Entropy change of ice-water mixture at 273K

[lowerlowerhk]

Homework Statement

A mixture of 1.773kg of water and 227g of ice is in an initial equilibrium state at 273K, in a reversible process, brought to a second equilibrium state where the water-ice ratio,by mass, is 1:1 at 273K. Calculate the entropy change of the system during the process.

Homework Equations

$$E=mL$$
$$\Delta S=\int\limits_{i}^{f}\frac{dQ}{T}$$
specific heat of fusion of water = 333000J/kgK

The Attempt at a Solution

Since final mass ratio is 1:1 the final mass is both ice and water is
m=(1.733+0.277)/2=1.025kg, which means 0.748 kg of water is turned to ice. Some water gives away energy and freezes to ice.
$$\Delta E=-0.748(333000)=-249084J$$
$$\Delta S=-249084/273=-912J/K$$
But that is not the answer.

 

[Andrew Mason]

Since final mass ratio is 1:1 the final mass is both ice and water is
m=(1.773+0.277)/2=1.025kg, which means 0.748 kg of water is turned to ice. Some water gives away energy and freezes to ice.
$$\Delta E=-0.748(333000)=-249084J$$
$$\Delta S=-249084/273=-912J/K$$
But that is not the answer.

Your method is correct. Use 334 J/g as the heat of fusion for water. I get -915 J/K

AM