[bu2012]

It takes 2.22 kbar to convert ice (0 degrees C) to liquid water (0 degrees C)
-Using the density of ice at the freezing point as 0.9167 g/mL, find the isothermal compressibility of ice.
- What mass of a person would be need to liquify ice near the freezing point of water if they are skating on ice skates that have blades that are 20cm long and 0.2mm wide?

I honestly am not sure where to start. I know the equation for isothermal compressibility:
k = -1/V (dV/dP)_T but I am unsure how to incorporate the densities into this equation. Do I need to convert these densities into volume using the MW of water? Please *delete me* And I don't even know what equation I would use to start the 2nd question.

[fledarmus]

For the second part - how much pressure are the skates exerting on the ice?

[bu2012]

It doesn't give that. This is all I know. But I did figure out how to get volume, I must have missed that it gives me all of the info I need in order to solve for volume. I am not sure if I need to calculate the pressure or just use a different method of solving for the second half.

[fledarmus]

Very well - what is the definition of pressure? And how does that correspond to anything you were given in the problem?

[bu2012]

Well I know that PV=nRT but that doesn't relate to any measurements like mm or anything. There's also P= F/A since I can find the area, but I don't know the force. I think I need to incorporate p= m/V somehow in order to find the mass where p = rho. Am I getting anywhere with this?

[fledarmus]

The force would be the weight of the person - the area is the area of the bottom of the skate blades

[jusy1]

For the first part, knowing the densities of the two phases (you can assume 1 g/mL for water) and using water's molar mass you find k.

For the second part, you know the pressure needed and the area of the blades, you can get the force (and so the person's mass).