Quasi-Static Processes: Does dQ = 0?

[manhattan_project]

We know that dQ = dE + dW for any system. However, in quasi-static processes, dW = -dE. Does this mean that dQ = 0 and no heat (Q) is absorbed or given off? If so, why is that?

 

[Chestermiller]

We know that dQ = dE + dW for any system.
However, in quasi-static processes, dW = -dE.

Who says that this is the case for a quasi-static process? What about an isothermal quasi-static process, where dQ is not equal to zero?

 

[manhattan_project]

I guess I am confused when dQ = 0.
In Reif's Fundamentals of Statistical and Thermal Physics, in Section 2.9 on quasi-static processes, it states that the work dW done by the system when it remains in a particular state r is defined as dW_r = -dE_r. I understood this as saying that dW = -dE for quasi-static processes

 

[Chestermiller]

I guess I am confused when dQ = 0.
In Reif's Fundamentals of Statistical and Thermal Physics, in Section 2.9 on quasi-static processes, it states that the work dW done by the system when it remains in a particular state r is defined as dW_r = -dE_r. I understood this as saying that dW = -dE for quasi-static processes

I don’t have a copy of that book. But, it dorsn’t sound correct to say that a system remains in a particular thermodynamic state if its internal energy changes. The equation you have written is correct only for an adiabatic change.