Calculate the change in entropy of the system

AI Thread Summary
The discussion focuses on calculating the change in entropy for a system involving two aluminum cans filled with water at different temperatures. The initial approach includes finding the average temperature and calculating heat transfer (Q) using specific heat capacity and mass. However, the correct method involves integrating the change in temperature, as the temperature varies during the heat exchange. The formula for entropy change is highlighted as ΔS = cm * ln(T_f/T_i), emphasizing the need for integration rather than simple division. The participant expresses frustration over repeated similar problems, indicating a common misunderstanding in applying the entropy calculation correctly.
squib
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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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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?
 
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:

\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
 
Thanks so much, I've been very frustrated with how to do those and couldn't find the answer ANYWHERE. Big thanks!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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