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

1. Nov 26, 2006

### Lisa...

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 seperately 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: Nov 26, 2006
2. Nov 26, 2006

### Lisa...

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 $$q= \Delta H = \frac{5}{2} \Delta T$$

3. Nov 26, 2006

### Lisa...

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: Nov 26, 2006
4. Nov 26, 2006

### OlderDan

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.

5. Nov 27, 2006

### Lisa...

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: Nov 27, 2006
6. Nov 27, 2006

### OlderDan

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