Carnot engine basics? A line in my textbook says.. ' if we employ any other process that is not adiabatic,  say an isochoric process,  to take the system from one temperature to another,  we shall need a series of reservoirs in the temperature range T1 to T2 to ensure that at each stage the process is quasi static ' 
What does this line mean?? 
I think it means that as volume is constant,  the pressure would be directly proportional to temperature,  and as pressure decreases,  (in the first path)  we'll have to decrease the temperature slowly to... 
But I'm not sure..  Can someone explain?
Thanks 
 A: 
' if we employ any other process that is not adiabatic, say an isochoric process, to take the system from one temperature to another, we shall need a series of reservoirs in the temperature range $T_1$ to $T_2$ to ensure that at each stage the process is quasi static '

This means if we want to have a reversible process between two temperatures, then we should remove all irreversibilities. One of them is heat transferring due to definite temperature difference. So, if we want a reversible process, then we need a series of reservoirs in the temperature range $T_1$ to $T_2$ and temperature difference between each reservoir and next one should be $\mathrm dT$.
A: There are two components in this discussion. When we say two systems are in equilibrium, the two systems have the same temperature. When a process is a reversible process, the system is changing in a sequence of equilibrium states.
Intuitively speaking, if we let the system change by 1 degree C from temperature A to temperature B and we want the change to be reversible, we cannot make this perfectly. What we can do may be to make a million steps, for each step, the temperature change is 1e-6 degrees. Then we can assume for each step, because the temperature change is so small, it is a reversible process. After 1 million steps, the system completes temperature change by 1 degrees. We say this can be approximated as a reversible process with strong confidence. 
The reservoir is large, so giving out heat to the system will not change its temperature. Again this is an approximation. 
Overall, you can understand there is no perfect reversible process and entropy will increase. A reversible process analysis gives us a bottom line conclusion. 
