Thermodynamics problem involving adiabatic vessels

[Titan97]

Homework Statement

Two adiabatic vessels containing $n$ moles of mono atomic and diatomic gas respectively are connected by a rod of length $l$, cross sectional area $A$ and thermal conductivity $k$. The surface of the rod is insulated.
The initial temperatures of the vessels are $T_1$ and $T_2$ respectively. Find the time taken for the temperature difference to be half the initial temperature difference.

Homework Equations

$$\Delta U=nC_v\Delta T$$
$C_{v1}=\frac{3nR}{2}$
$C_{v2}=\frac{5nR}{2}$

The Attempt at a Solution

The gases will undergo an isochoric process. The heat from one vessel moves to the other along the rod.

Heat lost by vessel 1 = heat gained by vessel 2

$$\frac{3}{2}nR(T_1-{T'}_1)=\frac{5}{2}nR({T'}_2-T_2)$$
If the heat energy lost by vessel 1 is $dQ$ in a time $dt$ and the temperature difference is $\Delta T$,
$$\frac{dQ}{dt}=\frac{\Delta T\cdot kA}{L}$$
But I am not able to continue.

 

[Chestermiller]

Here's a hint:
$$\frac{dU_1}{dt}=-kA\frac{(T_1'-T_2')}{L}$$
$$\frac{dU_2}{dt}=+kA\frac{(T_1'-T_2')}{L}$$