Change in Internal Energy of 1kg Water to Steam?

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Discussion Overview

The discussion revolves around calculating the change in internal energy when 1 kg of water is converted to steam at 1 atm and 100°C. Participants explore the relationship between internal energy, enthalpy, and the properties of water and steam, including volume considerations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states the equation for change in internal energy as dU = dQ - PdV, where dQ is the heat added and PdV accounts for work done by the system.
  • Another participant questions the clarity of the original question, suggesting it lacks organization.
  • Several participants express uncertainty about how to derive internal energy from the given enthalpy and the need for final volume calculations.
  • One participant mentions the relationship between enthalpy and internal energy, specifically dH = dU + PdV, and notes the need for density or specific volume of steam to proceed.
  • There is a discussion about calculating the volume of steam and liquid water at the given conditions, with one participant providing density values and calculations for volume.
  • Another participant points out a potential misunderstanding regarding the units of volume calculated for vapor, emphasizing the significant difference between the volumes of liquid water and water vapor.
  • Clarifications are made regarding the number of moles of water involved in the calculations.

Areas of Agreement / Disagreement

Participants express differing levels of understanding regarding the calculations needed to find the change in internal energy. There is no consensus on how to proceed with the calculations, and multiple viewpoints on the necessary information and steps remain present.

Contextual Notes

Participants note limitations in the provided information, particularly concerning the final volume of steam and the density of steam, which are crucial for the calculations. There are unresolved mathematical steps and dependencies on definitions that affect the discussion.

Who May Find This Useful

This discussion may be useful for students studying thermodynamics, particularly those interested in phase changes and energy calculations in closed systems.

Lagraaaange
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Homework Statement


Hvap = 22.6E5 J/Kg. R=8.315E3 Jkmol^-1K^-1. 1kmol water = 18kg
Keeping in mind that latent heats are enthalpies, what is the change in internal energy when 1kg water is converted to steam at pressure of 1atm and temperature of 100C.

Homework Equations


dU = dQ - PdV.

The Attempt at a Solution


dQ = mHvap
PdV = 1atm (Vfinal - Vinitial) Vfinal >>Vinitial -> PdV = 1atm(Vfinal).
Assume ideal gas-> V = nRT/P =18kilomoles * R*373 / 1atm?
 
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Is there an organized question here, or is it just stream of consciousness which you are writing down?
 
You have 1 kg of liquid water in a closed container with a piston at 100 C and 1 atm. You add heat until all the water has vaporized to water vapor at 100 C and 1 atm (holding the pressure constant). What is the change in enthalpy ΔH for the water?

Chet
 
Chestermiller said:
You have 1 kg of liquid water in a closed container with a piston at 100 C and 1 atm. You add heat until all the water has vaporized to water vapor at 100 C and 1 atm (holding the pressure constant). What is the change in enthalpy ΔH for the water?

Chet
The H is given as the heat of vaporization which is the change in enthalpy. How do I get internal energy here? I'm uncertain of the final volume.
 
SteamKing said:
Is there an organized question here, or is it just stream of consciousness which you are writing down?
I have to calculate the change in internal energy of the steam. My main problem is obtaining the final volume of steam.
 
Lagraaaange said:
The H is given as the heat of vaporization which is the change in enthalpy. How do I get internal energy here? I'm uncertain of the final volume.
What is the definition of ΔH in terms of ΔU and Δ(PV)?
 
Lvap = H = U + PV => dH = dU + PdV. I just don't see how the given information will allow me to compute this. I feel that I need the density of steam or specific volume. I don't see what I'm missing here. This is actually an old exam problem from another school so I'm just doing it for practice for my exam tomorrow.
 
Lagraaaange said:
Lvap = H = U + PV => dH = dU + PdV. I just don't see how the given information will allow me to compute this. I feel that I need the density of steam or specific volume. I don't see what I'm missing here. This is actually an old exam problem from another school so I'm just doing it for practice for my exam tomorrow.
From the ideal gas law, what is the volume of 1 kg of water vapor at 1 atm and 100 C?
What is the volume of 1 kg of liquid water at 1 atm and 100 C?
 
density of water = 1000kg/m^3 => V = 1E-3m^3.
vapor: V = nRT/P. n = 1kg/16kg kilomoles.
Then dU = Lm - P(Vvapor - Vwater)?
 
  • #10
Lagraaaange said:
density of water = 1000kg/m^3 => V = 1E-3m^3.
vapor: V = nRT/P. n = 1kg/16kg kilomoles.
Then dU = Lm - P(Vvapor - Vwater)?
Sure. That's right.

Incidentally, I don't understand your answer for the volume of the vapor. It should have units of m^3. I comes out to about 1.67 m^3. Notice that the volume of the liquid water is negligible compared to the volume of the water vapor.

Chet
 
  • #11
Chestermiller said:
Sure. That's right.

Incidentally, I don't understand your answer for the volume of the vapor. It should have units of m^3. I comes out to about 1.67 m^3. Notice that the volume of the liquid water is negligible compared to the volume of the water vapor.

Chet
I think I didn't make it too clear. I was just specifying what n would be; not volume. Thanks.
 
  • #12
Lagraaaange said:
I think I didn't make it too clear. I was just specifying what n would be; not volume. Thanks.
n = 1000gm/18 = 55.6 moles
 

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