Does a crushed ice cube and a regular ice cube have different potential energies?

Question

For a question on an heat transfer experiment on melting ice cubes in water and measuring the temperature decrease in water, there was a question that asked:

Suggest the effect, if any, of crushing the ice, on the final temperature of the water

I suggested that it does, since crushing the ice would require energy (endothermic) therefore we are breaking the intermolecular bond forces between the molecules, thus the crushed ice would have less potential energy due to the bonds. As a result, the water will not need to transfer as much energy compared to the regular ice cube (since less energy would be needed to break the ice bonds), and the final temperature of the water will be higher than the regular experiment (less heat loss from the water). My teacher annotated however, that the energy exchanged is the same, and crushing will have no effect.

Am I over complicating everything, or is this answer reasonable?

Answer 1

Crushing the ice will make it’s melting faster because the rate of heat transfer depends on the area through which heat passes. It will have no measurable effect on the amount of heat.

To be exact, thermodynamic properties depend not only on the amount of matter, but also on the interface area between different phases (in this case, the surface area of ice exposed to air or liquid). However, this area is important only for very small particles and droplets. For bigger chunks, like pieces of ice, the number of molecules on the surface is an infinitesimal fraction of the total number of molecules, hence no effect. Your argument is physically correct, it's just that in this case the effect is unimportant.