What Is the Entropy Change When Ice Melts in a Room at 20 Degrees Celsius?

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SUMMARY

The discussion focuses on calculating the net entropy change when a 15 kg block of ice at 0 degrees Celsius melts into liquid water at the same temperature within a room at 20 degrees Celsius. The correct approach involves recognizing that the problem only requires calculating the entropy change due to the phase transition of ice to water, without considering the subsequent warming of the water. The net entropy change is derived from the entropy increase of the melting ice and the corresponding decrease in the room's entropy, ultimately leading to a total entropy change of 1250 J/K.

PREREQUISITES
  • Understanding of the 1st Law of Thermodynamics
  • Knowledge of phase change and latent heat (Lf = 3.34 x 10^5 J/kg)
  • Familiarity with the concept of entropy (dS = Q/T)
  • Basic principles of thermodynamics and heat transfer
NEXT STEPS
  • Study the implications of the 1st Law of Thermodynamics on isolated systems
  • Learn about the calculation of entropy changes during phase transitions
  • Explore the relationship between temperature and entropy in thermodynamic processes
  • Investigate the concept of thermal equilibrium and its relevance in thermodynamic calculations
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone interested in understanding entropy changes during phase transitions in isolated systems.

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Homework Statement


A 15 kg block of ice at 0 celsius degrees melts to liquid water at 0 celsius degrees inside a large room that has temperature 20 celsius degrees. Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored.
a) Is the melting of the ice reversible or irreversible?
b) Calculate the net entropy change of the system during this process.


Homework Equations


dS = Q/T for constant temperature
Q = m*Lf, where Lf = 3.34 x 10^5 J/kg for water
dS = int(dQ/T) for non-constant temperature


The Attempt at a Solution


I've tried to calculate the phase change from ice to water using dS=Q/T, getting 18351 J/K and then adding the entropy change for the temperature change from 0 to 20 degrees using dS=int(dQ/T)=m*c*ln(T2/T1), getting 4443 J/K. Thats no where near the correct answer, which is supposed to be 1250 J/K.
PLEASE help
 
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Hello ks23,

Welcome to Physics Forums!

ks23 said:
I've tried to calculate the phase change from ice to water using dS=Q/T, getting 18351 J/K
So far so good. :approve:
and then adding the entropy change for the temperature change from 0 to 20 degrees using dS=int(dQ/T)=m*c*ln(T2/T1), getting 4443 J/K.
Hold on. The problem statement doesn't say anything about letting the liquid water warm up to 20 degrees. You can stop the calculations the moment the ice completely melts. There's no need in this particular problem to let the water and room reach thermal equilibrium. The problem statement only concerns itself with the ice melting.
Thats no where near the correct answer, which is supposed to be 1250 J/K.
PLEASE help
Don't forget about the entropy decrease of the room (ignoring the ice). The entropy of the ice==>water (ignoring the room) increases as it melts. The entropy of the room (ignoring the ice/water) decreases by some amount too because it is loosing energy that is ultimately transferred into the ice to make it water. How much energy does the room lose? (The 1st law of thermodynamics should help you with this one.) What is the temperature of the room? The overall entropy change is the sum of the ice/water's entropy change with the room's entropy change.
 

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