Two adiabatic vessels, as shown in the figure, each contain a mass...

Let $T_1$ and $T_2$ be the temperatures of the water in the two vessels at time $t$, with $T_1 > T_2$.

The rate of heat flow through the metal rod from the hotter vessel to the colder vessel is given by:
$\dfrac{dQ}{dt} = \dfrac{KA(T_1 - T_2)}{L}$

The heat lost by the first vessel causes its temperature to decrease, while the heat gained by the second vessel causes its temperature to increase. The rates of change of their temperatures are:
$\dfrac{dT_1}{dt} = -\dfrac{1}{ms} \dfrac{dQ}{dt}$
$\dfrac{dT_2}{dt} = \dfrac{1}{ms} \dfrac{dQ}{dt}$

Let the temperature difference be $\Delta T = T_1 - T_2$. The rate of change of this temperature difference is:
$\dfrac{d(\Delta T)}{dt} = \dfrac{dT_1}{dt} - \dfrac{dT_2}{dt} = -\dfrac{2}{ms} \dfrac{dQ}{dt}$

Substituting the expression for $\dfrac{dQ}{dt}$:
$\dfrac{d(\Delta T)}{dt} = -\dfrac{2KA}{msL} (T_1 - T_2) = -\dfrac{2KA}{msL} \Delta T$

Separating the variables and integrating from $t = 0$ to $t = t$, where the temperature difference drops from its initial value $\Delta T_0$ to $\dfrac{\Delta T_0}{2}$:
$\int_{\Delta T_0}^{\Delta T_0/2} \dfrac{d(\Delta T)}{\Delta T} = -\int_0^t \dfrac{2KA}{msL} dt$

$\ln\left(\dfrac{\Delta T_0/2}{\Delta T_0}\right) = -\dfrac{2KA}{msL} t$

$\ln\left(\dfrac{1}{2}\right) = -\dfrac{2KA}{msL} t$

$-\ln 2 = -\dfrac{2KA}{msL} t$

$t = \dfrac{Lms}{2KA} \ln 2$