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

In summary: Mathis attempted to solve this problem, and found that the temperature of the water decreases by 41.74 degrees Celsius.
  • #1
Lisa Marie
9
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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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  • #2
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!
 
  • #3
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.
 
  • #4
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
 

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

1. What is entropy?

Entropy is a measure of the disorder or randomness in a system. It is often referred to as the measure of the amount of energy that is unavailable to do work.

2. How does entropy change in a system?

The change in entropy of a system is determined by the difference between the entropy of the final state and the entropy of the initial state. If the final state has higher disorder, the change in entropy will be positive, and if the final state has lower disorder, the change in entropy will be negative.

3. What factors affect the change in entropy?

The change in entropy can be affected by changes in temperature, pressure, and the number of particles in a system. Additionally, chemical reactions and physical phase changes can also cause a change in entropy.

4. How is entropy related to energy?

Entropy is related to energy through the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time. This means that as energy is transferred or transformed, the amount of energy that is unavailable to do work increases, causing an increase in entropy.

5. Can entropy be reversed?

In a closed system, entropy cannot be reversed. This is due to the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time. However, in an open system, where energy can be exchanged with the surroundings, it is possible for local decreases in entropy to occur.

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