Solved thermodynamics exercise

Stretching an elastic wire

Exercise 12 · Lesson 4 — The First Law of Thermodynamics

generalized work
elastic wire
tension
first law

Statement

A wire of unstretched length $L_{0}$ is stretched slowly, without hysteresis, from $L_{A}$ to $L_{B}$ , with $L_{0} \leq L_{A} < L_{B}$ . In the range considered, its equilibrium tension obeys the law

F = K (L - L_{0}),

where $K > 0$ is a constant stiffness, independent of temperature. The external pulling force balances the tension of the wire at every stage. We adopt the internal-energy model

U (T, L) = C (T - T_{⋆}) + \frac{1}{2} K (L - L_{0})^{2} + U_{⋆},

where $C$ is a constant total heat capacity and $T_{⋆}, U_{⋆}$ are fixed reference values. Changes in kinetic energy and in gravitational potential energy are negligible.

Draw the wire, its fixed point, the movable end and the external pulling force. Write the elementary work received for a change $d L$ , then calculate $W_{A \to B}$ .
The stretching is isothermal. Calculate $Δ U$ and the heat $Q_{A \to B}$ received in this model. Does an isothermal transformation in general imply $Δ U = 0$ ?
The wire is brought back from $L_{B}$ to $L_{A}$ following the same tension law. Calculate the work received on this return path and over the round trip. Why does this result not mean that all work is a state function?

Hint

Use

δ W = F_{ext} d L

and integrate the variable force. Then compare with the change in internal energy given by the model, using

Δ U = Q + W

Detailed solution

Solution

Question 1. In the diagram, the left end is fixed. The right end is pulled towards the right; its displacement is

d L

. The work received by the wire is positive when it lengthens:

δ W = F_{ext} d L = K (L - L_{0}) d L .

The force is not constant, so we must calculate

W_{A \to B} = \int_{L_{A}}^{L_{B}} K (L - L_{0}) d L = [\frac{K}{2} (L - L_{0})^{2}]_{L_{A}}^{L_{B}} = \frac{K}{2} [(L_{B} - L_{0})^{2} - (L_{A} - L_{0})^{2}] > 0.

The stiffness is measured in $N m^{- 1}$ , so $K (L - L_{0})^{2}$ is expressed in $N m$ , that is, in joules.

Question 2. Since

T_{B} = T_{A}

, the thermal term does not change. The reference constants also cancel, so

Δ U = C (T_{B} - T_{A}) + \frac{K}{2} [(L_{B} - L_{0})^{2} - (L_{A} - L_{0})^{2}] = W_{A \to B} .

The first law then gives $Q_{A \to B} = Δ U - W_{A \to B} = 0$ . In this ideal model, the work received is stored as elastic energy. The temperature remains constant, but the internal energy increases: an isothermal process therefore does not imply $Δ U = 0$ . The result $Q = 0$ depends here on the model adopted and does not extend to all materials.

Question 3. Swapping the limits changes the sign of the integral:

W_{B \to A} = \int_{L_{B}}^{L_{A}} K (L - L_{0}) d L = [\frac{K}{2} (L - L_{0})^{2}]_{L_{B}}^{L_{A}} = - W_{A \to B} .

The work received over the round trip is therefore zero: the wire gives back the stored work. Here, the imposed law makes the work equal to the change in elastic energy, which is a function of $L$ . This is a property of this hysteresis-free model and of this family of transformations, not a general property of work transfers.