Find final temp of 2 substances once they have reached equilibrium

[hraghav]

Homework Statement

A quantity of Lerasium (mL=1.49kg) at a temperature of TL,i=1108.1∘C is placed into an isolated container with a quantity of Atium (mA=2.44kg) at a temperature of TA,i=33.6∘C.
Lerasium melts at TL,melt=56.4∘C and vapourizes at TL,vap=327.7∘C.
Lerasium has a heat capacity of cL,liquid=221.6J/kgK when liquid, and cL,vapour=256.4J/kgK when vapour. The latent heat of vapourization is LL,vapour=15139J/kg.
Atium melts at TA,melt=308∘C and vapourizes at TA,vap=1103.1∘C. Atium has a heat capacity of cA,solid=524.8J/kgK when solid, and cA,liquid=688.2J/kgK when liquid. The latent heat of fusion for Atium is LA,fusion=58350J/kg.
What is the final temperature of the two substances once they have reached equilibrium?

Relevant Equations

Q1 =m L⋅c L,liquid ⋅(T L,i −T L,melt)
Q2 =mA⋅cA,solid ⋅(TA,melt −TA,i)
Q3 =mA⋅LA,fusion
Q L = Q A

Cooling Lerasium to its melting point:
Q1 =m L⋅c L,liquid ⋅(T L,i −T L,melt)
Q1 = 1.49kg⋅221.6J/kgK⋅(1108.1−56.4)K
Q1 = 330.184 ⋅(1051.7) = 347254.512 J

Heating Atium to its melting point:
Q2 =mA⋅cA,solid ⋅(TA,melt −TA,i)
Q2 = 2.44kg⋅524.8J/kgK⋅(308−33.6)K
Q2 = 1280.512⋅(274.4) = 351372.493 J

Melting Atium:
Q3 =mA⋅LA,fusion
Q3 = 2.44kg⋅58350J/kg = 142374 J

Total Energy Changes:
QL,total = Q1 = 347254.512 J
QA,total = Q2+Q3 = 351372.493+142374 = 493746.493 J

Assuming all the energy from cooling Lerasium to its melting point goes into heating Atium then we have:
347254.512 J = Q2+mA⋅cA,liquid⋅(Tfinal−308∘C)
347254.512 J = 351372.493 + 2.44kg⋅524.8J/kgK ⋅(Tfinal−308∘C)
347254.512 J = 351372.493 + 1280.512⋅(Tfinal−308∘C)
347254.512 J = 352653.005⋅(Tfinal−308∘C)
0.98469 = Tfinal−308∘C

Tfinal = 308.985∘C

But this isn't correct and I am not sure where I am making an error. Could someone please look at this and let me know?

Thank you

 

[Lnewqban]

It seems to me that vapor is the initial condition of that 1.46 kg of Lerasium, since its temperature is above 327.7 °C.

 

[hraghav]

It seems to me that vapor is the initial condition of that 1.46 kg of Lerasium, since its temperature is above 327.7 °C.

Q1 = 1.49kg⋅256.4J/kgK⋅(1108.1−327.7)K
Q1 = 298140.8944

Q2 = 1.49kg⋅15139J/kg = 22557.11J

Q3 = 1.49kg⋅221.6J/kgK⋅(327.7−56.4)K
Q3 = 89578.9192

Q4 = 2.44kg⋅524.8J/kgK⋅(308−33.6)K
Q4 = 351372.4928

Q5 = 2.44kg⋅58350J/kg = 142374J

QL,total = Q1 + Q2 + Q3
QL,total = 298140.8944 + 22557.11 + 89578.9192
QL,total = 410276.9236

QA,total = Q4 + Q5
QA,total = 351372.4928 + 142374
QA,total = 493746.4928

Since QL,total < QA,total we get
QL,total = Q4 + mA⋅cA,liquid⋅(Tfinal−308)
410276.9236=351372.4928+(2.44)* (688.2)* (Tfinal-308)
Tfinal = 343.079 °C.

This is still not the correct answer. Is there something I am missing?

 

[Lnewqban]

The calculation of Q3 incorrectly assumes that the substance was cooled down all the way to the melting-solidification temperature (which is lower than the estimated equilibrium temperature of both substances).

In the same way, the calculation of Q4 assumes that the originally solid Atrium reached its melting point (which could be or not, depending on the accuracy of the balance temperature estimation).

 

[haruspex]

I would put it into a spreadsheet. Easiest for the formulas is to put temperatures on the x axis and heat transferred on the y axis. Where they cross is the answer.

 

[hraghav]

The calculation of Q3 incorrectly assumes that the substance was cooled down all the way to the melting-solidification temperature (which is lower than the estimated equilibrium temperature of both substances).

In the same way, the calculation of Q4 assumes that the originally solid Atrium reached its melting point (which could be or not, depending on the accuracy of the balance temperature estimation).

Yes that makes sense. Thanks I got it now

 

[Lnewqban]

As good reference for future problems, or for people visiting this thread, please see:

https://xaktly.com/ThermalEquilibrium.html

You are welcome.