STEM Problem Detail

Problem Statement

Prompt

Provide me step by step solution of this problem 5.19

Solution Rendering...

The difference of heat capacities of two phases is related to the difference of the temperature derivatives of their entropies by

\[
\Delta c_H = c_H^L - c_H^F = \frac{T}{m} \left[ \left( \frac{\partial S_L}{\partial T} \right)_H - \left( \frac{\partial S_F}{\partial T} \right)_H \right] = \frac{T}{m} \left( \frac{\partial (Q_{L,F}/T)}{\partial T} \right)_H.
\]

In the last step, we took into account that in coexistence, the difference of entropies is proportional to the latent heat of the transition $$Q_{L,F} = T(S_L - S_F)$$. The latent heat can be obtained like in Problem 5.16 or by realizing that the Clausius-Clapeyron equation is nothing but a Maxwell relation applied to phase coexistence. For a magnetic system, this relation reads $$(\partial S/\partial M)_T = -V\mu_0(\partial H/\partial T)_M$$, or

\[
\frac{S_L - S_F}{M_L - M_F} = \frac{Q_{L,F}}{T H_c(C/T - \chi_0)} = -V\mu_0 \frac{dH_c(T)}{dT}.
\]

By evaluating the derivative on the right-hand side, we see that the latent heat per unit mass is

\[
\frac{Q_{L,F}}{m} = \frac{\mu_0 H_0^2 (C - \chi_0 T)}{\rho T_0^2},
\]

where $$\rho$$ denotes the density of the compound, and thus

\[
\Delta c_H = -\frac{T \mu_0 \chi_0 H_0^2}{\rho T_0^2} = -11.4 \, \text{J/kgK}.
\]

(This approach can also be used in Problem 5.16.)

At 30 K, the critical magnetic field strength is $$H_c = 2.1 \times 10^7 \, \text{A/m}$$. Thus at $$H = 4 \times 10^7 \, \text{A/m}$$ phase $$\mathcal{F}$$ is indeed metastable. The heat exchanged at constant $$T$$ and $$H$$ during the $$\mathcal{F} \to \mathcal{L}$$ transition is equal to the enthalpy difference $$Q = \Delta(G + TS)$$. As this transition is strongly irreversible, $$Q < T \Delta S$$ and the second law cannot be used to calculate $$Q$$. Instead we have to derive the Gibbs free energies $$G_{\mathcal{L},\mathcal{F}}(T,H)$$ and the entropies $$S_{\mathcal{L},\mathcal{F}}(T,H)$$ of both phases like in Problem 5.16. We start by recalling that at phase coexistence at a given temperature, $$G_{\mathcal{F}}(T,H_c) = G_L(T,H_c)$$ and $$Q_{L,\mathcal{F}}/T = S_L(T,H_c) - S_F(T,H_c)$$. Moreover,

\[
\left( \frac{\partial G_{\mathcal{L}, \mathcal{F}}}{\partial H} \right)_T = -\mu_0 V M_{\mathcal{L}, \mathcal{F}} =
\begin{cases}
-V\mu_0 \chi_0 H, & \text{phase } \mathcal{F} \\
-V\mu_0 (C/T)H, & \text{phase } \mathcal{L}
\end{cases},
\]

and

\[
\left( \frac{\partial S_{\mathcal{L}, \mathcal{F}}}{\partial H} \right)_T = V\mu_0 \left( \frac{\partial M_{\mathcal{L}, \mathcal{F}}}{\partial T} \right)_H =
\begin{cases}
0, & \text{phase } \mathcal{F} \\
-V\mu_0 (C/T^2)H, & \text{phase } \mathcal{L}
\end{cases}.
\]

We can further write

\[
G_{\mathcal{L}, \mathcal{F}}(T, H) = G_{\mathcal{L}, \mathcal{F}}(T, H_c) + \int_{H_c}^H \left( \frac{dG_{\mathcal{L}, \mathcal{F}}}{dH} \right)_T dH
\]

and

\[
S_{\mathcal{L}, \mathcal{F}}(T, H) = S_{\mathcal{L}, \mathcal{F}}(T, H_c) + \int_{H_c}^H \left( \frac{dS_{\mathcal{L}, \mathcal{F}}}{dH} \right)_T dH.
\]

Imagine now a reversible path starting at $$H > H_c$$ in phase $$\mathcal{F}$$ whereby the sample reaches phase equilibrium at $$H_c$$ upon an isothermal decrease of the field strength, then undergoes a reversible $$\mathcal{F} \to \mathcal{L}$$ transition, and finally returns to the initial $$H$$ in phase $$\mathcal{L}$$ upon an isothermal increase of the field strength. This path can be used to calculate the exchanged heat given by the difference of enthalpies; note that all quantities involved—enthalpy, Gibbs free energy, and entropy—are functions of state. We find that the heat exchanged per unit mass is

\[
\frac{Q}{m} = \frac{\mu_0}{\rho} \left( \frac{\chi_0}{2} - \frac{C}{T} \right)(H^2 - H_c^2) + \frac{Q_{\mathcal{L}, \mathcal{F}}}{m}
\]

so that 0.34 kJ of heat is emitted in the process.

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Problem #21