Einstein temperature of aluminium M. H. Yip , L. E. Kjenstad a
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Institutt for fysikk, Norges Teknisk-Naturvitenskapelige Universitet, N-7491 Trondheim, Norway.
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Abstract
The weight and rate of evaporation of liquid nitrogen before and after adding a piece of aluminium was measured, this in order to find the Einstein temperature of said metal. It was found to be Θ E = 290 ± 38K . 1. Introduction Introduction
Measurements of heat capacities have been important for the development of quantum theory. According to classical mechanics, heat capacity should not be temperature dependent. dependent. In fact, as later experimen experiments ts showed, showed, this is only approximately true for non-extreme temperatures. At near absolute zero temperatures, the heat capacity is near zero. By using Planck’s Planck’s assumption assumption of quantizat quantization, ion, Einstein was able to come up with a theory that accounted for this. Einstein’s model, published in 1907, was the first model that showed a correlation between the heat capacity for solids and temperature. Einstein used quantum theory to explain how classical theory differed from experimental measurements. [3] 2. Theory Theory Equipartition Equipartition theorem 1. The theorem of equipartition of energy states that molecules in thermal equilibrium have the same average energy associated with each independent degree of freedom of their motion and any quadratic term in of energy form will contribute 12 kB T so much energy.
Ax2 =
1 kB T 2
For example example,, diamon diamonds ds has a crysta crystall structur structure. e. A molecule in a crystal has translational energy in three dimensions and potential energy also has three dimension and almost zero rotation energy. If we apply the equipartition theorem, then E = 62 RT per mole. It follows that cV m =
6 R, 2
this is called Dulong and Petits law, and applies for high temperatures only. At low temperatures, Einstein used quantum physics 1 −βE e model to find heat capacity. ρ(E ) = Z is statistic 1 total energy distribution. Here β = kB T , Z = 1−e1−β and E = E n = n, n = 1, 2, 3, 4 . . . ∞. = ν ν where is Planck’s constant and ν oscillator’s oscillator’s frequency. Preprint submitted to Veileder
By using statistical statistical normalisation normalisation and calculatio calculation, n, it yields E 1
E = e
E1 kB T
(1)
−1
then it follows that cV m = 3R
E 1 kB T
E1
2
e kB T
e
E1 kB T
−1
2
(2)
Equation (2 (2) tells us that c V m is not a constant. constant. When T is approaching to zero, c V m also approaches zero. The Einstein temperature is defined as ΘE =
E 1 . kB
Room temperature is relatively high temperature for liquid nitrogen. nitrogen. Latent Latent heat, denoted " L", is the heat per mass necessary for phase change of a substance. Usually defined positive from liquid phase to gas phase, which means that the heat per mass flows into the system. (3)
L∆m = ∆Q
We will use Gauss uncertainty rule to estimate uncertainty of Einstein temperature. ∆ΘE =
Θ (
∂ E ∂ ΘE ∆x1 )2 + . . . + ( ∆dxn )2 ∂x 1 ∂x n
(4)
3. Method Method and apparatus apparatus
An overview of the apparatus used in the experiment is presented presented in figure 1. The electronic electronic scale scale was used to measure the rate of evaporation of liquid nitrogen in room temperature. Two polystyrene cups stacked on top of each other were used as an insulating container for the nitrogen. The insulating material allows for a better control of heat flow through the liquid-air interface. The heat required to vaporise a liquid is proportional to the evaporate evaporated d mass. This rate of evaporat evaporation ion was measured by recording the weight at specific time intervals. October 29, 2015
Figure 2: Mass of nitrogen in a cup as a function of time. The small circles on the red line represents measurement points before the dropped sample was dropped, and the red line is the trend-l trend-line ine for the circles circles.. The small small circles circles on the blue line represents measurement points after evaporation rate seems returned to normal, and the blue line is the trend-line for these points.
Figure 1: Figure 1: A simple overview overview of the apparatus apparatus which was was used used during during the experime experiment nt.. Two polystyre polystyrene ne cup cupss are placed placed on top of an electroni electronicc scale. scale. The cups contai contain n liquid liquid nitrogen, and a sample of aluminium is lowered into it.
When a cup of liquid nitrogen is evaporating in room temperature, perature, the liquid nitrogen nitrogen will not get hotter. hotter. System System is defined to exclude the nitrogen gas, so the magnitude of heat gain can be determined by using latent heat " L". In the first part of the experiment the rate of evaporation was measured measured without the aluminium. Six observaobservations of weight weight were were made at one minute minute interva intervals. ls. After the sixth observation, a piece of aluminium was lowered into into the liqu liquid id.. This This caus caused ed an incr increa ease se in the the rate rate of evaporation. When the temperature of both aluminium and nitrogen had reached an equilibrium, a new series of measurements were were made. These observa observations tions forms the basis for calcucalculating the Einstein temperature of aluminium.
