Does Converting Water to Ice Increase the Entropy of the Universe?

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SUMMARY

The conversion of 10 kg of water at 20 degrees Celsius to ice at -10 degrees Celsius results in a net increase in the entropy of the universe. The heat capacities for water (Cp_h2o = 4180 J/(kg·K)) and ice (Cp_ice = 2190 J/(kg·K)), along with the heat of fusion (l_h2o = 3.35 x 10^5 J/kg), are critical for calculating the entropy changes. The total change in entropy is derived from the system and surroundings, where the surroundings gain heat from the system during the phase transition, leading to an overall positive entropy change.

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Entropy - Help! Water --> Ice

Here is my question:
10 kg of h2o at 20 degrees C is converted to Ice at -10 degrees C by being put in contact with a reservoir at -10 degrees C. The process takes place at constant pressure. The heat capacities at constant pressure of water (Cp_h2o) and Ice (Cp_ice) are
4180 and 2190 (J)/(kg)*(K)​
respectively. The heat of fusion of water, l_h2o, is
3.35x10^5 (J)/(kg).​
Calculate the change in entropy of the universe.

I know delta S_universe = delta S_system + delta S_surroundings, but i am a little stuck, any pointers?
 
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During the fusion process , the heat would be lost to the surroundings which would lead to increase in entropy. The reservoir is at a fixed temperature (-10) and would gain heat ...therefore entropy change for surroundings = Q/T.

For the system , the change in entropy would be due to heat lost to the surroundings As the transformation in from H2o to ice ... the net entropy change would be:

dS = Ca ln (T2/T1) + Cb ln(T2/T3)

BJ
 

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