The issue of shapes is going to depend on shape factors. There a bunch of different equations that deal with thermal conduction, which is all that I feel like bothering with with how late it is. I'm going to ignore your swirling since that is going to invoke convection and at steady state I'm assuming the liquid and solid are static, therefore conduction will be only the heat transfer that is relevant.

At this point we have the following equation for heat transfer for the system:

q = S*k*dT,

where S is the shape factor, k is thermal conductivity of water (in units of W/(m*K)), and dT is the change in temperature from the temperature of the ice to the temperature of the liquid. For future reference, the thermal conductivity of ice is 2.4 W/(m*K) per this source that I can't get to format properly...

http://www.its.caltech.edu/~atomic/snowcrystals/ice/ice.htm

Everything is similar between the systems except the shapes, as you have specified. For the sake of this exercise I'm just going to go ahead and calculate q1 and q2 with all of the variables. I had to look up the shape factors (you don't have to memorize these things), from this source.

We can then see that the shape factor for a spherical medium is S=(2*pi*D)/(1-D/(4*z)). D is the diameter of the sphere, z in this case is the distance between the center of the sphere to the surface of the glass. As seen in the diagram, the only restriction is that this distance is greater than the radius of the sphere. I will use a .05m diameter tumbler and a .025m diameter ice sphere for this example. Therefore z=.025m, and D=.025m (R=.0125 and our restriction is satisfied). Given this, the S factor is .209440m. This then results in q1=10.611 W for the ice sphere with T1=21.1111°C (70°F) and T2=0°C (32°F).

The cube is a bit more difficult and I had to do some more research on it. All of my findings and assumption are based upon a Heat Transfer book by Anthony Mills, page 143-146 of the version in Google Books. The S value of a cube is dependent upon heat transfer occurring on it's 6 sides, 8 corners, and 12 edges. This results in a couple of equations combining to form the following equation:

S = (6*W^2)/L + (12*.54*W)+(8*.15*L), therefore for a cube where W=L=D=.025m, S = .342m

this is for a rectangle so W=L for a cube. The coefficients should remain the same in this simplified shape. Given this shape factor, q2 = 17.328 W for the cube shaped ice.

This shows that more heat is being transferred to the cube shaped ice as opposed to the sphere shaped ice. This accounts for the faster melting time of the shapes and rate differences from shape to shape.

I'm not entirely confident that I did this problem properly, it's been a while since I've dealt with Heat Transfer, it's late at night, and I'm slightly drunk. If there are any issues please let me know.