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.

 

[rwisz]

Your calculations are correct, but there are a few things to consider. First, when you are calculating the heat absorbed by the ice, you need to take into account the fact that it is undergoing a phase change from solid to liquid. So the equation should be Q = m_ice * Lf + m_ice * C_ice * delta T. This takes into account the heat required for the phase change as well as the change in temperature.

Second, when you are calculating the heat released by the water, you need to use the negative value for the heat of fusion (since it is being released). So the equation should be Q = -m_water * Lf + m_water * C_water * delta T.

When you plug in the correct values, you should get a mass of water of approximately 11 grams, which is the answer given in the question. So it seems that you made a small calculation error somewhere. Double check your calculations and make sure you are using the correct signs for the heat of fusion and the heat released by the water.