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Friday, April 14, 2006
HW #7 - P #9 - Entropy Change for an Incompressible Substance - 5 pts
A 20-kg aluminum block initially at 200°C is brought into contact with a 20-kg block of iron at 100°C in an insulated enclosure. Determine the final equilibrium temperature and the total entropy change for this process.
8 comments:
Anonymous
said...
For the first step, to find T2, I'm using the fact that deltaU is zero for the system, and deltaU=integralCvdt, from T1 for each block to unknown T2. I converted to Kelvin and multiplied through by 20kg and am way off. What am I doing wrong?
Anon 6:50 It is important to keep track of the deltaU for the iron and the Aluminum independetly. 1st law, both blocks inside the system gives us: deltaUsystem = deltaUiron + deltaUaluminum = 0. Then, determine deltaU separately for iron and aluminum. Solve for T2 as you described. I suspect you added the Cp values.
for the DeltaS here, can we use the Incompressible liquid equation C*ln(T2/T1)? I am guessing not, sinse its not working. What definition of S do we use? Q/t isnt really working either... thanks
nikfoo 8:45 AM Yes, deltaS = m Cp Ln{T2/T1} for solids and incompressible liquids. But you must apply this separately to the Iron and then to the Aluminum. At the end you add the deltaSiron to the deltaSaluminum to get the total deltaS.
no... it is kJ/K use equation like dr.B said - the two K units cancel themselves inside the ln so you should have regular entropy units. ( the K in the denominator comes from the Cp unit) hope that helps
8 comments:
For the first step, to find T2, I'm using the fact that deltaU is zero for the system, and deltaU=integralCvdt, from T1 for each block to unknown T2. I converted to Kelvin and multiplied through by 20kg and am way off. What am I doing wrong?
Anon 6:50
It is important to keep track of the deltaU for the iron and the Aluminum independetly. 1st law, both blocks inside the system gives us: deltaUsystem = deltaUiron + deltaUaluminum = 0. Then, determine deltaU separately for iron and aluminum. Solve for T2 as you described. I suspect you added the Cp values.
for the DeltaS here, can we use the Incompressible liquid equation C*ln(T2/T1)? I am guessing not, sinse its not working. What definition of S do we use? Q/t isnt really working either...
thanks
nikfoo 8:45 AM
Yes, deltaS = m Cp Ln{T2/T1} for solids and incompressible liquids. But you must apply this separately to the Iron and then to the Aluminum. At the end you add the deltaSiron to the deltaSaluminum to get the total deltaS.
I'm pretty sure that the total entropy change should be in KJ and not in KJ/K as it says in the hint. Correct me if I'm wrong?
wow i was totally on the right track. thanks dr.B!
no... it is kJ/K
use equation like dr.B said - the two K units cancel themselves inside the ln so you should have regular entropy units. ( the K in the denominator comes from the Cp unit)
hope that helps
Anon & Nikfoo
Nikfoo is correct and did a good job of explaining why. deltaS [=] kJ/K
Let me know if I need to say more about this.
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