Help Solving Entropy Problem: Part A-D

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SUMMARY

The discussion revolves around solving an entropy problem involving an ice cube melting at 0 degrees Celsius in a kitchen at 25 degrees Celsius. The participant successfully calculated the change in entropy for the ice cube (0 J/K) and the water (11.0 J/K) but seeks assistance for parts C and D. Part C requires calculating the change in entropy of the kitchen as it loses heat to the melting ice, while part D involves determining the net change in entropy of the universe during the process and its implications.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically entropy.
  • Familiarity with the concept of heat transfer and temperature changes.
  • Knowledge of integral calculus for calculating changes in entropy.
  • Basic understanding of the laws of thermodynamics, particularly the second law.
NEXT STEPS
  • Calculate the change in entropy of the kitchen using the formula ΔS = Q/T, where Q is the heat lost by the kitchen.
  • Learn about the concept of reversible and irreversible processes in thermodynamics.
  • Study the implications of the second law of thermodynamics on entropy changes in isolated systems.
  • Explore examples of entropy calculations in phase changes and heat transfer scenarios.
USEFUL FOR

Students studying thermodynamics, educators teaching entropy concepts, and anyone interested in understanding heat transfer and entropy changes in physical processes.

jlmac2001
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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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