Question

In: Chemistry

Two 20.0-g ice cubes at –13.0 °C are placed into 295 g of water at 25.0...

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

Solutions

Expert Solution

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


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