How do I solve this Chemistry problem on the phase changes of water?

Question

A 20.0-g sample of ice at -10°C is mixed with 100.0 g of water at 80°C. Calculate the final temperature of assuming no heat loss to the surroundings. The heat of capacities of ice and water are 2.03 and 4.18 J/g•°C respectively. The enthalpy of fusion for ice is 6.02 kJ/mol.

Answer 1

In looking at this problem, realize that temperature is just a measure of the internal energy of a substance.  When mixed together energy will move from high to low, so energy will move from the liquid water to the ice until they reach equilibrium, at which point they will have the same temperature (internal energy) and (maybe) the same state. 

 

ΔE = mCΔT

 

If both substances were the same state then ΔE would be the same for each with opposite signs.  But there is a possible state change which must be added in.  So let's calculate the total amount of ΔE in the liquid water and see if it is enough to melt the ice.

 

liquid:  ΔE = 100g x 4.18J/g°C x (0 - 80)°C

            ΔE = -33440 J

solid:   ΔE = 20g x 2.03J/g°C x (0 - -10)°C

            ΔE = 406 J

melt:    ΔE = nΔH_fus

            ΔE = 20g x 1 mol/18.02g x 6.02kJ/mol

            ΔE = 6681.5 J

So for the ice to reach 0°C and melt requires (406 + 6681.5) 7087.5 Joules.  The liquid water must release 33440 Joules of energy before it will begin to freeze, so it will remain liquid.  So now we calculate the new temperature of the original liquid and realize that it will now give energy to the original ice but now liquid until they reach the same temperature.

 

liquid:  ΔT = ΔE/mC

            ΔT = -7087.5J/(100g x 4.18J/g°C)

            ΔT = -16.956°C

            new T = 80 - 16.956 = 63.044°C

 

ΔE_ice = -ΔE_liq

20g x 4.18J/g°C x (T_f - 0)°C = -100g x 4.18J/g°C x (T_f - 63.044)°C

20T_f = -100T_f + 6304.4

120T_f = 6304.4

T_f = 52.5°C