Find the mass of odium for both solid and liquid state

[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.
An extra 3.75kg of solid honorite is added with the same initial temperature, making the total honorite in the container 8.26kg. The combination is allowed to achieve a new thermal equilibrium. What is the mass of odium and honorite in each of the states, solid and liquid?

Relevant Equations

Q melt = m honorite × L honorite
Q cool = (m odium) × (c liquid odium) ×(T initial−T freeze )
(m odium, freeze) = L odium / Q excess

Mass of honorite initially: 4.51 kg at 261.2 K
Additional mass of honorite added: 3.75 kg at 261.2 K
total mass of honorite is 4.51 kg + 3.75 kg = 8.26 kg
Mass of liquid odium: 14.69 kg at 391.14 K

Q melt = m honorite × L honorite
Qmelt = 8.26kg × 15500J/kg
Qmelt = 128030 J

Q cool = (m odium) × (c liquid odium) ×(T initial−T freeze )
Qcool = 14.69kg × 431.1J/kgK ×(391.14K−331.44K)
Qcool = 378071.6823 J

Qexcess = Qcool − Qmelt
Qexcess = 378071.6823 J−128030J= 250041.6823 J

(m odium, freeze) = Q excess / L odium
(m odium, freeze) = 250041.6823/10700
(m odium, freeze) = 23.368 kg

Mass of liquid odium: 14.69 kg
Mass of solid odium = 23.368 kg

But both are incorrect. Could someone please let me know where is the error?

Thank you

 

[haruspex]

Have you tried the spreadsheet method I outlined in https://www.physicsforums.com/threa...-once-they-have-reached-equilibrium.1064836/?

 

[hraghav]

Have you tried the spreadsheet method I outlined in https://www.physicsforums.com/threa...-once-they-have-reached-equilibrium.1064836/?

I did try but it doesn't seem to work. Could you help me out with the steps for this as I am trying this method for the first time

 

[haruspex]

I did try but it doesn't seem to work. Could you help me out with the steps for this as I am trying this method for the first time

What do you have for the formula for the heat gained by the cooler body as a function of its temperature?

 

[hraghav]

What do you have for the formula for the heat gained by the cooler body as a function of its temperature?

I am not sure about this but this is what I got:
Qhonorite(Teq) = (mhonorite)⋅(Csolid_honorite)⋅(Tmelt_honorite − Tinitial_honorite) + (mhonorite) ⋅ (Lfusion_honorite) + (mhonorite)⋅(Cliquid_honorite)⋅(Teq−Tmelt_honorite)

 

[haruspex]

I am not sure about this but this is what I got:
Qhonorite(Teq) = (mhonorite)⋅(Csolid_honorite)⋅(Tmelt_honorite − Tinitial_honorite) + (mhonorite) ⋅ (Lfusion_honorite) + (mhonorite)⋅(Cliquid_honorite)⋅(Teq−Tmelt_honorite)

No, you need conditionals.
Something like (assuming we start with a solid):
m*if(t<tmelt,(t-ti)*csolid,

(tmelt-ti)* csolid+fusion +​

if(t<tevap,(t-tmelt)*cliquid,​

(tevap-tmelt)*cliquid+(t-tevap)*cgas​

)​

)​

 

[hraghav]

No, you need conditionals.
Something like (assuming we start with a solid):
m*if(t<tmelt,(t-ti)*csolid,

(tmelt-ti)* csolid+fusion +​

if(t<tevap,(t-tmelt)*cliquid,​

(tevap-tmelt)*cliquid+(t-tevap)*cgas​

)​

)​

Oh is it possible if you could help me with the algebraic method of finding the answers as my finals is in the pen paper form and I would not be able to use this. This is one of the sample final practice questions

 

[hraghav]

I found the correct answer for solid odium ie 8.157 kg but I am still struggling with the liquid mass. I also tried 14.69 - 8.157 = 6.533 for liquid but this is still not correct. Could someone please spot the error for me. Thanks

 

[haruspex]

Oh is it possible if you could help me with the algebraic method of finding the answers as my finals is in the pen paper form and I would not be able to use this. This is one of the sample final practice questions

Perhaps the least error prone is to simulate the process.

As the heat flows, each body alternates between a stage where its temperature is changing and a stage where its state is changing. By "energy boundary" I mean one of those stage transitions, such as starting to melt, or completing evaporation.

So long as they are at different temperatures, heat is flowing. Calculate which of three things will happen first: body A reaches an energy boundary, body B reaches an energy boundary, or the temperatures equalise. Calculate the new temperatures at that point and repeat the procedure.

 

[haruspex]

I found the correct answer for solid odium ie 8.157 kg but I am still struggling with the liquid mass. I also tried 14.69 - 8.157 = 6.533 for liquid but this is still not correct. Could someone please spot the error for me. Thanks

If you have correctly quoted the initial liquid mass and you result for final solid mass is correct then so is your result for final liquid mass.
But since the 14.69 only has two decimal places your answer should not have three. 6.53 would be better.