Change in entropy question

  • Thread starter Lisa Marie
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  • #1

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!
 

Answers and Replies

  • #2
vela
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You're assuming that all the ice is going to melt.

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
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1
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
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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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