Question

by Two 20 0g ice cubes at-13.0 ℃ are placed into 205 g of water at 25.0 ℃ Assuming no transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts heat capacity of H20(s) 37.7J(mol K) heat capacity of H20() 75.3 J(mol K) enthalpy of fusion of H2O 6.01 kJ/mol

Answer 1

Heat (Q) consumed by the ice to become water is the same heat absorbed by the water to decrease it temperature.

Q= n x DT x Cp

n= moles

mol ice= 20g/18g/mol= 1.11 mol

mol water= 205g/18g/mol= 11.39 mol

For the water:

Q= 11.39 mol x (Tf - 25) x 75.3 J/mol.K

For the ice, we have three processes, one involving the change of temperature (from -13 to 0, so DT=13), another involving the change of state to liquid water, and finally one involving the increase of temperature of the ice, now water, to the final T:

Q1= 1.11 mol x 13 x 37.7 J/mol.K

Q2= 1.11 mol x 6010J/mol

Q3= 1.11mol x (Tf-0) x 75.3 J/mol.K

Total Q ice= 1.11 mol x 13 x 37.7 J/mol.K + 1.11 mol x 6010J/mol + 1.11mol x Tf x 75.3 J/mol.K

Now, we have said that the heat released by the ice is gained by the water so:

Total Q ice= total Q water

1.11 mol x 13 x 37.7 J/mol.K+1.11 mol x 6010J/mol+1.11mol x Tf x 75.3 J/mol.K=11.39 mol x (25 -Tf) x 75.3 J/mol.K

Tf= 15.1 ºC