- #1

Robsta

- 88

- 0

## Homework Statement

A person (mass 57.8 kg) tries to penetrate a garden fork (mass 1 kg) into a melting block of ice by standing on it with all his/her weight. Assume that no heat flows from the fork to the ice. The fork has four prongs, and each prong has a square cross section of area 1 mm

^{2}. By how much must the temperature of the ice be lowered to resist penetration? [You may take the density of water and ice at 0 °C to be 1000 kg m

^{−3}and 916.7 kg m

^{−3}, respectively. The latent heat of fusion of ice is 333 × 103 J kg

^{−1}.]

## Homework Equations

The Clausius-Clapeyron Equation gives the gradient of the phase boundary between water and ice.

dp/dT = L/TΔV

## The Attempt at a Solution

I can work out what the pressure exerted on the ice is, that's fine. I just use the weight of the person and fork and divide it by the area. This pressure will melt the ice into water and penetrate if the ice is not too cold.

I can also calculate L/ΔV from information given at the end of the question. So the gradient of the phase boundary is just (L/ΔV)/T

Perhaps if I say the ice is solid at p = 10

^{5}Pa and T = 273 K and then assume that the phase boundary is linear, then I can say Δp/ΔT = dp/dT = L/TΔV and solve for T. Then I know the temperature at which the ice is on the point of melting for the given pressure. If the ice is colder than that, it won't be penetrated.

Does that sound right? It seems like I'm making a big assumption about the linearity of the phase boundary over a wide range of pressures.