Thermodynamics - obtaining quenching entropy

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  • #1
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I'm having a problem determining the total entropy change in a metal quenching problem. Here is the information I am given (subscript 'm' is for metal, 'w' is water):

mm = 20 kg
mw = 1000 kg
T1m = 800 C
T1w = 30 C
Cpm = .4 kJ/kg K
Cpw = 4.18 kJ/kg K

With this information, I found T2 to be 304.6 K.

How can I obtain the entropy change with this information? I have no pressure values, and all my entropy equations involve pressure, or specific volume which depends on pressure. Thanks for any help.
 

Answers and Replies

  • #2
Mapes
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Do you know of an equation that connects dS to Cp that you might be able to integrate?
 
  • #3
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I did find a relationship, but it is returning negative values for me, so something is amiss...

s2 - s1 = Cp ln(T2/T1)

I'm trying to find the total entropy change. Would that mean that I sum the entropy change in the quenched metal and the water?
 
  • #4
stewartcs
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Would that mean that I sum the entropy change in the quenched metal and the water?
Yes.

CS
 
  • #5
Mapes
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Yes. It's not unusual for entropy to decrease; it happens whenever a hot object cools down.
 
  • #6
stewartcs
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I did find a relationship, but it is returning negative values for me, so something is amiss...

s2 - s1 = Cp ln(T2/T1)
You'll need to consider the mass of each as well in your equation.

CS
 
  • #7
stewartcs
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Yes. It's not unusual for entropy to decrease; it happens whenever a hot object cools down.
Caveat:

But the total change in the universe will always increase.

CS
 
  • #8
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Wow, thanks for all the responses everyone :). OK so here is what I came up with, and it seems to be giving me the right answer, but just for future reference:

s2-s1 = mmCpm ln(T2/Tm1) + mwCpw ln(T2/Tw1)
 

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