First Law of Thermodynamics as a rate equation In a standard engineering thermodynamics textbook (Fundamentals of Thermodynamics, Sonntag, et. al, 6th ed) the First Law of Thermodynamics is differentiated with respect to time, and written as a rate equation. It is stated (p. 141): "In so doing we are departing from the strictly classical point of view, because basically classical thermodynamics deals with systems that are in equilibrium. However, since these rate equations are...used in many applications of thermodynamics, they are included in this book".
The authors have not provided a rationale. Under what conditions is it valid to write the First Law of Thermodynamics as a rate equation ?
[Equation referred to is: dU/dt=dQ/dt-dW/dt]
 A: I'm not sure exactly how you are using the phrase "first law of thermodynamics". In a strict treatment, that phrase equates to the statement
First law of thermodynamics: Energy is conserved.
One can also express the same law by:
First law of thermodynamics: The amount of work required to change the state of a closed and thermally isolated system depends solely on the initial and final states.
As I say, it is one or other of these statements in words which is correctly called the "first law of thermodynamics". 
Now I guess that the expression you are referring to in your question might be something like:
$$
dU = T dS - p dV
$$
which applies to a closed simple mechanical system. This expression is true for any situation in which the quantities are well-defined. In particular, in order for the temperature and pressure to be well-defined quantities, the state of the system will usually need to be an equilibrium state, and the small quantities are differences between properties of neighbouring equilibrium states. You can if you like divide such an equation by a small time $dt$ so as to obtain
$$
\frac{dU}{dt} = T \frac{dS}{dt} - p \frac{dV}{dt}
$$
but this would require careful interpretation, because if the movement between states takes place too fast then the system will not be passing through a sequence of thermal equilibrium states (the change will not be quasistatic) and in this case the quantities $T$ and $p$ may not be well-defined. If the movement between states is slow compared to the thermal relaxation time then the process is called quasistatic. To be precise, a quasistatic process is one in which the system moves through a sequence of equilibrium states; this is an idealization so might arguably be said to never apply exactly, but in practice relaxation to equilibrium involves exponential decay, such that the degree of approximation in assuming the idealization is acceptable for many real processes. After all, we never take everything into account when we apply science to experimental realities. The assumption of a quasistatic model may often be among the more precise aspects of a given treatment. 
A: It is virtually always valid, even locally (I.e., differentially), as long as you are dealing with small molecule ( non-polymeric) fluids with time scales for relaxation that are short compared to the time scale for the process.  In other words, if the time between collisions is short compared to the process time, the approach is valid.
A: I can’t speak to the validity of the approach, but I would think that any processes and the cycles they comprise if carried out at a finite rate must necessarily be irreversible.  Reversible processes need to be carried out quasistatically (infinitely slowly). 
Hope this helps 
