- #1
enjoi668
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Homework Statement
The first part of the problem says this figure shows a one-dimensional row of 5 microscopic objects each of mass 6e-26 kg, connected by forces that can be modeled by springs of stiffness 30 N/m (so the effective stiffness of the inner springs is 4ks = 120 N/m; approximate the end springs this way, too). These 5 objects can move only along the x axis.
http://www.webassign.net/csmech/10-52.jpg
(this is the picture of the system)
The first part of the question asks to calculate the entropy of the system at different quanta and I calculated the entropy of the system with 3 quanta to be 4.9089e-23 J/K and with 4 quanta to be 5.8659e-23 J/K, and got that part correct.
The part I need help with reads 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.)
Homework Equations
1/T = dS/dEint (definition of temperature) where S is the entropy and Eint is the internal energy.
Eint = q*hbar*sqrt(ks/m) where q is the number of quanta, hbar is the reduced planks constant, ks is the spring constant, and m is the mass.
The Attempt at a Solution
I figured in order to solve for T in the equation defining temperature, I had to take the difference in Entropy between 4 and 3 quanta which is simply 5.8659e-23 J/K - 4.9089e-23 J/K = 9.57e-24 J/K. Then find dEint and divide. My problem arrises when finding Eint for the system when it has 4 and 3 quanta respectively. In the equation Eint = q*hbar*sqrt(ks/m), q is obvious and hbar is a constant, but what value should I use for ks and m? I think I should simply add all of the masses up to get 5*6e-26=3e-25kg, but I am genuinely confused as to what to use for the spring constant of this system. Should I simply take ks and multiply it by the number of springs? The sentence at the end of the first part that reads "the effective stiffness of the inner springs is 4ks = 120 N/m; approximate the end springs this way, too" makes absolutely no sense to me, and I feel like it is an indication as to how to find the spring constant of the system.