Constancy of temperature in a closed system Consider a thermodynamical classical isolated system, made by a small subsystem  and a way large reservoir. The two could exchange heat.
Usually in such situation we say that the system is closed or is a $(N,V,T)$ system.
What perplexes me is that for the $(N,V,T)$ system, $N$ and $V$ are constant even if the system+reservoir are not yet at equlibrium. But that's not true for the temperature $T$.
I know that the larger reservoir imposes its temperature to the system. But this needs a step more.
Please, where I am wrong?
 A: The thermodynamic parameters $N, V, T$ are all mentioned at the equilibrium states only. So, when your system is in contact with a heat bath, there causes an exchange of energy between the system. According to the definition of a heat reservoir, its temperature is not affected by any slight exchange of heat. Hence the system will eventually come in thermal equilibrium at a temperature $T$ of the heat bath.   
The system is closed in this sense means that the combined system of system+reservoir is isolated from the rest of the universe.
A: It is a matter of boundary conditions: you are considering a subsystem with walls which are diathermal (they allow heat exchange) but impermeable (they don't allow matter exchange) and rigid (they don't allow changes in volume). 
Let $S$ be the large system and $s$ be the small subsystem. With this choice of boundary conditions, the generic initial state of $s$ will be $(N_s^i, V_s^i, T_s^i)$ and that of $S$ will be $(N_S, V_S, T_S)$. 
After thermal equilibrium is reached, the state of $s$ will be $(N_s^f=N_s^i, V_s^f=V_s^i , T_s^f = T_S \neq T_s^i)$ while the state of $S$ will be unchanged (we are here assuming that $S$ is so large compared to $s$ that its temperature $T_S$ remains unchanged during the reaching of thermal equilibrium).
You could also consider different boundary conditions: in this case you could have $N_s^i \neq N_s^f$ and/or $V_s^i \neq V_s^f$.
