posted by Anonymous .
An ice cube of mass 10g at -10C is added to the beaker containing 150mL of water at 88.5C. Assuming no heat is lost to the surrounding; calculate the equilibrium temperature for the liquid after the ice melts.(Heat of fusion for H2O = 6.01 kJ/mol. Heat of vaporization 40.67kJ/mol. Density of H2O = 1.0g/mL.
H2O(s) = 37.1 J/mol^-1/K^-1
H2O(l) = 75.6 J/mol^-1/K^-1
H2O(g) = 33.6J/mol^-1/K^-1
(heat to move ice from -10 to zero C) + (heat to melt ice) + (heat to move ice from zero C to final T) + (loss of heat to move water at 88.5 to final T) = 0
[(mass ice*sp.h.ice x (Tfinal-Tinitial)] + [(mass ice x heat fusion)] + [(mass H2O x sp. h liq H2O x (Tfinal-Tinitial)] + [(mass H2O x sp.h. x (Tfinal-Tionitial)] = 0
Solve for Tfinal. I think the answer is approximately 77 C.