Does freezing point depression also affect the speed at which something melts? If you have a pure water frozen, then it should begin to melt if you apply 33°F air to it.
If you have impure water that's frozen, it's melting temperature might have dropped to 30°F. 
If you apply 33°F air to it at the same time as the pure ice cube, will they both melt at the same rate or will the impure cube melt faster since it's melting point is lower?
Restated, does freezing point depression affect the entire amount of energy needed to melt it (eg. making it melt faster), or does it only seem to be a noticeable phenomena only at the melting point only? 
 A: For simplicity, let us assume the main heat flow contribution is from heat conduction.
In this case the heat flow from the environment to the ice cube will be proportional to the difference of temperature, i. e. $\dot{Q} \propto \Delta T$.
If the ice cubes are assumed to be equal in all relevant quantities (heat capacity, geometry, mass, heat of fusion, etc...) except melting point, then it is clear that the ice cube with the lower melting point will melt faster, as there is more heat transfer.

Concerning the second part of your question: does freezing point depression affect the entire amount of energy needed to melt it? - in this scenario I assumed the amount of energy is the same, but one ice cube melts faster nonetheless... in reality, it is possible that by adding impurities to the water changes the latent heat of fusion.
A: *

*The total heat energy needed to melt an object is given by $$q=\Delta H_f m$$ where $\Delta H_f$ is the heat of fusion and $m$ is the mass of the object in question.

*By taking the time derivative of the above equation and comparing it to the Stefan-Boltzmann law (concerning radiation) and convective transfer (also concerning conduction) we have (assuming the effect of the melted water is negligible)
$$\frac{dq}{dt}=\Delta H_f \frac{dm}{dt}=\rho\Delta H_f \frac{dV}{dt}=\left[\sigma\left(\epsilon_rT^4_{r}-\epsilon_{ice}T^4_{ice}\right)+h\left(T_{ice}-T_{air}\right)\right]A$$
Where $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the emmisivity of the substance, and the subscript $r$ denotes any radiative source (the air and effects from the surroundings).
For a cube we have $A=V^{2/3}$ so $$\frac{dV}{dt}=\frac{\sigma\left(\epsilon_rT^4_{r}-\epsilon_{ice}T^4_{ice}\right)+h\left(T_{ice}-T_{air}\right)}{\rho\Delta H_f}V^{2/3}$$
which may be solved to give
$$V(t)=V_0-\left[\frac{\sigma\left(\epsilon_rT^4_{r}-\epsilon_{ice}T^4_{ice}\right)+h\left(T_{ice}-T_{air}\right)}{3\rho\Delta H_f}t\right]^3$$
In other words, the time taken to melt completely is given by
$$t=\frac{3\rho\Delta H_f\sqrt[3]{V}}{\sigma\left(\epsilon_rT^4_{r}-\epsilon_{ice}T^4_{ice}\right)+h\left(T_{ice}-T_{air}\right)}$$
Unfortunately I am not familiar enough with the specific values of the given constants for many materials that would make the impure ice so I will not give a "rule of thumb" for which block will melt faster; however, the values may be looked up for a specific substance then plugged in. 
To be accurate it is important to change the values of $\rho$, $\Delta H_f$, $\epsilon$, and $h$ to account for the impure solution. If those changes are small, then the intuition that the block with a lower melting point will melt faster is correct.
A: As a starting point, assume that the freezing point of the impure water is lower than that of pure water due to colligative properties. Here is one discussion of colligative properties. This statement defines the thermodynamics or preferred direction of the process.
As a starting point, assume that we start with the systems of pure water and impure water both at their freezing point temperatures and both having the same shape (external area $A$ in units of m$^2$). The impure water will be colder than the pure water. Assume the surrounding air for both cases is at the same temperature just above the temperature of pure water.
The most significant heat transfer from the hotter air to the solid pure water or impure water will be by convection. Radiation is to be neglected to first order because the temperatures of system and surroundings are nearly the same and because the emissivities of air and ice are low. Conduction is THROUGH a material not TO a material and requires a temperature gradient that is fundamentally beyond the scope of the first order analysis to define precisely. So, conduction is also to be neglected to first order.
The rate of heat transfer by convection $\dot{q}$ (W) is proportional to temperature difference as $\dot{q} = h A (T_{surr} - T_{sys})$ with $h$ as the convection coefficient (W/m$^2\ ^o$C).
All else the same, the rate of heat transfer to the impure solid will be larger than that to the pure water because $(T_{surr} - T_{sys})$ will be larger for the former case.
Melting (fusion) is a kinetic process. The mass specific rate of fusion $\tilde{r}_{fus}$ (kg/s) is controlled by the rate of heat transfer and the specific fusion enthalpy $\Delta_{fus} \tilde{H}$ (J/kg).
$$\tilde{r}_{fus} = \frac{\dot{q}}{\Delta_{fus} \tilde{H}}$$
Colligative solutions may have different thermodynamic properties than their pure solvents. However, as stated here, we can to first order assume that the fusion enthalpy is the same for pure and impure water.
Conclusion: Under the given assumptions and with all else the same, the rate of fusion (melting) will be greater for impure water than for pure water.
The comparative differences in melting rates will otherwise depend primarily on the temperatures that we choose for the two systems and their surroundings.
A: 
Restated, does freezing point depression affect the entire amount of
  energy needed to melt it (eg. making it melt faster), or does it only
  seem to be a noticeable phenomena only at the melting point only?

