Entropy changes during transfer of heat between two bodies Consider two metal blocks, one at 500K and another at 300K, they are bought to contact with each other until they reach a common temperature.
Find the entropy change of each object and the total change considering there is no change in the volume of the metal blocks in the process and no the other interactions excluding the heat transfer between the two blocks.Also is this process irreversible, why?

also
$C_v= 10 J/K$ for both the metal blocks

My attempt:
This process is irreversible, as once it reaches the equilibrium temperature it cannot go back to the initial state without external intervention that is to say heat wont flow from cold body to hot body and also when two bodies are at the same temperature heat doesn't flow from one to the other to increase the temperature of one and to decrease the other.

now to compute the $\Delta S$ we can perform the same process very very slowly such that the two systems are always at equilibrium.
thus we can write $dS=\frac{\delta q_{rev}}{T}=\frac{dq}{T}=\frac{C_vdT}{T}$
thus for system 1: $\Delta S_1=C_v ln\frac{T_f}{T_i}$
and thus $\Delta S_{TOTAL}=C_v ln\frac{T_{f}}{T_{i1}}+C_v ln\frac{T_{f}}{T_{i2}}$

My questions :
I have a feeling my reasoning is not rigorous enough and my definitions are going in circles .
and my analytic solution is wrong as the i'm basically using $\delta q$ of the irreversible process and using the same information in the assumed reversible process.
This answer is correct.  After determining the final temperature Tf, you have gone back to the initial state of each block and, to get its change in entropy, you have subjected it to an alternate reversible process in which, instead of contacting it with the other block, you have contacted it with an infinite continuous sequence of constant temperature reservoirs running from its initial temperature to its final temperature.  At least, that is what your equation implies.  This is definitely the correct way to determine the entropy change.