Question

A mixture of equal masses of water and ice(m = mw = mice = 1 kg) is contained in a thermally insulated cylindrical vessel under a light piston. The pressure on the piston is slowly increased from the initial value p0 =105 Pa to p1 = 2.5 x 106 Pa. The specific heats of water and ice are cw = 4.2 kJ/(kg . K) and cice = 2.1 kJ/(kg . K), the latent heat of fusion of ice is λ = 340 kJ/kg, and the density of ice is ρice = 0.9ρw (where ρw is the density of water).

Determine the mass Δm of ice which melts in the process and the work A done by an external force if it is known that the pressure required to decrease the fusion temperature of ice by 1°C is p = 14 x 106 Pa, while the pressure required to reduce the volume of a certain mass of water by 1% is p' = 20 x 106 Pa.

(1) Solve the problem, assuming that water and ice are incompressible.

(2) Estimate the correction for the compressibility, assuming that the compressibility of ice is equal to half that value for water.

Answer 1

(1) Assuming that water and ice are incompressible, we can find the decrease in the temperature of the mixture as a result of the increase in the external pressure:

Such a small change in temperature indicates that only a small mass of ice will melt, i.e. Δm << mice.

We write the energy conservation law:

Let us estimate the work A done by the external force. The change in the volume of the mixture as a result of melting ice of mass Δm is

We have taken into account the fact that the density of water decreases as a result of freezing by about 10%, i.e.,

Therefore, we obtain an estimate

The amount of heat ΔQ required fot heating the mass m of ice and the same mass of water by ΔT is

Since A << ΔQ,. we can assume that λ Δm = ΔQ, whence

The change in volume as a result of melting ice of this mass is

Considering that for a slow increase in pressure the change in the volume ΔV, is proportional to that in the pressure Δp, we can find the work clone by the external force:

(2) We now take into account the compressibility of water and ice. The change in the volume of water and ice will be

where V0w, =10-3 m3 and V0ice = 1.1 x 10-3 m3 are the initial volumes of water and ice. The work A done by the external force to compress the mixture is

The total work of the external force is

Obviously, since we again have Atot << ΔQ, the mass of the ice that has melted will be the same as in case (1).