Why is Entropy expressed in terms of Enthalpy? Why does ${\Delta}S_{m(fusion)} = \frac{{\Delta}H_{m(fusion)}}{T}$ ?
I always thought ${\Delta}S = \frac{dQ}{T}$
In this case does it mean $dQ = {\Delta}H$ ??
Why is it so?
 A: I guess you have a special case of constant pressure, as in general you have
$$
dH = TdS + Vdp,
$$
if the number of particles is constant. Then if you assume constant pressure, i.e. $dp = 0$, as I believe was done in your problem, you get
$$
dH = TdS,
$$
and rearrange this to
$$dS = \frac{dH}{T}.$$
Hope this answered your question.
A: ${\rm d}Q$ is normally taken as heat into/out of a closed system.  ${\rm d}H$ is the change of energy in an existing system when adding something to the system.  Enthalpy takes into account the internal energy of that which you are adding, $T{\rm d}S$ and the work done in creating space in the system, $V{\rm d}P$.
It follows that where ${\rm d}P =0$ then the change in energy of the system is just that energy of the 'part' you have added $T{\rm d}S$.
A: Phase transitions are commonly studied at constant pressure and temperature. The right thermodynamic potential for these constraints (i.e. fixed pressure and temperature) is the gibbs free energy sometimes called the free enthalpy $G=H-TS$. 
Now, for a single component system, when two phases 1 and 2 are at coexistence in thermodynamic equilibrium the gibbs free energy per mol has to be the same in the two phases (I may recall here that this also makes sense because the Gibbs free energy happens to be proportional to the chemical potential of the component). This implies that
$\Delta_{12}G_m = \Delta_{12}H_m - T\Delta_{12}S_m = 0$
Overall, it turns out that one can then relate the latent heat $\Delta_{12}H$ to a change in entropy when the system changes from phase 1 to phase 2.
In the case of fusion (going from solid state to liquid state) for instance you get the formula 
$\Delta S_{m(fusion)} = \frac{\Delta H_{m(fusion)}}{T}$
A: Let me give a more physically motivated picture.  As you make a phase change we consider the two phases involved (liquid and solid). Each phase has a free energy which varies with temperature and at the phase transition temperature these free energy curves cross. This means the difference between the free energy of the solid state at T "just below" the phase transition is about the same as that of the liquid state "just above" the phase transition. Thus as we move from the solid to liquid we have very little change in free energy, but the distribution in terms of the enthalpic and entropic components change dramatically. As we increase the entropy going from the solid to liquid, we have to do so under the condition of no change in free energy, thus the enthalpy has to exactly match the entropic free energy TS.  
