Find the Mass of Water that Freezes: Entropy Change Calculations

AI Thread Summary
The discussion revolves around calculating the mass of water that freezes when 32 g of water at 10°C is mixed with 430 g of ice at -5°C. The user initially calculates the heat exchanges for both the ice and water but finds discrepancies when trying to balance the energy equations. After correcting a typo regarding the heat of fusion, the user arrives at a mass of approximately 9.4 grams of water that freezes, which does not match the expected answer of 11 grams. Other participants suggest that the user's calculations appear correct and imply the answer key might be erroneous. The conversation highlights the complexities of entropy change calculations in thermodynamic systems.
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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.
 
Last edited:
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The heat of fusion of water is 3.34*10^3 kJ/kg.
 
typo error. 3.34 *10^5 J/kg or 334 * 10^3 J/kg
I corrected the above typo error on the 1st post.
 
Your result, 9.4 grams, looks okay to me. Could be an error in the answer key.
 
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