= Solution
Take any fixed material subvolume $B_0$ of the <reference configuration>, with outward <normal vector> $N$, current particle <velocity> $v_i=\dot x_i$, and positive <temperature> $\theta$. The <first law of thermodynamics>, including <kinetic energy>, is
$$
\frac{d}{dt}\int_{B_0}\rho_0\left(u+\tfrac12v_iv_i\right)dX=\int_{B_0}\rho_0(r+g_iv_i)dX+\int_{\partial B_0}(N_IP_{Ii}v_i-N_Iq_I^0)\,dA_0.
$$
The heat-flux sign is negative because $N_Iq_I^0$ is outward. The <Second law of thermodynamics> supplies the corresponding integral <Lagrangian entropy inequality>:
$$
\frac{d}{dt}\int_{B_0}\rho_0\eta\,dX\geq\int_{B_0}\frac{\rho_0r}{\theta}\,dX-\int_{\partial B_0}\frac{N_Iq_I^0}{\theta}\,dA_0.
$$
The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.
Apply the <divergence theorem> and use the reference <momentum conservation> equation $\rho_0\dot v_i=P_{Ii,I}+\rho_0g_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_iP_{Ii,I}$ term. Since $v_{i,I}=\dot F_{iI}$, the remaining local <Lagrangian internal energy balance> and <Lagrangian entropy inequality> are
$$
\boxed{\rho_0\dot u=\rho_0r-q_{I,I}^0+P_{Ii}\dot F_{iI}},\qquad
\boxed{\rho_0\dot\eta\geq\frac{\rho_0r}{\theta}-\left(\frac{q_I^0}{\theta}\right)_{,I}}.
$$
They hold pointwise because $B_0$ was arbitrary. In particular,
$$
\rho_0\theta\dot\eta\geq\rho_0r-q_{I,I}^0+\frac{q_I^0\theta_{,I}}{\theta}.
$$
The specific <Helmholtz free energy> is $\psi=u-\theta\eta$. Substitution yields its balance
$$
\rho_0(\dot\psi+\eta\dot\theta+\theta\dot\eta)=\rho_0r-q_{I,I}^0+P_{Ii}\dot F_{iI},
$$
and elimination of the heat supply yields the <Clausius-Duhem inequality>
$$
P_{Ii}\dot F_{iI}-\rho_0(\dot\psi+\eta\dot\theta)-\frac{q_I^0\theta_{,I}}{\theta}\geq0.
$$
For $\psi=\psi(F,\theta,\xi_r)$, expand the derivative to obtain
$$
(P_{Ii}-\rho_0\psi_{,F_{iI}})\dot F_{iI}-\rho_0(\psi_{,\theta}+\eta)\dot\theta+\rho_0f_r\dot\xi_r-\frac{q_I^0\theta_{,I}}\theta\geq0,\qquad f_r=-\psi_{,\xi_r}.
$$
The usual <Coleman-Noll procedure> admits independent reversible choices of $\dot F$ and $\dot\theta$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. Thus
$$
\boxed{P_{Ii}=\rho_0\psi_{,F_{iI}},\qquad\eta=-\psi_{,\theta}},\qquad
\boxed{\rho_0f_r\dot\xi_r-\frac{q_I^0\theta_{,I}}\theta\geq0}.
$$
Internal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance gives
$$
\boxed{\rho_0\theta\dot\eta=\rho_0r-q_{I,I}^0+\rho_0f_r\dot\xi_r}.
$$
\b[The last printed equation is missing a factor $\rho_0$ on its internal-variable term.] The PDF defines $f_r=-\psi_{,\xi_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric <thermodynamic force conjugate to an internal variable> $\widehat f_r=-\rho_0\psi_{,\xi_r}$; the term is then $\widehat f_r\dot\xi_r$ with no extra density. The two conventions must not be mixed.
For <thermoelasticity with a temperature-dependent constraint>, admissible rates obey
$$
\phi_{,F_{iI}}\dot F_{iI}=h'(\theta)\dot\theta.
$$
The reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:
$$
P_{Ii}-\rho_0\psi_{,F_{iI}}=q\phi_{,F_{iI}},\qquad -\rho_0(\psi_{,\theta}+\eta)=-qh'(\theta).
$$
Therefore
$$
\boxed{P_{Ii}=\rho_0\psi_{,F_{iI}}+q\phi_{,F_{iI}},\qquad\eta=-\psi_{,\theta}+\frac{q}{\rho_0}h'(\theta)}.
$$
The reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.
For <linear thermoelasticity with prescribed thermal volume>, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta=\theta-\theta_0$. Hold the <internal variables> fixed in this reversible linearization, and evaluate all derivatives below at $(I,\theta_0)$ and zero reaction. The stress-free reference state has $\psi_{,F}=0$. Since $\partial\det F/\partial F=I$ there, first-order nominal and <Cauchy stress tensors> agree. Their constitutive laws become
$$
\boxed{\sigma_{ji}=C_{jilk}u_{k,l}+\beta_{ji}\vartheta+q\delta_{ji}},\qquad
\boxed{\rho_0(\eta-\eta_0)=C_e\vartheta-\beta_{ji}u_{i,j}+qh'(\theta_0)},
$$
where equality of mixed free-energy derivatives gives
$$
\boxed{C_{jilk}=\rho_0\psi_{,F_{ij}F_{kl}},\qquad\beta_{ji}=\rho_0\psi_{,F_{ij}\theta},\qquad C_e=-\rho_0\psi_{,\theta\theta}}.
$$
The linearized constraint is $u_{i,i}=h'(\theta_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent <incompressibility> corresponds to $h'=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric <heat capacity> at fixed deformation is $\theta_0C_e$.
Together with $\rho_0\ddot u_i=\sigma_{ji,j}+\rho_0g_i$, the first-order thermal balance is $\theta_0(C_e\dot\vartheta-\beta_{ji}\dot u_{i,j}+h'(\theta_0)\dot q)=\rho_0r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.
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