How Does Energy Transfer Affect Entropy in a Water-Ice System?

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The discussion focuses on calculating the change in entropy when 100 J of energy is transferred from 5 kg of water at 60 degrees Celsius to 1 kg of ice at 0 degrees Celsius. The initial attempt at the solution suggests that all the ice melts, which raises questions about the feasibility of this assumption given the limited energy transfer. Participants highlight the need to use the integral form of the entropy change equation, ΔS = ∫ dQ/T, due to the changing temperature of the water. There is also ambiguity regarding the thermal contact between the water and ice, which affects the problem's complexity. Clarifying these aspects is essential for accurately determining the entropy change in the system.
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Homework Statement


5 kg of water at 60 degrees are put in contact with 1 kg of ice at 0 degrees and are thermally isolated from everything else. The latent heat of ice is 3.3x105 J/kg

What is the change of entropy of the universe when 100J of energy are transferred from the water to the ice

2. Homework Equations

Q=+/- mL
Q=mcΔT
ΔS=|Q|/Tc-|Q|/TH

The Attempt at a Solution


mcΔT+mL+mcΔT=0
(5)(4200)(Tf-60)+(1)(3.3×105)+2000(1)Tf=0

23000Tf=96000
Tf=41.74 C

ΔS=100/(41.74+273)-100/(273)=-0.0486

I'm not getting the right answer for this one. Help would be greatly appreciated!
 
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You're assuming that all the ice is going to melt.
Lisa Marie said:

Homework Statement


5 kg of water at 60 degrees are put in contact with 1 kg of ice at 0 degrees and are thermally isolated from everything else. The latent heat of ice is 3.3x105 J/kg

What is the change of entropy of the universe when 100J of energy are transferred from the water to the ice

2. Homework Equations

Q=+/- mL
Q=mcΔT
ΔS=|Q|/Tc-|Q|/TH

The Attempt at a Solution


mcΔT+mL+mcΔT=0
(5)(4200)(Tf-60)+(1)(3.3×105)+2000(1)Tf=0
What does the last term represent? I figure it has something to do with the ice, but I'm not sure.

This expression also assumes that all of the ice melts. Is this possible if only 100 J of energy is transferred from the warm water to the ice? Can you describe in words what's going to happen?

23000Tf=96000
Tf=41.74 C

ΔS=100/(41.74+273)-100/(273)=-0.0486

I'm not getting the right answer for this one. Help would be greatly appreciated!
 
Firstly you can not use the relation $$\Delta S= \Delta Q/T$$ to find out entropy change for water, because here T is changing. Instead you should use the general relation $$\Delta S= \int dQ/T$$ to find out the change in entropy. This process should work out.
 
There is a bit of ambiguity regarding this problem statement. It is not clear whether the water and ice are in intimate contact, or whether they are separated by a diathermal barrier, and that, once the 100 J are transferred, the diathermal barrier is removed and replaced by an insulation barrier. The version with the barriers is simpler to solve. If 100 J is removed from the 5 kg of water, how much does its temperature drop?

Chet
 
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