Solving Thermo Problem: $\Delta U$, $\Delta K_{cm}$, $\Delta I$, $W_{ext}$, $Q$

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In summary: U equation the change in kinetic energy as seen from c.m is work done by ext forces,work done by internal conservative forces,work done by internal non conservative forces on the particles minus the work done by external forces on c.m but what while change in internal energy is negative of work done by internal conservative forces .then while adding they cancel out then what's the use of defining them.also now consider a gas in a cylinder the usual example with a piston.the pressure P of the gas be say less than P_{ext}then the net force on c.m due to these should be P_{ext}-P.P due to
  • #1
pardesi
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we know that $\Delta U$ $=$ $\Delta K_{cm}$ $+$ $\Delta I$
where $\Delta K_{cm}$ is the kinetic energy from cm frame of refrence and the other part is the change in change in potential energy due to various conservative forces.

also by first law of thermodynamics $\Delta U$ $=$ $W_{ext} + Q$
can we define $W_{ext}$ and $Q$ in terms of the other variables kinetic energy and potential energy
 
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  • #2
pardesi said:
we know that

[tex]\Delta U =\Delta K_{cm} + \Delta I [/tex]
where [itex]\Delta K_{cm}[/itex] is the kinetic energy from cm frame of refrence and the other part is the change in change in potential energy due to various conservative forces.

also by first law of thermodynamics

[tex]\Delta U = W_{ext} + Q [/tex]

can we define [itex]W_{ext}[/itex] and Q in terms of the other variables kinetic energy and potential energy?

I changed all of your dollar signs to [ tex ] [ /tex ] tags (don't put the spaces) so that the LaTeX would work and the post would be readable. Now let me see if it can be understood...
 
  • #3
work energy theorem

[tex] W_{ext} = \Delta KE + \Delta U[/tex]

first law of thermo stated that way is just a restatement of this
 
  • #4
ice109 said:
work energy theorem

[tex] W_{ext} = \Delta KE + \Delta U[/tex]

first law of thermo stated that way is just a restatement of this
but how do you define [tex]W[/tex] and [tex]Q[/tex] using these
 
  • #5
pardesi said:
but how do you define [tex]W[/tex] and [tex]Q[/tex] using these

you don't, it's the same formula, Q=ke and w=w
 
  • #6
ice109 said:
you don't, it's the same formula, Q=ke and w=w
what is [tex]k,e,w[/tex]
 
  • #7
pardesi said:
what is [tex]k,e,w[/tex]

it's simple
[tex]Q=KE[/tex] and [tex]W_{ext}=W[/tex]

your statement of the first law of thermo is the work-energy theorem.
 
  • #8
ice109 said:
it's simple
[tex]Q=KE[/tex] and [tex]W_{ext}=W[/tex]

your statement of the first law of thermo is the work-energy theorem.
i don't think that's true because consider an ideal gas let it undergo adiabatic compression although the heat gained is [tex]0[/tex] the temp. of body which is measured by change in kinetic energy changes
 
  • #9
pardesi said:
i don't think that's true because consider an ideal gas let it undergo adiabatic compression although the heat gained is [tex]0[/tex] the temp. of body which is measured by change in kinetic energy changes
I think the mistake I've made is the the U from the thermo equation equals KE from the work energy theorem and the Q from the thermo equation equals the U from the work energy theorem.

I would love to confirm or deny though.
 
  • #10
ice109 said:
I think the mistake I've made is the the U from the thermo equation equals KE from the work energy theorem and the Q from the thermo equation equals the U from the work energy theorem.

I would love to confirm or deny though.
actually the problem while doing that is we know as in the original [tex]U[/tex] equation the change in kinetic energy as seen from c.m is work done by ext forces ,work done by internal conservative forces,work done by internal non conservative forces on the particles minus the work done by external forces on c.m but what
while change in internal energy is negative of work done by internal conservative forces .then while adding they cancel out then what's the use of defining them.

also now consider a gas in a cylinder the usual example with a pison.the pressure P of the gas be say less than [tex]P_{ext}[/tex]then the net force on c.m due to these should be [tex]P_{ext}-P[/tex].P due to lower wall by Newtons third law,and [tex]P_{ext}[/tex] due to external force above by piston
 
  • #11
someone please...
 

1. What is the definition of $\Delta U$?

$\Delta U$ is the change in internal energy of a system, which is the sum of the kinetic and potential energy of all particles within the system.

2. How is $\Delta K_{cm}$ calculated?

$\Delta K_{cm}$, or the change in kinetic energy of the center of mass, is calculated by taking the difference between the final and initial kinetic energies of the system's center of mass.

3. What does $\Delta I$ represent?

$\Delta I$ represents the change in moment of inertia, which is a measure of an object's resistance to rotational motion.

4. What is the significance of $W_{ext}$ in thermodynamics?

$W_{ext}$ is the work done by external forces on a system, which can affect the system's internal energy and overall thermodynamic state.

5. How is $Q$ related to the other variables in this problem?

$Q$ represents the heat added or removed from a system, and it is related to $\Delta U$, $\Delta K_{cm}$, and $\Delta I$ through the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings.

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