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Entropy heat bath block

  • Thread starter quietrain
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  • #1
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Homework Statement


Find entropy change of total system of heatbath + water when water is placed in heat bath from 20 C to 50 C, then 50C to 80C. assume equilibrium is reached first in both cases.
water = 1000g , C = 4.2J/C/G

The Attempt at a Solution



the entropy for water Sw = CdT / T + Cdt / T , where first integral from 293K to 323K, second from 323K to 353K.

bath entropy Sb = -C(353 - 293) / 353

Sw=780.4
Sb=-713.8

S-total = 66.6 J/K

but ans is 35.6J/K .

apparently, the 2 step process of heating should get me a smaller entropy change, but the value i calculated 66.6 is around the same as if i did a 1 step heating from 20 to 80 straight.

so where did i go wrong? thanks!
 

Answers and Replies

  • #2
SammyS
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[tex]\int_{293}^{323} \frac{C}{T}\,dT=C\cdot\ln\left(\frac{323}{293}\right)[/tex]
 
  • #3
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[tex]\int_{293}^{323} \frac{C}{T}\,dT=C\cdot\ln\left(\frac{323}{293}\right)[/tex]
yea using that, i got the answer to a 1 time heat bath heating from 293 to 353 straight

but if i use that expression twice from 293 to 323 plus 323 to 353, i essentially get the same value which is wrong, the ans is around twice less.
 
  • #4
Andrew Mason
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yea using that, i got the answer to a 1 time heat bath heating from 293 to 353 straight

but if i use that expression twice from 293 to 323 plus 323 to 353, i essentially get the same value which is wrong, the ans is around twice less.
You are assuming one heat bath at 80C. This is not what you are given. There are two heat baths (because it says that the water reaches equilibrium in both cases).

AM
 
  • #5
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both heat baths will always be at 80C right? so which part of my workings are wrong? you mean for the denominator of the bath entropy, i have to use

-C(323 - 293) / 323 -C(353 - 323)/353
?

but then that would mean one heat bath is at 80C and the other is maintained at 50C?
 
  • #6
Andrew Mason
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both heat baths will always be at 80C right?
If that was the case, how could the 1000 g of water reach equilibrium at 50C?

so which part of my workings are wrong? you mean for the denominator of the bath entropy, i have to use

-C(323 - 293) / 323 -C(353 - 323)/353

but then that would mean one heat bath is at 80C and the other is maintained at 50C?
Correct.

AM
 
  • #7
654
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i see thankyou
 

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