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Topic: entropy change of ice to water  (Read 5279 times)

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Offline soupastupid

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entropy change of ice to water
« on: March 08, 2012, 04:51:34 AM »
Calculate dS for the overall process of 10g of ice, initally at -15oC melting in 30g of liq water at 60oC and reaching thermal equilibrium.
assume process occurs in an isolated system

these are my thoughts:
thermal eq: dq=0

heating ice + melting ice +heating cold water - cooling hot water = 0
mCdT + mol*dH + mCdT - mCdT = 0

from here i assume that T_f is the same for cold and hot water
T_f = [  - mCdT (heating ice) - mol*dH (melting ice) + mCT_i (initial cold water) - mCT_i (initial hot water)  ]  /  C_water*(m_coldwater - m_hotwater)

the constants i used are:
Heat of fusion= 6.0 kJ/mol
Csp(s)=2.01 J/gK
Csp(l)=4.18 J/gK

The T_f i get is 133oC
i know this is clearly wrong but i cant seem to find mistake

Offline cakaro13

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Re: entropy change of ice to water
« Reply #1 on: March 27, 2012, 12:51:12 PM »

heating ice + melting ice +heating cold water - cooling hot water = 0
mCdT + mol*dH + mCdT - mCdT = 0


Your initial equation is a little bit off.

heating ice + melting ice + heating cold water(from melted ice) = -cooling water

essentially qice=-qwater

everything you should factor into q is there, you just have some sign issues I think. Take a closer look at it and keep in mind that heat energy entering a system should be positive and heat energy leaving a system should negative.

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