Help Solving Entropy Problem: Part A-D

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The discussion revolves around solving a thermodynamics problem related to entropy changes during the melting of an ice cube. Parts A and B have been attempted, with the calculated entropy change for the ice cube being zero and the change for the melted water rising from 0 to 25 degrees Celsius being approximately 11.0 J/K. The user seeks assistance specifically for parts C and D, which involve calculating the entropy change of the kitchen and the net entropy change of the universe. The urgency is emphasized as the assignment is due soon. The overall tone reflects frustration at the lack of support received.
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I need help finishing this problem. I did part A and B but don’t know how to do C and D. I don’t know if A and B are correct.

Problem: An ice cube (mass 30g) at 0 degrees Celsius left sitting on the kitchen table where it gradually melts. The temperature in the kitchen is 25 degrees Celsius.

A. Calculate the change in the entropy of the ice cube as it melts into water at 0 degrees Celsius. (Don’t worry about the fact that the volume changes somewhat).

Answer:
C=Q/delta T Q= 30 cal/K or 126 J/K so C = 126 J/K / 0 degrees Celsius = 0 so delta S= 0 (this is the entropy)


B. Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from 0 degrees Celsius to 25 degrees Celsius.

Answer:
Delta S = Sf -Si= (integral from Ti to Tf) Cv/T dT= 126 J/K (integral from 273K to 298K) 1/T dT = (126J/K) ln (298/273) = 11.0 J/K



C. Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water. (How would I do this one?)



D. Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative or zero? Is this what you would expect? (Need help with this one also, Please!)
 
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will someone help please?

I just need someone to help me. It's due tomorrow. Please
 
thanks for not helping me

SorryI asked for help.
 
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