Find the final temperature of the combined honorite and odium?

[hraghav]

Homework Statement

Solid honorite has a melting temperature of 261.2K, and a heat capacity of 462.2J/kgK. When it is liquid, its heat capacity is 580.2J/kgK. The latent heat for the transition from solid to liquid is
15500J/kg.
Liquid odium has a freezing temperature of 331.44K and a heat capacity of 431.1J/kgK. When solid, its heat capacity is 335.1J/kgK. The latent heat for the transition from solid to liquid is
10700J/kg.
A 4.51kg block of honorite at 261.2K is immersed in 14.69kg of liquid odium at 391.14K and left to achieve thermal equilibrium in an isolated, insulated container.
What is the final temperature of the combined honorite and odium?

Relevant Equations

Qmelt, honorite = (m honorite)⋅(Lfusion, honorite)
Qcool, odium = (modium)⋅(Cliquid, odium)⋅(391.14K−331.44K)
Qfreeze, odium = (m odium)⋅(L fusion, odium)
Qheat, honorite = (mhonorite)⋅(Cliquid, honorite)⋅(Tf−261.2K)
Qsolid, odium = (m odium)⋅(Csolid, odium)⋅(331.44K−Tf)

Qmelt, honorite = (m honorite)⋅(Lfusion, honorite)
Qmelt, honorite = 4.51kg⋅15500J/kg = 69905J

Qcool, odium = (m odium)⋅(Cliquid, odium)⋅(391.14K−331.44K)
Qcool, odium = 14.69kg⋅431.1J/kgK⋅(391.14K−331.44K)
Qcool, odium = 378071.6823

Qfreeze, odium = (m odium)⋅(L fusion, odium)
Qfreeze, odium = 14.69kg⋅10700J/kg = 157183J

(Qmelt, honorite) + (Qheat, honorite) = (Qcool, odium) + (Qfreeze, odium) + (Qsolid, odium)
69905 + 2616.702*(Tf-261.2) = 540177.3013 + 4922.619*(331.44−Tf)
Tf = 369.437474K

This is incorrect. Could someone please help me figure this out?

Thanks

 

[Chestermiller]

There are three possibilities:
1. The final temperature is below 335.1
2. The final temperature is equal to 335.1
3. The final temperature is above 335.1

You need to consider each of these three possibilities.

 

[haruspex]

There are three possibilities:
1. The final temperature is below 335.1
2. The final temperature is equal to 335.1
3. The final temperature is above 335.1

You need to consider each of these three possibilities.

Equal to 261.2K is another distinct case which, in principle, is possible.

 

[Steve4Physics]

Tf = 369.437474K

This is incorrect. Could someone please help me figure this out?

Expanding on what @Chestermiller and @haruspex have said...

Note that using suitable symbols can make things easier to read and check, so let ‘H’ be honorite and let ‘D’ be odium. (The letter ‘O’ looks too much like zero.)

H is initially solid and D is initially liquid. In general there are various conceivable final states:
none of H melted, none of D solidified
none of H melted, some of D solidified
none of H melted, all of D solidified
some of H melted, none of D solidified
some of H melted, some of D solidified
some of H melted, all of D solidified
all of H melted, none of D solidified
all of H melted, some of D solidified
all of H melted, all of D solidified

You can eliminate some of the above immediately, given the initial conditions here, but even so, you are left with a range of possibilities.

A useful (IMO) approach is to do some preliminary work and calculate the following quantities:

1. Amount of energy needed to completely melt H (note that H is initially already at its melting temperature).

2. Amount of energy needed to then raise H to D’s melting temperature.

3. Amount of energy which needs to be removed to cool D to its melting (freezing) temperature.

4. Amount of energy which then needs to be removed to freeze D.

Using these 4 values and some logic, you can work out what the final state of the system will be. Then you can calculate the termperature of this state.

By the way, final answer should have an appropriate number of significant figures. ‘369.437474K’ is inappropriate!

 

[Lnewqban]

What is up with the name of these non-existing substances?
In a previous thread we had Lerasium and Atium, which are names used only in science fiction.
Honorite and Odium this time.

@hraghav
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html

 

[Lnewqban]

Qmelt, honorite = (m honorite)⋅(Lfusion, honorite)
Qmelt, honorite = 4.51kg⋅15500J/kg = 69905J

Qcool, odium = (m odium)⋅(Cliquid, odium)⋅(391.14K−331.44K)

I believe that those 69 905 J of needed melting heat transferred into the melting solid Honorite are not able to cool down the liquid Odium that much.

Immediately after all the initially solid Honorite is melted, that cold liquid mass will steal more energy (at a different rate) from the hotter liquid mass of Odium, tending to a temperature balance point (unless Odium freezes before reaching that point).