= Solution
The reference to part “(c)(i)” in the source is circular; the relevant modes are those verified in part (b)(i). <Linearity> of the <heat equation> and its homogeneous <Dirichlet boundary conditions> allows <superposition> of those modes:
$$
F_n(x)=\sin\frac{n\pi x}{L},\qquad \lambda_n=\kappa\frac{n^2\pi^2}{L^2},\qquad T(x,t)=\sum_{n=1}^{\infty}b_nF_n(x)e^{-\lambda_nt}.
$$
Match the initial uniform interior <temperature> $T_0$ using the <orthogonality> of the <Fourier sine series>:
$$
b_n=\frac2L\int_0^LT_0\sin\frac{n\pi x}{L}\,dx=\frac{2T_0}{n\pi}\{1-(-1)^n\}.
$$
Thus \b[$b_n=0$ for <even integers> $n$ and $b_n=4T_0/(n\pi)$ for <odd integers> $n$], giving
$$
T(x,t)=\frac{4T_0}{\pi}\sum_{j=0}^{\infty}\frac{\sin((2j+1)\pi x/L)}{2j+1}\exp\left[-\frac{(2j+1)^2\pi^2\kappa t}{L^2}\right].
$$
For every $t>0$, the exponential factor makes the series and the derivatives needed in the <heat equation> converge uniformly on later time intervals. At $t=0$, the <Fourier sine series> equals $T_0$ at interior points but zero at the endpoints: the instantaneous cooling boundary condition is incompatible with the pre-immersion endpoint values. This corner mismatch is physically the idealization of an instantaneous bath contact and does not invalidate the solution for $t>0$.
\Image[/past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2023/paper-1-cooling-profile.png]
{title=Cooling profiles in a zero-temperature bath and the mean temperature retained after insulation}
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