The entire energy needed to melt the pure cube will be greater than that required to melt the impure cube, and the impure cube will melt faster (start to melt sooner and, when it does melt, at a faster rate). Consider the following:
We start with the following assumptions.


*

*Both cubes have exactly the same mass, surface area, geometry, etc.  That is their physical properties are identical and how they physically interact with the surrounding air is identical.

*The specific heats $C$ of both the pure and impure cubes are the same.  

*The latent heats of fusion $h_f$ are the same for both cubes.

*The overall heat transfer coefficients, $h_C$ are the same for both cubes.

*The air is considered a 33 C thermal reservoir, i.e., its bulk  temperature is constant during the heating of the cubes.
Regarding assumptions 2-4 I was not able to locate any data differentiating these properties based on the purity of water.  These assumptions may not be correct, but I doubt any differences would be significant enough to alter the conclusions.
We start out with both cubes being at any arbitrary temperature less than the lower of the two melting temperatures, that is,  less than 30 C.  Up to the point where the temperature of the impure cube reaches 30 C, the rate of heat transfer and the total heat transfer to each cube is the same.
When the two cubes reach 30 C, phase transition begins for the impure cube. The amount of heat required to melt the impure cube, where $h$ is the latent heat of fusion, is 
$$Q_{impure}=mh_f$$
The rate at which heat is added is 
$$\dot Q_{impure}=h_{c}A(33^{o} C  - 30^{o} C)$$
Where $h_c$ is the overall heat transfer coefficient of the ice/water mixture. 
During the time melting of the impure cube begins, the air is still raising the temperature of the pure cube. The heat required to bring the pure cube to its melting point of 32 C, is
$$Q_{pure}=mC_{ice}(32-30)^{0}C$$
This heat is over and above that which was required to bring the pure and impure cubes to the melting temperature of the impure cube. This means that, overall, more heat was required to melt the pure cube than the impure cube.
Once the pure cube reaches its melting temperature, the heat required to convert it from ice to water is 
$$Q_{pure}=mh_f$$
Since the latent heat is assumed the same as the impure cube, this heat is the same as needed to melt the impure cube. Or $Q_{pure}=Q_{impure}$. 
The rate at which heat is added to convert the pure cube ice to water is 
$$\dot Q_{pure}=h_{c}A(33^{o} C  - 32^{o} C)$$
Where $h_c$ is the overall heat transfer coefficient of the ice/water mixture, assumed to be the same for the pure and impure ice/water phase. 
Under the assumption that $h_c$ is the same for the pure and impure cube, the heat transfer rate during the phase transition for the pure cube will be less than the impure cube because of the smaller temperature difference of 1 C for the pure ice vs. 3 C for the impure ice. 
Given the above, if the assumptions are correct, I would conclude


*

*More heat is required to melt the pure ice than the impure ice, the difference being the additional heat needed to bring the pure ice to its melting point compared to the overall heat required to melt the impure ice. This additional heat is $$Q_{pure}=mC_{ice}(32-30)^{0}C$$

*The impure ice will melt faster/sooner than the pure ice for two reasons: (A) the impure ice will begin to melt before the pure ice since its melting temperature is reached first and (B) during the transition phases (ice to water), the impure ice will melt faster than the pure ice because the temperature difference between the air and the impure ice during melting of 3 C, was greater than the temperature difference between the air and the pure ice during melting of 1 C.
Hope this helps.  
