Find the Mass of Water that Freezes: Entropy Change Calculations

[loosely]

1. At 10°C, 32 g of water is mixed with 430 g of ice at -5°C. (The heat capacity of water is 4190 J/(kg * °C), that of ice is 2090 J/ (kg *°C), and the heat of fusion of water is 3.34 * 10^5 J/kg)
a) What mass of water freezes ?
b) What is the change in entropy of the system?

What I did on part (a)

ice : -5°C -> 0°C
Q = m_ice C_ice delta T
Q = (0.43 kg) (2090 J/ (kg *°C) ) (0°C-(-5°C))
Q = + 4493.5 J

water: 10°C -> 0°C
Q = m_water C_water delta T
Q = (0.032 kg) (4190 J/ (kg *°C) ) (0°C-(10°C))
Q = - 1340.8 J

Qice + Qwater = 0
but it doesn't equal to zero.

Not sure what to do next.
Qice + Qwater + Qleftover = 0
Qleftover = m_Water Lf
m_Water = Qleftover / Lf
= 3152.7 J / (3.34 * 10 ^5 J/kg)
=0.00 9439 kg which is approximately 9.4 grams
Doesn't worked out to be the answer of 11 grams.

Where did I gone wrong?
Your help is appreciated.
Thanks.

 

[SteamKing]

The heat of fusion of water is 3.34*10^3 kJ/kg.

 

[loosely]

typo error. 3.34 *10^5 J/kg or 334 * 10^3 J/kg
I corrected the above typo error on the 1st post.

 

[gneill]

Your result, 9.4 grams, looks okay to me. Could be an error in the answer key.