Thermodynamics: is heat transferred the same as change in enthalpy? If there is a closed, isothermal system, is the heat transferred to the system equal to the change in enthalpy? The specific problem is below. 

 A: Look at the differentials of the energy and enthalpy
\begin{align*}
\mathrm{d}E 
&= T \,\mathrm{d}S - P \,\mathrm{d}V \tag{energy} \\
\mathrm{d}H
&= T \,\mathrm{d}S + V \,\mathrm{d}P \tag{enthalpy} \;.
\end{align*}
The first term in each ($T \,\mathrm{d}S$) is the differential heat. The total change in either one can be the heat, provided that the second term is zero.1
So the change in energy is all heat when $P \,\mathrm{d}V$ is zero (i.e. when $\Delta V = 0$), and the change in enthalpy is zero when $\Delta P = 0$.
Given that we do most chemistry (at least bench-top chemistry) exposed to the atmosphere so that it is at constant pressure, that latter case comes up a lot.
The question you have to answer is "Is this a constant pressure situation?".

1 OK, we actually have to have the chemical work $\mu\,\mathrm{d}N$ and the magnetic work and the electric work and other terms usually neglected in simple treatments zero as well. You're suppose to accept the mechanical work as a stand-in for all the kinds of macroscopic work that might come up when doing textbook thermo.
A: In this constant-pressure system (phase change at constant temperature), the work done on the surroundings is $W=P\Delta V$.  From the first law, for a closed system, we know that $$\Delta U=Q-W=Q-P\Delta V$$Rearranging this equation gives, $$\Delta U+P\Delta V=Q$$But, at constant pressure, $$\Delta U+P\Delta V=\Delta H$$  Therefore, for this process, $Q=\Delta H$
