A friend recently asked “what is the difference between an ice sphere and one (large) ice cube?”

Well, let’s do some math. First we have to make some simplifying assumptions, which honestly really would need to be verified. But let’s not let lack of evidence get in the way of doing some fun math!

The idea behind the sphere of ice is that it cools your drink while presenting the minimum surface area, and (presumably) therefore minimizing drink dilution. A sphere has the minimal surface area for a given volume, which is why soap bubbles generally form spheres.

Let’s assume that the total cooling capacity of an ice cube is related to its volume. If we have a cube with side length **a**, then the volume of that cube will be:

V_{c} = **a**^{3}

Similarly, if we have a sphere of radius **r**, its volume will be:

V_{s} = (4/3)π**r**^{3}

Since we want the same cooling capacity in both the cube and the sphere, we need to set the two volumes equal, and then we can determine what the cube side-size **a** would be for the same volume as a sphere of radius **r**:

V_{c} = V_{s}

**a**^{3} = (4/3)π**r**^{3}

**a** = [(4/3)π**r**^{3}]^{(1/3)}

**[1] a** = [(4/3)π]^{(1/3)}•**r**

Equation [1] tells us what size cube we need (side length **a**) in terms of a sphere of radius **r**, so that the volume of the cube and the volume of the sphere are the same.

Now let’s look at surface area. Surface area of a sphere of radius **r**:

S_{s} = 4π**r**^{2}

Surface area of a cube of side length **a**:

S_{c} = 6**a**^{2}

which, using equation [1] to get that in terms of r becomes:

S_{c} = 6[(4/3)π]^{(2/3)}**r**^{2}

and therefore the ratio of the surface area of the cube, S_{c}, to the sphere, S_{s}, becomes:

S_{c}/S_{s} =6[(4/3)π]^{(2/3)}**r**^{2} / 4π**r**^{2}

=6[(4/3)π]^{(2/3)} / 4π

=3[(4/3)π]^{(2/3)} / 2π

= 1.24 (approx)

So the cube with the same total volume as a sphere will have about 24% more surface area. Still way better than having multiple small cubes of ice in your drink. Enjoy!