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Old Feb 27th 2019, 04:28 AM   #1
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Internal Energy in isolated system

Hello,
I have an isolated system (no Heat, Work exchange with environment). The
1. Law of Thermodyn says deltaU = 0 for such a system.

Then I consider NaOH (solid) and water in this isolated system. The solving process creates entropy (entropy production=irreversible). Temperature rises.

delta U shall stay constant, but S and T increase.

What I am missing??


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Lothar
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Old Feb 27th 2019, 05:20 AM   #2
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The law of thermodynamics says that internal energy will not change for an isolated system assuming that the net heat injection into the system is zero.

If there is a chemical reaction in the control volume that injects an amount of heat $\displaystyle q_{in}$ into the system, then the internal energy will increase, but because the system is isolated, no exchange of heat will occur with its environment and the system must increase its temperature instead.
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Last edited by benit13; Feb 27th 2019 at 05:31 AM.
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Old Feb 28th 2019, 09:32 AM   #3
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Hello,
you write:
...then the internal energy will increase...
Internal Energy in isolated system can not increases, additional Energy can not come "from inside the system". That's energy conversion.
How will the internal energy increase, without transferring if from the outside?

Last edited by LotharSchuh; Feb 28th 2019 at 09:37 AM.
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Old Feb 28th 2019, 09:35 AM   #4
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I just want to make sure:
What are your definitions of:
U, S & T.
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Old Mar 1st 2019, 02:36 AM   #5
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Originally Posted by LotharSchuh View Post
Hello,
you write:
...then the internal energy will increase...
Internal Energy in isolated system can not increases, additional Energy can not come "from inside the system". That's energy conversion.
Sure it can. The energy conversion occurs from a type that is not tracked within the system (e.g. chemical potential energy) and is converted to a type that is tracked (thermal energy). Therefore, the energy injection into the system from the chemical reaction cannot be ignored.

In general:

$\displaystyle \Delta Q = Q_{gen} + Q_{in} - Q_{out}$

"isolated" means that $\displaystyle Q_{in} - Q_{out} = 0$ but it doesn't mean that $\displaystyle Q_{gen} = 0$.

This is why the fact "The internal energy of an isolated system cannot increase" assumes no energy conversion taking place within the control volume.

How will the internal energy increase, without transferring if from the outside?
Because of the injection from the generation process.

$\displaystyle \Delta U = \Delta Q = Q_{gen}$

If the energy exchange is sensible,

$\displaystyle Q_{gen} = m c_p \Delta T$

or if latent:

$\displaystyle Q_{gen} = m l$

Note that $\displaystyle Q_{gen}$ can also be negative
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Old Mar 1st 2019, 05:47 AM   #6
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Hello,
I got the point you made.

You do not include the potential energy of/between the atoms/ions as part of the internal energy U.

Right?
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Old Mar 1st 2019, 10:12 AM   #7
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Originally Posted by LotharSchuh View Post
Hello,
I got the point you made.

You do not include the potential energy of/between the atoms/ions as part of the internal energy U.

Right?
Right. Just to be specific...

Include:
- Kinetic energy (translation and rotation)
- Vibrational kinetic energy

Do not include:
- Gravitational potential energy
- Elastic potential energy
- Chemical potential energy
- Nuclear potential energy
- Electric/magnetic field potential energy
... etc.

If there are conversions from any of those forms of potential energy to kinetic energy, then be careful... sometimes it's not appropriate to include. For example, an exothermic chemical reaction will inject some energy into the control volume which will raise the temperature of the gas. However, if a packet of gas undergoes bulk motion due to gravity (free-fall) it will not necessarily cause an increase in temperature. GPE is the classic case where care had been made, but it's true for all sources of potential energy.
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