specific enthalpy change during phase change consider 1 kg of water in a sealed cylinder and piston, whic exert 1 atm pressure on the system. Also water is saturated liquid at T=100$^\circ$ C. Then we give the system heat, so part of the system change to saturate vapor, if we assume the process reversible, then we have:
$$du=\delta q - Pdv \rightarrow du+Pdv=\delta q \rightarrow dh=\delta q$$
and because $\delta q>0$ then  $dh>0$.  But we know that enthalpy is a state function and, $$dh=(\frac{\partial h}{\partial P})_TdP+(\frac{\partial h}{\partial T})_PdT$$ and because $dP=0$ and $dT=0$, during the process, so $dh=0$. 
How this paradox could be explained? any response would be appreciated.
 A: The problem is that the expression
$$dU = \delta Q - pdV$$
is wrong for the case of systems that consist of more than one type of substance that can transform into the other kind of substance. For instance, it doesn't work if there exists more than one phase, or more than one chemical species, etc.
One way to see this is exactly what you have just done. There is clearly an inconsistency. As user31748 noted, there is a latent heat of vaporization. That is, the energy put into the system as heat in this case does not go into changing the temperature. Rather, it goes into converting a chunk of the liquid into vapor.
For the saturated system you're considering, we have the following:


*

*Thermal equilibrium: $T_{\textrm{v}} = T_{\textrm{l}}$ (v for vapor, l for liquid).

*Mechanical equilibrium: $p_{\textrm{v}} = p_{\textrm{l}}$.

*Phase equilibrium: $\mu_{\textrm{v}} = \mu_{\textrm{l}}$.
Here, $\mu$ is the chemical potential, and it, roughly speaking, governs how much of each of the two phases we have. The correct thermodynamic relation is then
$$dU = \delta Q - pdV + \mu_{\textrm{v}} dN_{\textrm{v}} + \mu_{\textrm{l}}dN_{\textrm{l}}$$
Roughly speaking (again), the energy that goes in as heat goes into decreasing $N_{\textrm{l}}$ and increasing $N_{\textrm{v}}$. The amount of energy required to change a unit mass of liquid into vapor (if we're on the saturated vapor pressure curve) at constant pressure is called the latent heat.
For the enthalpy, we have
$$dH = \delta Q + vdP + \mu_{\textrm{v}} dN_{\textrm{v}} + \mu_{\textrm{l}}dN_{\textrm{l}}$$
and so you can see that in this case $dH$ is not just $\delta Q$. 
Note. In general, $U$ is not a convenient quantity for dealing with multiple substances in phase equilibrium. For those types of systems, you want to look at something like the Gibbs free energy instead.
