Thermodynamics and heat capacity

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SUMMARY

The discussion focuses on calculating the time required for a 10.0 g sample of solid platinum to fully sublimate when heated at a rate of 10.0 W. Key thermodynamic properties include a molar mass of 195 g/mol, a latent heat of sublimation of 363 kJ/mol, and a triple point temperature of 1550°C. The energy required for sublimation is calculated using the formula Q = nLS, while the energy needed to heat the platinum from 25°C to 1550°C is calculated using Q = mcΔT. The lack of specific heat capacity data for platinum complicates the calculation, although the Dulong-Petit rule is mentioned as a potential approach.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically sublimation and latent heat.
  • Familiarity with the equations Q = nLS and Q = mcΔT.
  • Knowledge of the Dulong-Petit rule for estimating specific heat capacities.
  • Basic concepts of heat transfer and power calculations.
NEXT STEPS
  • Research the specific heat capacity of platinum and its implications for thermodynamic calculations.
  • Explore the Dulong-Petit rule and its applicability to different materials.
  • Study the sublimation curve of platinum and its behavior under varying pressures.
  • Learn about heat transfer methods and their impact on phase changes in materials.
USEFUL FOR

Students studying thermodynamics, particularly those focusing on phase changes and heat transfer, as well as educators and professionals in materials science and engineering fields.

struggles
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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.
 
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struggles said:

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.
 
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!
 
struggles said:

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.
 
rude man said:
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 I am not fully getting how this helps simplify the problem!
 
struggles said:
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 I am 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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