= Solution
{c}
For any fixed subregion $\Omega$, with no internal <heat> generation, <conservation of energy> says that the rate of stored thermal <energy> equals the incoming <heat flux>:
$$
\frac{d}{dt}\iiint_\Omega\rho cT\,dV=-\iint_{\partial\Omega}\mathbf q\cdot\mathbf n\,dA.
$$
The minus sign is needed because $\mathbf q\cdot\mathbf n>0$ denotes outgoing <heat flux>. Substitute <Fourier's law> $\mathbf q=-k\nabla T$ and apply the <divergence theorem>:
$$
\iiint_\Omega\rho cT_t\,dV
=\iint_{\partial\Omega}k\nabla T\cdot\mathbf n\,dA
=\iiint_\Omega\nabla\cdot(k\nabla T)\,dV.
$$
Here differentiation under the <integral> uses a fixed region and sufficient regularity of $T$. Since the identity holds for every $\Omega$, continuity of the integrands gives the local <heat equation>
$$
\boxed{\rho cT_t=\nabla\cdot\bigl(k(\mathbf x)\nabla T\bigr).}
$$
For spatially varying <thermal conductivity>, the <product rule> gives $\nabla\cdot(k\nabla T)=k\nabla^2T+\nabla k\cdot\nabla T$; the second term cannot generally be discarded.
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