Thermodynamics: calculate q, w, E,H,S for a 5 step process

Click For Summary

Homework Help Overview

The original poster seeks to calculate q, w, ΔE, ΔH, and ΔS for a thermodynamic process involving the heating of ice from -30.0 °C to 140.0 °C at constant pressure. The process is divided into five steps, including heating solid ice, phase changes, and heating liquid and gaseous water. Relevant heat capacities and enthalpies are provided.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster outlines a multi-step approach to calculate thermodynamic properties for each phase of the process. They express uncertainty about calculating work (w) and internal energy change (ΔE) for the initial steps, particularly during phase transitions. Some participants suggest considering the work done during the transition from solid to gas and question the need for density values to calculate volume changes.

Discussion Status

Participants are exploring various methods to approach the calculations, with some suggesting that ΔE could be zero during phase changes at constant temperature. There is a recognition of the need to account for volume changes and work done in different phases, but no consensus has been reached on the best approach to calculate these values.

Contextual Notes

Participants note the importance of using density values for different phases to calculate volume changes and work done. The discussion emphasizes the constraints of the problem, including the requirement to calculate properties at constant pressure and the need for specific heat capacities and enthalpies.

Lisa...
Messages
189
Reaction score
0
I need to calculate q, w, \Delta E, \Delta H and \Delta S for the process of heating a sample of ice weighing 18.02 g (1 mole) from -30.0 °C to 140.0°C at constant pressure of 1 atm.
Given are the temperature independent heat capacities (Cp) for solid, liquid and gaseous water: 37.5 J/K/mol, 75.3 J/K/mol, 36.4 J/K/mol respectively. Also, the enthalpies of fusion and vaporization are 6.01 kJ/mol and 40.7 kJ/mol respectively. Assure ideal gas behavior.


I thought of this process as 5 steps:

I) Solid water of -30°C is heated to 0°C
II) Solid water of 0°C melts to give liquid water at 0°C
III) Liquid water of 0°C is heated to 100°C
IV) Liquid water of 100°C is vaporized to give gaseous water at 100°C
V) Gaseous water of 100°C is heated to 140°C

Then I figured I needed to calculate q, w, \Delta E, \Delta H and \Delta S for each step separately and sum them to give the values of q, w, \Delta E, \Delta H and \Delta S for the whole process.

=> So q is calculated for I),III) and V) by q=n Cp\Delta T with the Cp values of respectively solid, liquid and gaseous water. For II) and IV) q= n H with values of H of respectively fusion and vaporization enthalpies.

=> Because the process is carried out at constant pressure, all the q's equal the H's.

=> Entropies are calculated for I), III) and V) by \Delta S = n C_p ln \frac{T_2}{T_1} and for II) and IV) with \Delta S =\frac{\Delta H}{T} with T the melting/boiling point and delta H the fusion/ vaporization enthalpy.

=> Now the point at which I got stuck: calculating w and \Delta E

I know that \Delta E = w + q and w = -p \Delta V. For V) I can calculate w with p= 1 atm and by using the ideal gas law to find delta V and with the delta E formula + known q delta E can be obtained...

But what about the work that is done in the other four steps? I don't know the changes in volume. Could somebody please please please explain to me how I'd calculate w and delta E for the first 4 steps?

EDIT: Now I figured delta E = 0 for II and IV therefore w=-q, because \Delta E = n C_v \Delta T thus it only depends on the temperature, which remains constant during II and IV, so \Delta T =0 = \Delta E . Is that a correct way of thinking? And now how would I tackle calculation of delta E & w of step I, III and V?
 
Last edited:
Physics news on Phys.org
Or perhaps it is necessary to calculate w and E directly for the transition (solid) => (gas) with:

w = -p \Delta V = - n R \Delta T and
\Delta E = \frac{3}{2} R \Delta T (=q +w)
and q= \Delta H = \frac{5}{2} \Delta T
 
On a second thought, I don't think my edit is such a brilliant idea, because then q= n Cp delta T should be 0 too for step II and IV which can't be cause heat is neaded to melt the ice/vaporize the water.

I know I need densities of -30,0°C ice (solid), 0°C ice/water (solid/liquid), 100°C water (liquid). Then I can calculate delta V for all the steps (the volume of gaseous water can be calculated from the ideal gas law) and thus find w and E... but how would I get that far?
 
Last edited:
Lisa... said:
On a second thought, I don't think my edit is such a brilliant idea, because then q= n Cp delta T should be 0 too for step II and IV which can't be cause heat is neaded to melt the ice/vaporize the water.

I know I need densities of -30,0°C ice (solid), 0°C ice/water (solid/liquid), 100°C water (liquid). Then I can calculate delta V for all the steps (the volume of gaseous water can be calculated from the ideal gas law) and thus find w and E... but how would I get that far?
Since the whole process takes place at constant pressure, do you need to break the process down into steps to find the total PΔV? The ice maybe expands a bit on warming and then melts to become denser water, which first becomes more dense (to 4°C) and then expands and finally evaporates and expands some more, but the total ΔV from start to finish is the difference between the volume of ice at -30°C and the volume of gas at 140°C. The density of ice is reasonably constant at .92g/mL, so the volume is really small compared to the ~22L of gas at the end. I don't think you need to worry about a small uncertainty in ice density.
 
Thank you so very very much! :) Now I can finish the problem! You're great :D

Oh btw and I guess that delta E can be calculated with this total work and the TOTAL q right?
 
Last edited:
Lisa... said:
Thank you so very very much! :) Now I can finish the problem! You're great :D

Oh btw and I guess that delta E can be calculated with this total work and the TOTAL q right?

I believe that is correct. Conservation of work/energy. The heat either results in some work being done, or it stays in the system
 

Similar threads

Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K