Potential "Thread Topics"

Protected One

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It seems (to me at least) that there have been few, if any, "fresh topics" to discuss, outside of a few in the LOUNGE. Most are some slight variation on a topic frequently discussed. So, for lack of anything better to do at the moment (LOL) I thought...why not post a thread asking what topics the forum members might come up with that would be perhaps different and of interest to others?

The number of "Likes" could be the deciding factor in what topics are of most interest. :)


Let's hear your ideas, folks. :D
 
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The first law of thermodynamics

Can you find the glaring flaw?? Then please discuss the ramifications of the error...

The first law of thermodynamics may be stated in several ways :

The increase in internal energy of a closed system is equal to total of the energy added to the system. In particular, if the energy entering the system is supplied as heat and if energy leaves the system as work, the heat is accounted for as positive and the work as negative.
{\textstyle \Delta U_{\rm {system}}=Q-W} {\textstyle \Delta U_{\rm {system}}=Q-W}
In the case of a thermodynamic cycle of a closed system, which returns to its original state, the heat Qin supplied to the system in one stage of the cycle, minus the heat Qout removed from it in another stage of the cycle, plus the work added to the system Win equals the work that leaves the system Wout.
{\displaystyle \Delta U_{\rm {system\,(full\,cycle)}}=0} {\displaystyle \Delta U_{\rm {system\,(full\,cycle)}}=0}
hence, for a full cycle,
{\displaystyle Q=Q_{\rm {in}}-Q_{\rm {out}}+W_{\rm {in}}-W_{\rm {out}}=W_{\rm {net}}} {\displaystyle Q=Q_{\rm {in}}-Q_{\rm {out}}+W_{\rm {in}}-W_{\rm {out}}=W_{\rm {net}}}
For the particular case of a thermally isolated system (adiabatically isolated), the change of the internal energy of an adiabatically isolated system can only be the result of the work added to the system, because the adiabatic assumption is: Q = 0.
{\displaystyle \Delta U_{\rm {system}}=U_{\rm {final}}-U_{\rm {initial}}=W_{\rm {in}}-W_{\rm {out}}} {\displaystyle \Delta U_{\rm {system}}=U_{\rm {final}}-U_{\rm {initial}}=W_{\rm {in}}-W_{\rm {out}}}
 
Advance thinking -- 5 seconds worth

1 The quality of todays television shows.

2 Professional versus college sports on television.

3 The complexity of today's automobile versus that high school car.

4 Longevity of marriage and your quality of life.

5 You are a parent. Were you successful?

6 Do you need another gun? How many are enough?

7 Your opinion of your military service if it was less than 6 years.

This list has the makings of banishment. :rolleyes:
 
Can you find the glaring flaw?? Then please discuss the ramifications of the error...

The first law of thermodynamics may be stated in several ways :

The increase in internal energy of a closed system is equal to total of the energy added to the system. In particular, if the energy entering the system is supplied as heat and if energy leaves the system as work, the heat is accounted for as positive and the work as negative.
{\textstyle \Delta U_{\rm {system}}=Q-W} {\textstyle \Delta U_{\rm {system}}=Q-W}
In the case of a thermodynamic cycle of a closed system, which returns to its original state, the heat Qin supplied to the system in one stage of the cycle, minus the heat Qout removed from it in another stage of the cycle, plus the work added to the system Win equals the work that leaves the system Wout.
{\displaystyle \Delta U_{\rm {system\,(full\,cycle)}}=0} {\displaystyle \Delta U_{\rm {system\,(full\,cycle)}}=0}
hence, for a full cycle,
{\displaystyle Q=Q_{\rm {in}}-Q_{\rm {out}}+W_{\rm {in}}-W_{\rm {out}}=W_{\rm {net}}} {\displaystyle Q=Q_{\rm {in}}-Q_{\rm {out}}+W_{\rm {in}}-W_{\rm {out}}=W_{\rm {net}}}
For the particular case of a thermally isolated system (adiabatically isolated), the change of the internal energy of an adiabatically isolated system can only be the result of the work added to the system, because the adiabatic assumption is: Q = 0.
{\displaystyle \Delta U_{\rm {system}}=U_{\rm {final}}-U_{\rm {initial}}=W_{\rm {in}}-W_{\rm {out}}} {\displaystyle \Delta U_{\rm {system}}=U_{\rm {final}}-U_{\rm {initial}}=W_{\rm {in}}-W_{\rm {out}}}

Wow, I was just thinking about this.
 
Can you find the glaring flaw?? Then please discuss the ramifications of the error...

The first law of thermodynamics may be stated in several ways :

The increase in internal energy of a closed system is equal to total of the energy added to the system. In particular, if the energy entering the system is supplied as heat and if energy leaves the system as work, the heat is accounted for as positive and the work as negative.
{\textstyle \Delta U_{\rm {system}}=Q-W} {\textstyle \Delta U_{\rm {system}}=Q-W}
In the case of a thermodynamic cycle of a closed system, which returns to its original state, the heat Qin supplied to the system in one stage of the cycle, minus the heat Qout removed from it in another stage of the cycle, plus the work added to the system Win equals the work that leaves the system Wout.
{\displaystyle \Delta U_{\rm {system\,(full\,cycle)}}=0} {\displaystyle \Delta U_{\rm {system\,(full\,cycle)}}=0}
hence, for a full cycle,
{\displaystyle Q=Q_{\rm {in}}-Q_{\rm {out}}+W_{\rm {in}}-W_{\rm {out}}=W_{\rm {net}}} {\displaystyle Q=Q_{\rm {in}}-Q_{\rm {out}}+W_{\rm {in}}-W_{\rm {out}}=W_{\rm {net}}}
For the particular case of a thermally isolated system (adiabatically isolated), the change of the internal energy of an adiabatically isolated system can only be the result of the work added to the system, because the adiabatic assumption is: Q = 0.
{\displaystyle \Delta U_{\rm {system}}=U_{\rm {final}}-U_{\rm {initial}}=W_{\rm {in}}-W_{\rm {out}}} {\displaystyle \Delta U_{\rm {system}}=U_{\rm {final}}-U_{\rm {initial}}=W_{\rm {in}}-W_{\rm {out}}}


"Q=0" is unobtainable.

Larry
 
"Q=0" is unobtainable.

Larry

Or:
Combining these principles leads to one traditional statement of the first law of thermodynamics: it is not possible to construct a machine which will perpetually output work without an equal amount of energy input to that machine. Or more briefly, a perpetual motion machine of the first kind is impossible.
 
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