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Heat Thermodynamics Questions

 
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Jan28-10, 07:51 AM   #18
 
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Heat Thermodynamics Questions


Quote by Jobrag View Post
OK lets wrap this up. The hydrogen and oxygen turn into water in its gaseous phase (steam). The energy released by the chemical reaction provides the heat to raise the temperature and pressure, nothing needs to enter the system as energy is changed from one state to another.
Ok thank you very much. So as in this (http://www.physicsforums.com/showthr...92#post2552092) thread says, there is the chemical energy that Resnick-Halliday never mentioned.
Jan28-10, 09:21 AM   #19
 
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Quote by mmmboh View Post
1. A mixture of hydrogen and oxygen is enclosed in a rigid insulating container and exploded by a spark. The temperature and pressure both increase. Neglect the small amount of energy provided by the spark itself.
A) Has there been a flow of heat into the system.
B)Has any work been done by the system.
C)Has there been any change in internal energy U of the system?

....

1.A)There is no heat flow because the system is insulated.
B)Yes adiabatic work is done. (I'm not sure why though)
C)Yes the change in internal energy is equal to the adiabatic work done (I think it's an increase in energy.
A) There is heat flow into the system. The post-explosion system is definitely at a higher temperature than the pre-explosion system (higher internal energy) and there is no work done (volume remains constant). Apply the first law to see that Q cannot be 0: [itex]\Delta Q = \Delta U + W[/itex]

B) If there is no change in volume, how can there be any work done?\\

C) The change in energy is equal to the heat flow.


AM
Jan28-10, 09:43 AM   #20
 
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Quote by Andrew Mason View Post
A) There is heat flow into the system. The post-explosion system is definitely at a higher temperature than the pre-explosion system (higher internal energy) and there is no work done (volume remains constant). Apply the first law to see that Q cannot be 0: [itex]\Delta Q = \Delta U + W[/itex]

B) If there is no change in volume, how can there be any work done?\\

C) The change in energy is equal to the heat flow.


AM
Dear Andrew Mason,
I think you are wrong by saying that there was a heat flow. I don't know if you have read the previous posts, in case of yes, please correct Jobrag and I. And I also suggest you to read http://www.physicsforums.com/showthr...92#post2552092 where Matterrave suggest to use a more generalized form of the firs law to explain the increase of temperature of the gas when both the work done by the gas and the heat flow are null.
Jan28-10, 11:39 AM   #21
 
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Quote by fluidistic View Post
Dear Andrew Mason,
I think you are wrong by saying that there was a heat flow. I don't know if you have read the previous posts, in case of yes, please correct Jobrag and I. And I also suggest you to read http://www.physicsforums.com/showthr...92#post2552092 where Matterrave suggest to use a more generalized form of the firs law to explain the increase of temperature of the gas when both the work done by the gas and the heat flow are null.
The heat flow from chemical reaction is referred to as the [itex]\Delta H[/itex] or change in Enthalpy of the reaction and is equal to the change in internal energy + work done by the thermodynamic system. This is just an application of the first law of thermodynamics to chemical reactions. The law is unchanged. There is no more general statement of the first law. The thread you referred to is incorrect if it suggests that [itex]\Delta Q = \Delta U + W[/itex] is not sufficient. For chemical reactions, one has to take into account the number of molecules before and after and the enthalpy of the reaction when determining the internal energy, of course. But one is still measuring [itex]\Delta U \text{ and } \Delta Q[/itex]. See: http://en.wikipedia.org/wiki/Enthalpy and http://en.wikipedia.org/wiki/Chemical_thermodynamics

AM
Jan28-10, 02:12 PM   #22
 
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Quote by Andrew Mason View Post
The heat flow from chemical reaction is referred to as the [itex]\Delta H[/itex] or change in Enthalpy of the reaction and is equal to the change in internal energy + work done by the thermodynamic system. This is just an application of the first law of thermodynamics to chemical reactions. The law is unchanged. There is no more general statement of the first law. The thread you referred to is incorrect if it suggests that [itex]\Delta Q = \Delta U + W[/itex] is not sufficient. For chemical reactions, one has to take into account the number of molecules before and after and the enthalpy of the reaction when determining the internal energy, of course. But one is still measuring [itex]\Delta U \text{ and } \Delta Q[/itex]. See: http://en.wikipedia.org/wiki/Enthalpy and http://en.wikipedia.org/wiki/Chemical_thermodynamics

AM
Ok thanks a lot for the information (I think it was necessary, since we were all wrong before you came it seems!). I didn't realize the number gas molecules would go down.
Also, I'd like to understand better how it is possible that there's a heat flow into the system if the system is insulated.
What about if I imagine a system with nothing out of it, i.e. the system is the whole universe?
Jan28-10, 04:33 PM   #23
 
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Quote by fluidistic View Post
Ok thanks a lot for the information (I think it was necessary, since we were all wrong before you came it seems!). I didn't realize the number gas molecules would go down.
Also, I'd like to understand better how it is possible that there's a heat flow into the system if the system is insulated.
The system is only insulated from outside heat flows. It is not insulated from internal heat sources. The internal heat sources are the molecules of gas, which generate heat from a chemical reaction. The additional internal (kinetic) molecular energy did not exist before the chemical reaction occurred. It resulted from the conversion of electromagnetic potential energy into molecular kinetic energy. This could be thermodynamically equivalent to sticking a battery and a switched heating element inside a container filled with an inert gas and then closing the switch. The fact that the container is insulated from the outside does not mean there is no heat flow into the gas.

AM
Jan28-10, 04:39 PM   #24
 
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Quote by Andrew Mason View Post
The system is only insulated from outside heat flows. It is not insulated from internal heat sources. The internal heat sources are the molecules of gas, which generate heat from a chemical reaction. The additional internal (kinetic) molecular energy did not exist before the chemical reaction occurred. It resulted from the conversion of electromagnetic potential energy into molecular kinetic energy. This could be thermodynamically equivalent to sticking a battery and a switched heating element inside a container filled with an inert gas and then closing the switch. The fact that the container is insulated from the outside does not mean there is no heat flow into the gas.

AM
Thank you very much. You've been extremely helpful.
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dissipative, flow, heat, physics, thermodynamic

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