Physics Help Forum Thermodynamis Help Please!! Heat flow, work done, etc.

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Jan 25th 2010, 01:02 PM #1 Junior Member   Join Date: Jan 2010 Posts: 4 Thermodynamis Help Please!! Heat flow, work done, etc. Alright so my friends and I have been trying really hard on our new assignment, and there are some problems that we still can't do. It's like no matter how hard we try the book doesn't explain it well enough for us to be confident in our answers. So Pleeease help! 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? 2.The water in a rigid. insulated cylindrical tank is set in rotation and left to itself. It is eventually brought to rest by viscous forces. The tank and the water constitute the system. A)Is any work done during the process in which the water is brought to rest? B)Is there a flow of heat? C)Is there any change in the internal energy U? 3.Compressing the system represented in the figure along the adiabatic path a-c requires 1000J. Compressing the system along b-c requires 1500J but 600J of heat flow out of the system. A)Calculate the work done, the heat absorbed, and the internal energy change of the system in each process and in the total cycle a-b-c-a. B)Sketch this cycle on a P-V diagram. C)What are the limitations on the values that could be specified for process b-c given that 1000J are required to compress the system along a-c 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. 2.A)Yes there is dissipative work caused by the viscous forces. B)There is no flow of heat because the system is insulated. C)Yes the change in internal energy equals the dissipative work (decrease in energy?) 3.A)For a-c I wrote there is 1000J work done, no heat flow, and 1000J change in energy. For b-c there is 1500J work done, 600J heat flow, 2100J change in energy. For a-b-c-a there is no heat absorbed, 600J work done, no change in internal energy. I haven't yet sketched the diagram but I am not sure what to do for the limitations on the values that could be specified for b-c given that 1000J are required to compress the system along a-c. Diagram for 3: Thanks for the help
Jan 26th 2010, 07:27 AM   #2
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 Originally Posted by mmmboh Alright so my friends and I have been trying really hard on our new assignment, and there are some problems that we still can't do. It's like no matter how hard we try the book doesn't explain it well enough for us to be confident in our answers. So Pleeease help! 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? 2.The water in a rigid. insulated cylindrical tank is set in rotation and left to itself. It is eventually brought to rest by viscous forces. The tank and the water constitute the system. A)Is any work done during the process in which the water is brought to rest? B)Is there a flow of heat? C)Is there any change in the internal energy U? 3.Compressing the system represented in the figure along the adiabatic path a-c requires 1000J. Compressing the system along b-c requires 1500J but 600J of heat flow out of the system. A)Calculate the work done, the heat absorbed, and the internal energy change of the system in each process and in the total cycle a-b-c-a. B)Sketch this cycle on a P-V diagram. C)What are the limitations on the values that could be specified for process b-c given that 1000J are required to compress the system along a-c 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. 2.A)Yes there is dissipative work caused by the viscous forces. B)There is no flow of heat because the system is insulated. C)Yes the change in internal energy equals the dissipative work (decrease in energy?) 3.A)For a-c I wrote there is 1000J work done, no heat flow, and 1000J change in energy. For b-c there is 1500J work done, 600J heat flow, 2100J change in energy. For a-b-c-a there is no heat absorbed, 600J work done, no change in internal energy. I haven't yet sketched the diagram but I am not sure what to do for the limitations on the values that could be specified for b-c given that 1000J are required to compress the system along a-c. Diagram for 3: Thanks for the help
Hi and welcome to PHF!
Generally it's one problem by thread - See the forum's rule number 15. -
1)a) I think you are right. At least I'm sure there was no heat flow from the exterior of the system into the system.
b)What is the formula that gives the work done by a gas?
c)What is the formula that gives the internal energy of a gas? It depends on only 1 variable, what is it?

3) I didn't check in details the problem. However my instinct tells me to use the first law of thermodynamics.
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If the problem is too hard just let the Universe solve it.

Jan 26th 2010, 01:02 PM   #3
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 Originally Posted by arbolis Hi and welcome to PHF! Generally it's one problem by thread - See the forum's rule number 15. - 1)a) I think you are right. At least I'm sure there was no heat flow from the exterior of the system into the system. b)What is the formula that gives the work done by a gas? c)What is the formula that gives the internal energy of a gas? It depends on only 1 variable, what is it? 3) I didn't check in details the problem. However my instinct tells me to use the first law of thermodynamics. I will wait for others to help you further.
1.b) W = integral of P dv. Does this mean that work absolutely cannot be done by a system if there is either no change in pressure OR volume?
1.c) change in internal energy of a gas = Q - W. If Q is zero, and W is zero, then the change in U would be zero also. I just have a hard time making sense of this...

Jan 26th 2010, 04:37 PM   #4
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 Originally Posted by philn 1.b) W = integral of P dv. Does this mean that work absolutely cannot be done by a system if there is either no change in pressure OR volume?
In the integral, P is multiplied dV. As V is constant in your case, what is worth dV? What does this tells you about the value of W?

 Originally Posted by philn 1.c) change in internal energy of a gas = Q - W. If Q is zero, and W is zero, then the change in U would be zero also. I just have a hard time making sense of this...
Here I'd look at the formula at Ideal gas - Wikipedia, the free encyclopedia.

So I think you can disregard the first law of thermodynamics for part c). I'm not 100% sure it applies since the system is changing (due to a chemical reaction).
I'd like to hear r.samanta or any other Helpers on this.
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 Jan 26th 2010, 04:50 PM #5 Junior Member   Join Date: Jan 2010 Posts: 3 So the work is zero and the internal energy increases as a result of an increase in temperature. This makes intuitive sense... Wikipedia mentions here that internal energy may also change due to "energy added by "other" processes": Internal energy - Wikipedia, the free encyclopedia. Any ideas for #3?
Jan 26th 2010, 04:59 PM   #6
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 Originally Posted by philn So the work is zero and the internal energy increases as a result of an increase in temperature. This makes intuitive sense... Wikipedia mentions here that internal energy may also change due to "energy added by "other" processes": Internal energy - Wikipedia, the free encyclopedia. Any ideas for #3?
Yes.
And you're right about the wikipedia article. I never seen the first law under this form. I guess I will learn about it in a more advanced course.
For 3), I think you made at least an error. You said that for B-C there is a change of energy of (1500+600) J. But the 600 J are leaving the system. Shouldn't it be (1500-600) J of change of internal energy?
I didn't check the rest, I have no time now.
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If the problem is too hard just let the Universe solve it.

 Apr 11th 2017, 10:05 AM #7 Junior Member   Join Date: Apr 2017 Posts: 1 A) yes A) sorry for my English *YES. There is a quantity of heat that is transferred to the system.* Quimical energy is transferred , raising the temperature and consequently the internal energy. That's why the expression of the internal energy is proportional to the temperature. You can think that the internal energy already take into account the amount of quimical energy , but it doesn't. It is only refered to the kinetic and potential energy. So , based on the first law of thermodynamics , the answer is YES.
 Apr 11th 2017, 10:16 AM #8 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 479 This thread is more than 7 years old. What is your question today?

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