Calculating Entropy Derivative of a Temperature-Dependent Spring

In summary, the conversation discusses the calculation of the derivative of entropy with respect to length at constant temperature for a spring with temperature dependent k and heat capacity at constant length. The participants question why the entropy should depend on the length and discuss the properties of polymers in elastic materials. They also mention the possibility of cooling the spring while stretching it to maintain constant temperature. Ultimately, they suggest that stretching a rubber spring would lead to a decrease in entropy due to the creation of order on a microscopic level.
  • #1
painfive
24
0
A spring has a temperature dependent k of aT2(L-L0) and a heat capacity at constant length of CL=bT, a, L0, and b are constants. I need to calculate the derivative of entropy with respect to length at constant temperature. I can't figure out why the entropy should depend on the length. If dS=dQ/T, and when you stretch the spring you don't add any heat, why would the entropy rise? I'm not 100% sure I'm wrong, but based on the rest of the problem it doesn't seem likely that the answer actually is zero.
 
Physics news on Phys.org
  • #2
I really need help on this soon. I have to turn this in this afternoon.
 
  • #3
Well I had to hand it in, and I just said there is no change in entropy, but I don't think I was right. Nobody has any ideas on this? It seems like a pretty basic thermodynamics question. I mean, I'm one to talk, but I'm just suprised nobody here has responded.
 
  • #4
I don't understand thermodynamics too well, but here's what I figure: A rubber band is made up of polymers, long chainlike molecules that can twist in a number of different directions at each bond. With three mer units, the molecule can't bend too much, but with tens of thousands of mer units the polymer's basically just like a piece of string. For each polymer, there's a finite number of states it can be in. There's much, much more crumpled states than organized states. An organized state would be one where the rubber band is stretched so that its polymers line up. The random vibrations of the polymers cause it to change states frequently, and since there are more crumpled states than aligned states, the polymers tend to a crumpled state and the rubber tends to be unstretched. When you stretch the rubber band, you're fighting its polymers' tendencies to remain in a crumpled state, so you feel a force. If you heat rubber (not hot enough to cause a chemical change), its polymers vibrate more strongly, and they have an even stronger tendency to return to a crumpled state, causing a higher restoring force for hot rubber than cooler rubber, the opposite it would be for metals.

If we're talking about a rubber spring, then the answer seems pretty clear: when you stretch it, you create order on a microscopic level, which I assume means the entropy of the system is lowered. There's probably a similar answer that applies to all elastic materials, but I don't know what it is.
 
  • #5
Painfive:
Sorry I didn't see this Friday ...
When you stretch your spring you do Work on it
so its temperature rises as length increases.
they gave you c_v , you find c_p gas analogy.
Otherwise you can cool (Qout) as you stretch it
to keep it as constant T.
 

1. How do you calculate the entropy derivative of a temperature-dependent spring?

The entropy derivative of a temperature-dependent spring can be calculated using the formula:

dS/dT = C/T

Where dS/dT represents the entropy derivative, C is the heat capacity of the spring, and T is the temperature.

2. What is the significance of calculating the entropy derivative of a temperature-dependent spring?

Calculating the entropy derivative of a temperature-dependent spring allows us to understand the relationship between its temperature and its entropy, which is a measure of its disorder or randomness. This information can be useful in various fields such as thermodynamics and material science.

3. Can the entropy derivative of a temperature-dependent spring be negative?

Yes, the entropy derivative of a temperature-dependent spring can be negative. This indicates that as the temperature of the spring increases, its entropy decreases. This is because the spring becomes more ordered and less random as its temperature increases.

4. How does the heat capacity of a spring affect its entropy derivative?

The heat capacity of a spring plays a crucial role in determining its entropy derivative. A higher heat capacity means that the spring can absorb more heat before its temperature increases, resulting in a smaller change in entropy. Therefore, a spring with a higher heat capacity will have a smaller entropy derivative compared to a spring with a lower heat capacity.

5. Are there any limitations to calculating the entropy derivative of a temperature-dependent spring?

Yes, there are limitations to calculating the entropy derivative of a temperature-dependent spring. This formula assumes that the spring is in thermal equilibrium with its surroundings and that its heat capacity remains constant with temperature. In reality, these conditions may not always be met, leading to potential errors in the calculated entropy derivative.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
763
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
366
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
2K
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
884
Back
Top