What Is the Final Equilibrium Temperature Between Aluminum and Iron Blocks?

[needy]

a 20 Kg Aluminum block initially at 200 C is brought into contact with a 20 Kg block of Iron at 100 C. I need the final equilibrium temperature and the total entropy change for this process.

I have an idea on how to find the chango of entropy, what I'm struggling with is finding the final temperature.

Help Please

 

[Mapes]

Hi needy, welcome to PF. If this is a homework question, it's probably going to be moved. Also, you should discuss what equations you've found and tried so far so we know where to start.

You've probably figured out that the Al block is going to heat the Fe block until the temperatures are equal. But the relationship between energy and temperature is different between Al and Fe. Have you gotten this far?

 

[JANm]

Specific heat of aluminium =0,88 J / kg K.
Define temperature of the iron block as zero how much energy has the aluminium to divide over the aluminium- and iron block together.

 

[Abninfamy]

a 20 Kg Aluminum block initially at 200 C is brought into contact with a 20 Kg block of Iron at 100 C. I need the final equilibrium temperature and the total entropy change for this process.

I have an idea on how to find the change of entropy, what I'm struggling with is finding the final temperature.

Help Please

Closed system, control mass.

What you're going to want to do is model both objects as incompressible.
ca=specific heat aluminum
ci=specific heat iron
m=mass
Tf=temp final
Ti=temp Fe
Ta=temp Al

Imcompressible 1st Law
U=Q-W Q=0, W=0
U=0
U=m*c*(T2-T1)
so
Ui+Ua=0
then
mi*ci*(Tf-Ti)+ma*ca*(Tf-Ta)=0

For incompressible substances, Change in S (entropy) is S= m*c*(ln(T2/T1))

Your second law equation is of course S2-S1=Q/T+o(entropy production) Q/T=0 (no boundary heat loss)

so S2-S1=o (Treat S2-S1 as change in S)

mi*ci*ln(Tf/Ti)+ma*ca*ln(Tf/Ta)=o

Double check for any typing errors, etc.
Hope that helps!

 

[JANm]

Ok the system is almost fully defined. What is needed is the contact surface O, and Needy I hope that you agree in cilindersymmetry. Then there are to heights hi and ha.
So the volume of aluminium block Va=O*ha and for the Ferro block: Vi=O*hi. Then we can define de temperature on the line -ha<= x <=+hi, in which the point x=0 means a point on the contact surface. It is logical to hypothesize that the final state Tf is reached there first, and that at that moment no heat-transport trough O takes place.
At that moment the two blocks can be separated and calculated how the remaining heattransfer takes place in them