Calculate the change in entropy of the system

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An aluminum can, with negligible heat capacity, is filled with 485 g of water at 0°C and then is brought into thermal contact with a similar can filled with 517 g of water at 48.5°C. Calculate the change in entropy of the system if no heat is exchanged with the surroundings.

1) I find avg temp, which turns out to be: ~298
2) I calc change in temp for both sides, and find Q. (deltaT)(4.184)(mass)
3) Divide each Q (they're the same, with one being negative) by each starting temperature
4) I come up with .163599 J/K, which is wrong

Where am I messing up? I have 4-5 problems identical to this but can't figure any of them out!!!
 
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  • #2
squib said:
An aluminum can, with negligible heat capacity, is filled with 485 g of water at 0°C and then is brought into thermal contact with a similar can filled with 517 g of water at 48.5°C. Calculate the change in entropy of the system if no heat is exchanged with the surroundings.

1) I find avg temp, which turns out to be: ~298
2) I calc change in temp for both sides, and find Q. (deltaT)(4.184)(mass)
3) Divide each Q (they're the same, with one being negative) by each starting temperature
4) I come up with .163599 J/K, which is wrong

Where am I messing up? I have 4-5 problems identical to this but can't figure any of them out!!!

Can you show your calculations?
 
  • #3
Andrew Mason
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squib said:
An aluminum can, with negligible heat capacity, is filled with 485 g of water at 0°C and then is brought into thermal contact with a similar can filled with 517 g of water at 48.5°C. Calculate the change in entropy of the system if no heat is exchanged with the surroundings.

1) I find avg temp, which turns out to be: ~298
2) I calc change in temp for both sides, and find Q. (deltaT)(4.184)(mass)
3) Divide each Q (they're the same, with one being negative) by each starting temperature
4) I come up with .163599 J/K, which is wrong

Where am I messing up? I have 4-5 problems identical to this but can't figure any of them out!!!
Since the temperature changes as the heat is absorbed/lost, you have to integrate:

[tex]\Delta S = \int_{T_i}^{T_f} dQ/T = cm\int_{T_i}^{T_f} dT/T = cm* ln(\frac{T_f}{T_i})[/tex]

AM
 
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Thanks so much, I've been very frustrated with how to do those and couldn't find the answer ANYWHERE. Big thanks!!
 

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