Question

Metal elasticity is due to energy, not entropy. Experiments show that the retractive force \(f\) of a metal rod as a function of temperature \(T\) and extension \(L\) relative to undeformed length \(L_{0}\) is given by \(f(T, L)=\) Ea \(\Delta L / L_{0}\), where \(\Delta L=L\left[1-\alpha\left(T-T_{0}\right)\right]-L_{0}=L-L \alpha(T-\) \(\left.T_{0}\right)-L_{0} . a\) is the cross-sectional area of the rod, \(E\) (which has the role of a spring constant) is called Young's modulus, and \(\alpha \approx 10^{-5}\) is the linear expansion coefficient. Compute \(H(L)\) and \(S(L)\). Is the main dependence on \(L\) due to enthalpy \(H\) or entropy \(S\) ?

Short answer

The main dependence on L is due to enthalpy H(L), not entropy S(L).

Step-by-step solution

Understanding the given equation

The retractive force of a metal rod as a function of temperature and extension is given by the equation: \[ f(T, L) = Ea \frac{\Delta L}{L_0} \]where \( \Delta L = L \left[1 - \alpha (T - T_0) \right] - L_0 \)].

Substitute ΔL and simplify

Substitute \( \Delta L \) into the force equation: \[ f(T, L) = Ea \frac{L - L \alpha (T - T_0) - L_0}{L_0} \]Simplify the expression: \[ f(T, L) = Ea \left( \frac{L}{L_0} - \frac{L \alpha (T - T_0)}{L_0} - 1 \right) \]\[ f(T, L) = Ea \left( \frac{L - L \alpha (T - T_0) - L_0}{L_0} \right) \].

Calculate the enthalpy

To compute the enthalpy, use the relationship between force and energy. Since force is related to the change in energy with respect to length, we can find the enthalpy, \(H(L)\), as: \[ H(L) = f(T, L) \times L \]Substituting for \( f(T, L) \): \[ H(L) = Ea \frac{L - L \alpha (T - T_0) - L_0}{L_0} \times L \]

Calculate the entropy

The entropy change \( S(L) \) is typically due to thermal factors and is not directly dependent on the length of the rod for this given context. Thus, we can represent it as: \[ S(L) \approx 0 \] figure out more about temperature dependence, if given.

Determine the main dependence

Given the enthalpy \( H(L) \) is dependent on both the Young's modulus \(E\), extension \(L\), and temperature parameter \( T-T_0 \), while the entropy \(S(L) \) is approximately 0 for this scenario, the main dependence on \(L\) is due to the enthalpy \(H\), not the entropy \(S\).

Key concepts

enthalpy

Enthalpy, denoted as H, is a measure of total energy in a thermodynamic system. It includes internal energy and the product of pressure and volume. In the context of metal elasticity, enthalpy helps us understand how internal energy changes with length under varying temperatures. For our problem, the retractive force is involved in describing how the metal rod behaves when it extends or contracts. Through the relationship \[ H(L) = f(T, L) \times L \], we established that the enthalpymainly depends on the properties of the rod (Young's modulus E and its cross-sectional area a), and the temperature difference (T - T_0). This provides a crucial understanding that the changes in the length of the rod are energetically driven by enthalpy variations, rather than entropy. Hence, enthalpy captures the energy contributions due to mechanical stretching and temperature changes.

Young's modulus

Young's modulus, represented by E, dramatically influences the elasticity of materials. It is a measure of the stiffness of a solid material. Essentially, it quantifies the relationship between stress (force per unit area) and strain (proportional deformation) in a material. In our exercise, we see Young's modulus playing a central role in the equation of the retractive force \[ f(T, L) = Ea \frac{\Delta L}{L_0} \].
The higher the Young's modulus, the stiffer the material, leading to a greater retractive force for the same amount of deformation. Metals with high Young's modulus don't easily deform, meaning more force is needed to achieve the same elongation compared to metals with lower Young's modulus. This translates to more significant changes in enthalpy (H) for given changes in length, emphasizing the prominent role of Young's modulus in determining metal elasticity.

entropy

Entropy, symbolized as S, is a measure of the randomness or disorder in a system. Unlike enthalpy, entropy is more related to thermal factors and distribution of energy states. In the exercise, we see that the entropy change \[ S(L) \approx 0 \] doesn't significantly depend on the length of the metal rod. This underlines that the primary dependence of the rod's behavior isn't on the disorder but rather on energy contributions (enthalpy). Even though temperature variations can influence entropy, they are not the main factor determining the length changes and force exertion in the metal rod in this scenario.
Essentially, this makes clear that for the given problem, entropy doesn't contribute significantly to understanding elongation and exertive forces, pointing to the dominance of enthalpy in explaining metal elasticity in thermodynamics.