Verified answer

This is an energy balance problem. With a perfectly insulated container, all you're doing is adding up the energy at the first state and setting it equal to the second state where all of your water is at a constant temperature.

Judging by the relative masses and temperatures of the ice and water, the end product will be in the liquid phase.

We can simply solve for the final temperature, T, by setting the heat lost from water to equal to the heat gained by the ice

Heat gained by ice = heat to bring ice to 0 C + heat to melt ice + heat to bring cold water to final temperature

Heat lost by water = heat to bring water to final temperature

First, calculate the heat required to bring the ice to 0 degrees C: q = m*cp*dT = (21.85 g ice/(18 g/mol water))*(37.5 J/K/mol)*(14.8 K) = 673.7 J

Then, the additional heat to melt it is q = m*dH = (21.85 g ice/(18 g/mol water))*(6010 J/mol) = 7295.5 J

{So the total energy to bring the ice to liquid water at 0 C is 7969.2 J

As a check of the assumption that the end product is liquid, subtract that energy from the liquid water:

q = -7969.2 J = m*cp*dT = (65.08 g / 18 g/mol)*(75.3 J/mol/K)*dT; dT = -29.27 K : it will be liquid}

For a final temperature of Tf, the heat to raise the cold water is q = m*cp*dT = (21.85 g ice/(18 g/mol water))*(75.3 J/K/mol)*(Tf - 0)

The heat lost by the warm water, in total, is q = m*cp*dT = (65.08 g / 18 g/mol)*(75.3 J/mol/K)*(93.0 - Tf)

Final equation is (21.85 g ice/(18 g/mol water))*(37.5 J/K/mol)*(14.8 K) +(21.85 g ice/(18 g/mol water))*(6010 J/mol) + (21.85 g ice/(18 g/mol water))*(75.3 J/K/mol)*(Tf - 0) = (65.08 g / 18 g/mol)*(75.3 J/mol/K)*(93.0 - Tf)

Substituting from above, 7969.2 J + 91.4*Tf = 25319.4 J - 272.25*Tf

Tf = 47.7 C

State 1) h = sum(mass*specific heat*temperature) = (21.85 g ice/(18 g/mol water))*(37.5 J/K/mol)*(273.15-14.8 K) + (65.08 g water/(18 g/mol water))*(75.3 J/K/mol)*(273.15 + 93.0)

h_1 = 11760 J + 99685 J =