Physics: Specific heat capacity of a solid

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SUMMARY

The discussion focuses on the calculation of specific heat capacity in a one-dimensional system of five microscopic objects, each with a mass of 4.10e-26 kg and connected by springs with a stiffness of 26 N/m. Using the Einstein model, participants calculated the entropy for varying quanta of energy, with specific values provided for each quantum state. The average absolute temperatures for energy ranges of 3 to 4 quanta and 4 to 5 quanta were determined to be approximately 561.89 K and 662.61 K, respectively. The final challenge involved determining the classical heat capacity per object, which was concluded to be 1.38e-23 J/K/object.

PREREQUISITES
  • Understanding of the Einstein model of solids
  • Familiarity with quantum mechanics concepts, specifically quanta of energy
  • Knowledge of thermodynamic principles, including entropy and heat capacity
  • Proficiency in using relevant equations such as C = ΔE/ΔT
NEXT STEPS
  • Study the Einstein model of solids in detail
  • Learn about the derivation and implications of the heat capacity formula C = ΔE/ΔT
  • Explore the differences between one-dimensional and three-dimensional systems in thermodynamics
  • Investigate the implications of quantum mechanics on heat capacity at low temperatures
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Students and professionals in physics, particularly those studying thermodynamics and quantum mechanics, as well as anyone involved in research related to specific heat capacity in solid-state systems.

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Homework Statement


The diagram below shows a one-dimensional row of 5 microscopic objects each of mass 410-26 kg, connected by forces that can be modeled by springs of stiffness 26 N/m (so each object can be modeled as if it were connected to a single spring of effective stiffness 4ks = 104 N/m -- neglect any possible differences for objects near the ends). These 5 objects can move only along the x axis.Use these precise values for the constants:
hbar = 1.054610-34 J · s (Planck's constant divided by 2)
k = 1.380710-23 J/K (the Boltzmann constant)What is one quantum of energy for one of these objects?
Delta E = 5.37743e-21 correct check mark J

Using the Einstein model, calculate the entropy of this system for total energy of 0, 1, 2, 3, 4, and 5 quanta.
q = 0: S = 0 correct check mark J/K

q = 1: S = 2.22215e-23 correct check mark J/K

q = 2: S = 3.73900e-23 correct check mark J/K

q = 3: S = 4.90887e-23 correct check mark J/K

q = 4: S = 5.86590e-23 correct check mark J/K

q = 5: S = 6.67745e-23 correct check mark J/K

Calculate to the nearest degree the average absolute temperature of the system when the total energy is in the range from 3 to 4 quanta. (You can think of this as the temperature when there would be 3.5 quanta of energy in the system, if that were possible.)
T3 to 4 = 561.88730 correct check mark K

Calculate to the nearest degree the average absolute temperature of the system when the total energy is in the range from 4 to 5 quanta. (You can think of this as the temperature when there would be 4.5 quanta of energy in the system, if that were possible.)
T4 to 5 = 662.61229 correct check mark K

Calculate the heat capacity per object when the total energy is 4 quanta. (Think of this in terms of increasing from 3.5 quanta of energy in the system to 4.5 quanta of energy in the system, if that were possible.)
C4 = 1.067745e-23 correct check mark J/K/object

If the temperature were raised very high, classically what would we expect the heat capacity per object to be for this one-dimensional system? Give a numerical value.
Chigh T = ? wrong check mark J/K/object
(One reason for the discrepancy is that the high-temperature limit assumes that the number of oscillators is large (N >> 1), which is not the case in this tiny system.)

Homework Equations



C= deltaE/deltaT

classical limit of 3k = 4.210-23 J/K/atom

The Attempt at a Solution



I got all the questions except the last one which I've already tried searching for clues in the internet.
I tried 3kN= 3*1.380710e23*5 but got it wrong.
I thought it was just supposed to be 3k like in the relevant equation but that's wrong maybe because N=5??
I also thought that 3 in the 3k equation has something to do with 3 springs, so I tried 2*k instead of 3*k but got it wrong too.
Please help.
 
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I don't think you need to multiply it by 5 if you're already multiplying by 3. Since you do 3k for a three-dimensional system, i figured that you could just do 2k, since only two springs are attached to each 'atom'. This was also wrong. I also tried 5k, but that didn't work. Apparently 3k is also wrong.

Now that we've eliminated 5 options, anyone else have ideas?
 
I didn't know that 3k was for a three-dimensional system. If that's the case, then how about trying 1k since this is a one-dimensional system.

Ok. I got the answer. The answer is 1k= 1.3807e23.

Thanks for joining the discussion, phys2211. And Thanks a lot for that important point you gave out.
 
glad i could be of service :)
 
Hey how do u search for T3 to 4 on T4 to 5?
Thanks
 

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