A camera was used to improve accuracy of each measurement. Laten Latentt heat heat of liquid liquid nitrog nitrogen en is 2 · 105 J/kg J/kg.. This This yields ∆Q = 888.1184 ± 54.6J. Molar mass mass of aluminium . is 26 981538 981538g/m g/mol, ol, which which yields yields the mol molee numbe numberr n = 0.2124 ± 0.000037 000037mol mol.. We alread already y had initial initial temperatemperature T i = 296.15 ± 1K, final temperature: T f 77K K and f = 77 gas constant: R = 8.314472 314472J/(K J/(K mol) Matlab yields a graphical solution, see figure 3 3.. ΘE = 290K, 290 K, uncertainty due to time different ∆Θ different ∆ΘE,∆Time = 38 38K K and uncertainty due to mole number and initial temperature ∆Θ ture ∆ΘE, ∆n+∆T i = 4K, 4 K, Gauss uncertainty rule (4 ( 4) in our case, it would be better to express like this;
4. Results Results and discussio discussion n
∆ΘE =
4.1. Results Results
(∆Θ ) + 1
2
. . . + (∆Θn )2
(5)
It yields ∆ΘE = 38.1K. Howeve However, r, uncertain uncertainty ty is usually imprecise determined, so 38 ≈ 38.1 Then the result ΘE = 290 ± 38 38K K was found.
The aluminium sample weights 5 weights 5 .732 ± 0.001 001g g and the room temperature was 296 .15 ± 1K. The aluminium sample was dropped into the nitrogen after 7 minutes and 4 seconds (424 seconds). This caused an increase in the evaporation rate. The increased evaporation rate continued until 8 minutes and 4 seconds (484 seconds) when it returned to normal This means that the aluminium sample needs around 1 minute before its temperature to change from 296 .15 ± 1K (room temperature) to 77 to 77K K (liquid nitrogen temperature). Matlab yields figure 2. 455 seconds seconds is the mean value between the start and end of the increased evaporation rate. The time at 455 second is chosen to get ∆ mnitrogen , time at 425 and 485 second are chosen to get uncertainty of ∆ ∆ mnitrogen . Figure 2 Figure 2 shows ∆ shows ∆ mnitrogen = 4.44 ± 0.27 27g. g.
4.2. Discussion Discussion
∆ΘE,∆Time = 38 38K K is relatively big and undesirable. But what causes this big uncertain uncertainty? ty? In figure 2 shows that that these these two slopes slopes are quite quite differe different nt.. These These slopes represents the average evaporation rate of liquid nitrogen. The smaller slopes difference between red line and blue line, the smaller uncertainty. The actual slops is changing slowly slowly with time, but we treat it like constant constant.. The absolute value of the slope of the blue line is smaller than the red line. What is the reasons reasons to cases that these two two slopes are different? 2
5. Conclusi Conclusion on
The Einstein temperature was estimated to be ΘE = 290 ± 38K . This is a relatively large uncertainty that can be explained by the large time gap between the two series, in particular the time between evaporation returned to normal and the next measuremen measurement. t. The non-constan non-constantt cross section of the cup and ice forming on the sides of the cup may also have been a significant source of uncertainty. References [1] K. Razi Naqvi. Laborotorium Laborotorium i emnene emnene TFY4165/FY100 TFY4165/FY1005 5 Termisk fysikk, NTNU, 2014. [2] E. Lillestøl, O. Hunderi og J. R. Lien. Generell Generell fysikk for univeruniversiteter og høgskoler. Bind 2 Varmelære og elektromagnetisme. Universitetsforlaget, 2001. [3] A. Douglas Stone. Einstein Einstein and the Quantum. Princeton Princeton University versity Press, 2013, p. 146 [4] [4] Labh Labhef efte te TFY41 FY4165 65 Termis rmisk k fys fysikk, ikk, Høste østen n 2015 2015,,
This figure figure shows shows that the solutio solution n point point and its Figure Figure 3: This uncertaint uncertainty y. The horizontal horizontal red line represents represents ∆Q which is calculated by using ∆ mnitrogen , and these horizontal blue lines represe represent ntss the uncertain uncertainty ty of the red line due to ∆Time = massive evaporation evaporation maintains maintains for 60 seconds, seconds, then 30s. The massive creates a big uncertain uncertainty ty.. The anothe anotherr red ∆Time = 30. It creates line represents how much heat have been taken from our aluminium sample to bring down the temperature from 296.16K to 77K at different different Einsteins Einsteins temperature. temperature. Again, Again, these blue lines represents the uncertainty of the red line due to ∆n and ∆T .
http://home.phys.ntnu.no/brukdef/undervisning/tfy4165_ lab/orientering/termisk_labhefte_2015.pdf
The first reason reason may be the geometry geometry of the cups. The cups we used have different cross section area at different heights. With The lower liquid nitrogen level in the cups, the smaller cross section area. This could be improved by having a cross section that is non-dependent of the height of the cup, i.e. a box. The second reason is that ice begins to form on the exterior exterior of the cups. This slowly slowly decreases decreases the heat flow between between liquid nitrogen nitrogen and the environm environment ent.. And more importantly, the ice also have weight, but we could not exclude it in our measurement. This may be the reason why the blue line have smaller absolute value of slope in figure 2. This could be improved by having a large temperature reservoir with good thermal contact with the exterior surface of the cup, or just start the experiment after the ice is fully developed. developed. However, we’re still not sure how much these reasons can effects the two slopes in figure 2 until we do a new experiment and compare it. Some errors were not considered when we estimated the uncertaint uncertainty y. The evaporation evaporation temperature temperature of liquid nitrog nitrogen en may have have a uncert uncertain ainty ty.. The weigh weightt of ice is difficult to take into calculation, because it forms during the experiment. Labhefte [4 [4] at page 28 yields ΘE = 284 ± 6K. ComComparing our result to Labhefte [4 [ 4], Θ ], Θ E = 284 ± 6K is inside the interval ΘE = 290 ± 38 38K, K, but we have much bigger uncertainty. 3
