknown material? Can you identify a possible material?
Explain your answers.

This problem covers the concept of cal geometry and from the concept of the call Dimitri, the heat lost by the calorie meter and the water is a cuba, into the heat change by the Albania. So from this weekend, right, the mass of water plaster, ah the mass of water into the specific heat of water plus the mass of copper into the specific it of copper and two. The changing temporary show that is the uh initial temperature of calorie meter minus the equilibrium temperature equals the mass of L. A. Mini. Um into the specification of Allah medium into uh changing temperature, that is the equilibrium temperature. Minister initial temperature of the alumina. So from this we can write the specific heat of L. A. Medium equals the mass of water into the specific heat of water plus the mass of copper into the the mass of calorie meter, into the specifics of kilometer, into the initial temperature of the kilometer. Minister equilibrium temperature upon the mass of tele medium. And to the equilibrium temperature minus the initial temperature of technology. So from this we can secure the value So, C equals the mass of water is um 0.4 kg, 0.4 kg into the specific heat of water is 4186 tool paul kilogram body Celsius plus the mass of copper that is zero point the massive calorie meter. That is 0.04 kg. And do the specific Heat of the gallery meter. That is 630 jules spoke, low grandpa degree Celsius, into the change in temperature That is 70 70 -66.3 degrees Celsius Upon the mass of the l. A medium. That is 0.2 kg Into the changing temperature that is 66.3 -3 27°. C are the specific heat of the Al. My name is 0.8 jules or kilogram poor big resources. Now, if you take the issue with the actual value, so the specific it of the the experimental experimental value upon the actual value of the L. a medium equals 0.8 Upon 0.9. Or you can see the experimental value calculated is 0.89 times the actual value of tele mania. So The experimental value is 11% lower than the actual value. So we can see the experimental value value of the specific heat of L. A. Mini um is 11% lower then the actual value. Okay. And the error is less than 15%.

Did you know that the heat absorbed by all object you speak with tau minus negative off heat Lost by what object? So in this given case, the heat absorbed by l ammonium must week or two heat lost by yeah workers less heat lost by caliber Mater's We can write this in terms off mask especially could capacity and genital picture is Massa film union especially kid capacity of bologna into changing temperature off evening you is equal toe negative off massive water specific heat capacity of water Less massive kilometers Specific heat capacity of perimeter into changing temperature Oh, look it we can put up at the same temperature you should no where it's absolutely videos this massive zero point 200 Casey specific it capacity of aluminum is unknown Change in temperature for minimum 39.3 degrees Integrate is it captain minus mass off water is 0.4 k g into specific heat Capacity of water is 4186 John Perkins party disintegrate less mass off kilometer is 0.4 k g in the specific it capacity of kilometers is in the thirties all over. Can you bear degrees integrate and returned to Petra. What is minus 3.7 degrees in tickets. When we solve this for specific diversity of union, we get 6.29 Multi that. I think it power. ST. John divided by 7.86 Casey Syndicate. When you further saw this, we get specific. It capacity of water from this experiment is equal to 800 zone per k G. Where do use integrate This is experiencing one and the from table. We find that the specific it capacity of unit is equal to 900 child park. Edgy where? Degrees in Tikrit. So the percentage error is equal to 900. Manager changes. 100 100 divide by 900. Multiply 100 is if we get 11% which is acceptable, and this is the answer of the question.

For this problem on the topic of heat, we are given a set of data from a calamity experiment and we want to use this data to calculate the specific heat of alum in ium. Now we know the heat for the cold objects is equal to minus the heat energy for the hot objects. Or the heat gained by the alum in ium is equal to minus the heat loss for the water, which is Q. W plus the heat loss for the cholera emitter, QC. And so the massive aluminium times the specific heat of L. A. Mini. Um see A L times it's change in temperature, T F minus T I. For the L A. Mini um must equal minus the mass of water, times the specific heat of water plus the mass of the calorie emitter. And the specific heat of the calorie meter times the change in temperature for the pay T f minus t I W. And so using the values given in the table, we have This to be a zero two kg. Sometimes the specific heat for aluminum, which you want to solve for C A L. And this is multiplied by a temperature change of 39.3° C is equal to minus 0.4 Kg times the specific heat for water. 4100 and 86 jules per kg degrees Celsius Plus the mass of the calorie meter. zero zero for Yeah, kg times its specific heat, which is 630 jewels per kg degrees Celsius. All of this multiplied by the temperature change minus three 0.7 degrees Celsius. So rearranging, we get the specific heat for L. Ammonium cl To be 6.29 Times 10 to the power three jules Divided by 7.86 kg degrees Celsius. So we get the specific heat for alimony um to be 800 jewels per kg degrees Celsius.

No, that's the water. The aluminum water calorie meter system is thermally isolated from the environment, and so the heat transfer equation would be the mass of the aluminum times the specific heat of the aluminum multiplied by the final temperature minus the initial temperature of the aluminum. This would be equaling to the negative mass of water specific heat of the water times temperature final minus the attempted the initial temperature of the water minus the mass of the calorie meter. Specific heat of the calorie meter multiplied by the final temperature minus the initial temperature of the calorie meter. Now we can then say that thespians, if IQ heat of the aluminum is going to be equaling massive water specific heat of the water plus the mass of the calorie meter, specific heat of the calorie meter multiplied by the initial temperature of the water minus the final temperature of the system. This would be divided of the mass of the aluminum multiplied by the final temperature, minus the initial temperature for the aluminum, and we can then solve so the specific eat capacity of aluminum is equaling. This would be 0.400 kilograms multiplied by 4186 Jules per kilogram career degree Celsius plus 0.4 rather 0.0 for kilograms multiplied by 630 Jules per kilogram Curd degree Celsius multiplied by 70 minus 66.3 degrees Celsius. And this would all be divided bye. 0.200 kilograms multiplied by 66.3, minus 27 0.0 and again degree Celsius. So the specific heat of aluminum is found to be, we could say approximately 800 jewels per kilogram per degree Celsius, so this would be our value solved from the experiment and from table 11.1. The percent variation would be equaling the essentially the variation. So we have the absolute value of 800 minus 900 900 being of course, the actual specific heat of aluminum Jules per kilogram per degree Celsius and then, of course, divided by 900 Jules per kilogram per degree Celsius times 100% and we find that the percent variation is equaling 11.1% and of course, 11.1% is less than 15%. So we are in. We are within, uh but then the 15% tolerance. Now this this variation is probably the source of this variation is most likely the fact that this is not perfectly thermally isolated. So some he energy or thermal energy is being lost to the environment. Inherently it this is, ah, the laboratory set up, so we cannot be perfect. That's why that's probably one probable source of air that is giving us a such a high percent variation. That is the end of the solution. Thank you for watching.


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