Assumption of specific enthalpy

Given an adiabat turbine with state 1 fluid data as:

$$h_1 = a$$
$$p_1 = b$$
$$t_1 = c$$
and state 2 data fluid data as:

$$h_2 = unknown$$
$$p_2 = e << b$$
$$t_2 = f < c$$
$$h_2′ = g$$
$$h_2” = h$$

With kinetic and potential energy small enough to be ignored, a steam content of $x_2 = 0.965$

What is the technical work $w_{t,12}$

With the equation:

$$h_2 = h_2′ + x_2(h_2”-h_2′)$$

The specific enthalpy of $h_2$ can be easily solved. Now taking the first Thermo D. law, and using an energy balance, the technical work can be solved via:

$$W_{t,12} = dot{m_{in}}(h_1 + frac{c_{in}^2}{2}) – dot{m_{out}}(h_2 + frac{c_{out}^2}{2})$$

Whereby $dot{m_{in}} = dot{m_{out}} = dot{m}$

Atleast, this is what I had thought though I am missing $c_{in} text{ and } c_{out}$, when checking the solution, it is suggest the specific work is solved "simply"

$$w_{t,12} = h_1 – h_2$$

Whereby the mass velocity and fluid velocity are unneeded.

My assumption in this solution, is that the author assumes $c_{in} text{ and } c_{out}$ are equivalent thus equal zero in the equation, but I fail to see how $W_{t,12}$ can be divided by the mass velocity $dot{m}$.

What am I missing compared to my assumed solution and the authors, where should I have understood that $dot{m}$ is not required.