# Melting ice inside water by increasing pressure

An ice cube is floating in water inside a container and the system is in thermal equilibrium. And let say we somehow immerse the ice cube slowly into deeper region inside the container (we may pull a string which is attached to bottom of the ice cube). From my understanding since the pressure on the ice cube increases, it starts to melt in order to decrease the volume it encloses.

My question is that, does the energy which melts the ice come from the work done by pressure over the change of volume (P.dV) or does it come from surrounding water as heat energy so that the water also freezes at the same time? Or do they both mean the same thing?

Of course when we pull the ice into water , we must have been doing external work on the system. I'd be glad if you help me figure these things out.

At the most basic level, the ice which is at a $$P$$ too high for its $$T$$ wants to melt and will just do so spontaneously without any regard for what its neighbours are up to. As each tiny parcel of ice does this, it will shrink (causing the surrounding water to do move inwards, extracting some of the stored gravitational potential energy to do work on the melting ice) but it will also cool (since the work done on the parcel will not provide enough energy to melt it). Immediately after each parcel of ice melts, it will be colder than the surrounding water, and so heat will flow from the surrounding water to the newly-melted water.
How this continues depends on the constraints on your system. If the system is kept at constant $$T$$ (e.g. by keeping it in a water bath) then the slightly-colder water surrounding the newly-melted water will ultimately be heated back to the target $$T$$ by heat transfer from the surroundings. If the system is insulated, then the slightly-colder water surrounding the newly-melted water will cool its surroundings and so on. This could progress to a point where liquid water on the surface is cooled below its freezing point and a layer of ice forms on the surface.
Note that I have assumed that the system is closed (material can't enter or leave) and at constant $$P$$. If the system is at constant $$V$$ (e.g. in a rigid container with no room for air) then things get a bit weird because the overall liquid-to-solid ratio can't change (as that would change $$V$$).