In: Chemistry
Two 20.0-g ice cubes at –13.0 °C are placed into 295 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
Heat capacity of H2O(S) is 37.7 J/(mol X K); Heat Capacity of H2O(l) is 75.3 J/(mol X K); Emthalpy of Fusion of H2O is 6.01 kJ/mol
ice mass = 2 x 20 = 40 g ,
ice moles = mass of ice / molar mass of H2O = 40 /18.01528 = 2.220337
Heat required to increase ice temp from -13 to 0 = Molar heat capacity of ice x temp change x moles
= 37.7 x 2.220337 x ( 13) = 1088.2 J
Heat absorbed by ice to melt at 0C = enthalphy of fusion x moles 6.01 x 2.220337= 13.344 KJ = 13344 J
Now let final temp be T
Then heat absorbed by this water at 0C to rise to T = heat capacity of H2O liquid x moles x temp change
= 75.3 x 2.220337 x T
moles of liquid water = 295 /18.01528 = 16.375
heat released by water at 25 to become final temp T = 75.3 x 16.375 x ( 25-T)
now equating heat released by H2O at 25 = heat absorbed for changing ice temp from -13 to 0
+ heat required to melt ice at 0C + heat required to rise to temp T
75.3 x 16.375 x ( 25-T) = 1088.2 + 13344 + ( 75.3 x 2.220337 x T)
30825.9375 -1233.0375 T = 14432.2 + 167.1914 T
T = 11.7 C