Calculating Enthalpy and Internal Energy of Melting Ice

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The densities of ice and water at 0*C are 0.9168 and 0.9998g/cm3. The latent heat of ice is 80cal/g. What is "delta"H and "delta"U (enthalpy and internal energy, respectively) when 1kg of ice is melted?

I'm really confused on which formulas to use and if the densities are even relevant to this question. One of the formulas I was looking at are "delta"U = m x Cv x "delta"T (but because "delta"T is zero, this would make the variation of internal energy 0, which doesn't make much sense, does it?)

Any help would be greatly appreciated! Thanks in advance.

EricADDITION:
I just read the FAQ sticky note for this section and I'm sorry for not including my attempts thus far. I'm new to the forum and wasn't completely aware of what's expected. So here goes:

One thing I tried was using the formula described above but it doesn't seem to make sense that the internal energy wouldn't change.
Another formula I tried was the following:
H = U + pV

Where V = 0.9168g/cm3 / 1000g = 9.168 x 10^-4 cm3
But now I still don't have V and the question doesn't specify the pressure (I could assume we're at atmospheric pressure, but I'm still no further ahead)
I feel like the answer is really simple and I'm just making it complicated. Any insight would be great!
 
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The key is H = U + pV, and from that ΔH = ΔU + Δ(pV), which given p is constant becomes, ΔH = ΔU + pΔV. Unless otherwise stated, one would assume ambient or atmospheric pressure (~1 atm).

From the problem statement, "latent heat of ice is 80cal/g" or heat of fusion, or change in enthalpy from ice to water requires 80 cal/g. One is also given the mass.

BTW, volume = m/ρ. One can find the change of volume of ice to volume of water. The densities are given for both.