Will a block of ice that is 12"x12"x12" melt slower than a block sized 20"x20"x20"

So in the end, the melt rates will be proportional to the surface areas, which start out in the ratio
(20:12)² = 25:9
but as each block shrinks, so does its surface area.

In this simplified model, each block will lose side-length at constant and equal rates, making melt-time proportional to L.

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The rate of change in heat dq = Cp m (6l^2) dT for the smaller block of mass m. Note the surface area is a = 6l^2 where l = 12 " at the start. And,

dQ = Cp M (6L^2) dT for the big block. dT = 29 deg C, where the ice temp is 0 deg C. L = 20 " at the start.

So we have dQ/dq = M/m * (L/l)^2 and dQ = dq (M/m)*(L/l)^2 and there you are.

As M/m > 1 and L /l > 1, dQ > dq and the big block's heat transfer will be greater for a given interval of time dt. That is, dQ/dt > dq/dt. The big block will melt faster, but probably take longer to melt completely as there is M > m more ice to melt. QED.

No. The smaller block melts faster.

Just imagine a giant block 1mile x 1 mile x 1mile and a tiny block 1mm x 1mm x 1mm.
Which do you think takes longer to melt? The problems are equivalent.

All other conditions being equal, the larger block will take longer to melt.