Thermodynamics and heat capacity

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Homework Statement


  1. A 10.0 g sample of solid platinum is placed in a large, sealed vessel at a sufficiently low pressure that the platinum is able to sublimate directly to a gas. Some thermodynamic properties of platinum at low pressure are given in the following table.
Molar mass, M - 195 mol-1
Latent heat of sublimation, LS 363 kJ mol-1
Triple point pressure, pTP 3.50 Pa
Triple point temperature, TTP 1550'C

The platinum is heated at a rate of 10.0 W. If the initial temperature of the platinum is 25.0 ◦C, and it sublimates at the triple point temperature, how long will it take for the platinum to fully sublimate?


Homework Equations


Q = nLS
Q = mCT[/B]


The Attempt at a Solution


[/B]
My thoughts so far:
Time = Energy required/power
Energy required to sublimate at 1550"C = Q = nLS
Energy required to heat from 25 - 1550 = Q = mcΔT. This is where i have a problem as there is no data given for the specific heat capacity of platinum. I was considering using the dulong-petit rule that all solids have Cv = 3R but the volume is not kept constant so I'm not sure how to work out the energy required to heat the platinum.
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


  1. A 10.0 g sample of solid platinum is placed in a large, sealed vessel at a sufficiently low pressure that the platinum is able to sublimate directly to a gas. Some thermodynamic properties of platinum at low pressure are given in the following table.
Molar mass, M - 195 mol-1
Latent heat of sublimation, LS 363 kJ mol-1
Triple point pressure, pTP 3.50 Pa
Triple point temperature, TTP 1550'C

The platinum is heated at a rate of 10.0 W. If the initial temperature of the platinum is 25.0 ◦C, and it sublimates at the triple point temperature, how long will it take for the platinum to fully sublimate?


Homework Equations


Q = nLS
Q = mCT[/B]


The Attempt at a Solution


[/B]
My thoughts so far:
Time = Energy required/power
Energy required to sublimate at 1550"C = Q = nLS
Energy required to heat from 25 - 1550 = Q = mcΔT. This is where i have a problem as there is no data given for the specific heat capacity of platinum. I was considering using the dulong-petit rule that all solids have Cv = 3R but the volume is not kept constant so I'm not sure how to work out the energy required to heat the platinum.
You're in luck.

Someone has made a detailed study of the thermodynamic properties of platinum and published a paper here:

http://www.technology.matthey.com/article/49/3/141-149/

The paper may be downloaded free of charge, and it contains a formula which calculates the specific heat of solid platinum which is valid from room temperature all the way to the M.P. and beyond.
 
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Thanks for looking that up for me. However I'm revising and this is an old exam question so I would need to be able to solve it in exam conditions with just the information available in the exam so finding the heat capacity wouldn't be possible. Thank you anyway though!
 
  • #4
rude man
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Homework Statement


  1. A 10.0 g sample of solid platinum is placed in a large, sealed vessel at a sufficiently low pressure that the platinum is able to sublimate directly to a gas. Some thermodynamic properties of platinum at low pressure are given in the following table.
Molar mass, M - 195 mol-1
Latent heat of sublimation, LS 363 kJ mol-1
Triple point pressure, pTP 3.50 Pa
Triple point temperature, TTP 1550'C
The platinum is heated at a rate of 10.0 W. If the initial temperature of the platinum is 25.0 ◦C, and it sublimates at the triple point temperature, how long will it take for the platinum to fully sublimate?

Homework Equations


Q = nLS
Q = mCT
3. The Attempt at a Solution [/B]
My thoughts so far:
Time = Energy required/power
Energy required to sublimate at 1550"C = Q = nLS
Energy required to heat from 25 - 1550 = Q = mcΔT. This is where i have a problem as there is no data given for the specific heat capacity of platinum. I was considering using the dulong-petit rule that all solids have Cv = 3R but the volume is not kept constant so I'm not sure how to work out the energy required to heat the platinum.
Since pressure in the range 25C - 1550C is not given you could perhaps assume the pressure corresponds to the platinum always resting on the sublimation curve. If the pressure is assumed at 3.50 Pa at all temperatures then you'd need the specific heat of the metal. In the former case the problem is pretty trivial.
 
  • #5
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Since pressure in the range 25C - 1550C is not given you could perhaps assume the pressure corresponds to the platinum always resting on the sublimation curve. If the pressure is assumed at 3.50 Pa at all temperatures then you'd need the specific heat of the metal. In the former case the problem is pretty trivial.
I'm not quite sure what you mean. Surely you still have to heat the metal up to the triple point temp as the question states that that is where it sublimates? I think i might be misunderstanding what you are saying because although I can visualise the sublimation curve on a graph Im not fully getting how this helps simplify the problem!
 
  • #6
rude man
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I'm not quite sure what you mean. Surely you still have to heat the metal up to the triple point temp as the question states that that is where it sublimates? I think i might be misunderstanding what you are saying because although I can visualise the sublimation curve on a graph Im not fully getting how this helps simplify the problem!
I think I'd better opt out of this problem. I'm not sure I see how the experiment is performed. At least I need to think more about it. Someone else will probably rescue you in the meantime, hopefully.
 

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