= Solution
{c}
In a steady state the <heat equation> says $(kT_s')'=0$, so the positive-$x$ <heat flux> $q=-kT_s'$ is the same constant $Q$ in both materials. Assume perfect thermal contact: temperature is continuous at $x=\ell$, and <conservation of energy> requires continuous <heat flux> there. Integrating from the right endpoint gives
$$
\boxed{T_s(x)=\begin{cases}
\displaystyle\frac{Q\ell}{k_2}+\frac Q{k_1}(\ell-x),&0\leq x\leq\ell,\\[6pt]
\displaystyle\frac Q{k_2}(2\ell-x),&\ell\leq x\leq2\ell.
\end{cases}}
$$
The two formulas agree at the interface. Their slopes differ, but $-k_1T_s'(\ell^-)=-k_2T_s'(\ell^+)=Q$. The total temperature drop is $Q\ell(1/k_1+1/k_2)$, the sum of the two layers' contributions in <steady heat conduction through layers>.
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