Specific Heat of a nonlinear, temperature dependent spring

Infernorage
Messages
24
Reaction score
0

Homework Statement


A nonlinear spring has a temperature dependent force law,

F = -\frac{K}{T}(L-L_o)^3

At a temperature T = T_o and length L = L_o the specific heat at a constant length is C_L = C_o. What is the specific heat at T = T_o when the spring is stretched to length 2L_o?

Homework Equations





The Attempt at a Solution


I am really not sure where to start. Am I supposed to use the general dU=dQ-dW equation and somehow work the spring into that using the force equation and then solve for a specific heat equation? Any advice would be great. Thanks.
 
Physics news on Phys.org
Since the force is the negative gradien of its potential energy, in a one dimensional case as this we can write,

F=-\frac{dU}{dL} or

U=-\int FdL

Once obtained the potential energy, use the defenition of the heat capacity at constant length,

C_L=(\frac{dU}{dT})_L

With the given information you should be able to express your result in terms of C_o.
 
K space said:
Since the force is the negative gradien of its potential energy, in a one dimensional case as this we can write,

F=-\frac{dU}{dL} or

U=-\int FdL

Once obtained the potential energy, use the defenition of the heat capacity at constant length,

C_L=(\frac{dU}{dT})_L

With the given information you should be able to express your result in terms of C_o.

If you do that doesn't C_L=(\frac{dU}{dT})_L come out to be C_L = -\frac{K}{4T^2}(L-L_o)^4? How can you have a negative specific heat?
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top