Question 4.10: An state of an elastic rod is described by the state variabl...

a) Applying the cyclic rule (4.81) to the force F (T, L) we obtain,

\frac{∂x (y, z)}{∂y} \frac{∂y (z, x)}{∂z} \frac{∂z (x, y)}{∂x} =-1 .                            (4.81)

\frac{∂f }{∂T} \frac{∂T }{∂L} \frac{∂L }{∂f} = -1 .

and thus

\frac{∂f }{∂T} = -\frac{∂L }{∂T} \frac{∂f }{∂L} = −α A E.

Since the cross section A is independent of the temperature, the stress in the rod varies with temperature as,

\frac{∂τ }{∂T} = −α E.

b) At constant temperature T, the infinitesimal heat transfer is written as,

δQ = T dS(T, L) = T \frac{∂S }{∂L} dL.

Thus, after integration, we obtain the heat transfer for an isothermal process from an intial state i to a final state f,

Q_{if} = T \frac{∂S }{∂L} ΔL_{if}.

The differential of the free energy is,

dF = −S (T, L) dT + f (S, L) dL.

The Schwarz theorem applied to free energy F (T, L) yields,

\frac{\partial}{\partial L} \biggl(\frac{\partial F }{\partial T }\biggr) = \frac{\partial}{\partial T} \biggl(\frac{\partial F }{\partial L }\biggr).

which leads to the Maxwell relation,

– \frac{\partial S (T, A) }{\partial L} = \frac{\partial f (S, L) }{\partial T}.

Using the Maxwell relation and the cyclic rule (4.81), the heat transfer can be recast as,

\frac{∂x (y, z)}{∂y} \frac{∂y (z, x)}{∂z} \frac{∂z (x, y)}{∂x} =-1 .                (4.81)

Q_{if} = −T \frac{\partial f  }{\partial T} ΔL_{if} = T \frac{\partial L  }{\partial T} \frac{\partial f  }{\partial L} ΔL_{if} = α T A E ΔL_{if} .

c) For an abiabatic process, we need to determine the derivative of the length L (S, T) with respect to temperature when the entropy is kept constant. Using the cyclic rule (4.81),

\frac{\partial L  }{\partial T} = – \frac{\partial T  }{\partial S}\frac{\partial S  }{\partial L} .

When identifying the expressions of the heat transfer Q_{if} obtained above, we find,

\frac{\partial S  }{\partial L} = α A E.

Thus,

where

c_L = T \frac{∂S (T, L) }{\partial T} .