Calculate the change in entropy of the system

[squib]

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!

 

[apchemstudent]

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?

 

[Andrew Mason]

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:

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

AM

 

[squib]

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