Question

100. g of ice at 0 degrees C is added to 300.0 g of water at 60 degrees C. Assuming no transfer of heat to the surroundings, what is the temperature of the liquid water after all the ice has melted and equilibrium is reached?

Specific Heat (ice)= 2.10 J/g C

Specific Heat (water)= 4.18 J/g C

Heat of fusion = 333 J/g

Heat of vaporization= 2258 J/g

Answer 1

Heat of fusion of water = 333 J/g
100 g ice x 333 J/g = 33300 J (energy to melt 100 g ice at 0 deg C)
300.0 g water at 60 deg C must supply the heat energy to melt the ice
33300 J = 300 g x 4.184 J/g C x (60 - Tf)
33300 = 75312 - 1255.2 Tf
Tf = 33.47 deg C
Now you have 300 g H₂O at 33.47 deg C and 100 g H₂O at 0 deg C
300 g x 4.184 J/g C x (33.47 - Tf) = 100 g x 4.184 J/g C x (Tf - 0)
Cancel 4.184 and get-
41923.7 - 1255.2 Tf = 418.4 Tf
41923.7 = 1673.6 Tf
Tf = 25.1 deg C
Final temperature of 400 g H₂O = 25 deg C