Two 20.0-g ice cubes at - 11.0 °C are placed into 205 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts.

Heat capacity of H2O (l) = 75.3J / (mol*K)

Heat capacity of H2O (s) = 37.7J / (mol*K)

Enthalpy of Fusion of H2O=6.01kJ/mol

7.0 °C

We have 40.0 g of ice at - 11.0°C. Since our heat capacity constants are per mole, let's convert all our measurements to moles. Start by calculating the molar mass of H2O

Atomic weight oxygen = 15.999

Atomic weight hydrogen = 1.00794

Molar mass hydrogen = 15.999 + 2 * 1.00794 = 18.01488 g/mol

Moles ice = 40.0g / 18.01488 g/mol = 2.220386703 mol

Moles water = 205 g / 18.01488 g/mol = 11.37948185 mol

Let's first calculate how much energy needs to be absorbed from the environment to get that ice to a temperature of 0.0°C.

2.220386703 mol * 11 K * 37.7 J / (mol*K) = 920.7943655 J

Now let's melt that ice.

2.220386703 mol * 6010 J/mol = 13344.52408 J

So the total energy to melt the ice is:

13344.52408 J + 920.7943655 J = 14265.31845 J

Now calculate the amount of mole degrees that had to be extracted to get that much energy.

-14265.31845 J / 75.3 J / (mol*K) = - 189.4464601 mol*K

Divide by the amount of water being cooled.

-189.4464601 mol*K / 11.37948185 mol = - 16.64807437 K

So the temperature dropped by a bit under 17 degrees K. So

25.0 °C + - 16.64807437 °C = 8.351925631 °C

We now effectively have 2 masses of water. A 40 gram mass at 0 °C and a 205 gram mass at 8.351925631 °C. Obviously that isn't stable, so we need to bring both masses to the same average temperature. So let's do a weighted average:

(40 * 0 + 205*8.351925631) / (40+205)

= (0 + 1712.144754) / 245

= 6.988345936

Rounding to 1 decimal place gives a final temperature of 7.0 °C