# Homework Help: Heat Thermodynamics Questions

1. Jan 25, 2010

### 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 :D

2. Jan 25, 2010

### mmmboh

Anyone?

3. Jan 26, 2010

### Jobrag

My old mechanics teacher always used to tell us "draw a box round it" , look at your problems draw a boundry round them and ask "does anything cross the boundry?"

4. Jan 26, 2010

### mathman44

Can anyone help clarify these?

5. Jan 26, 2010

### Jobrag

Draw a boundry round the the container, does anything pass across the boundry?
If nothing crosses the boundry going in no heat can have entered the system.
If nothing crosses the boundry going out the system has not done any work.
If the system has performed no work and no heat has entered the system can the internal energy have changed?

6. Jan 26, 2010

### mmmboh

Ok got it thanks a lot!. I figured out the third question, but in the case of the "boundary", that would make my second answer correct right?

7. Jan 26, 2010

### fluidistic

I'm confused about this. Because the problem states that there is an increase of temperature. Plus, the internal energy of an ideal gas is directly proportional to the temperature. Hence an increase of temperature should mean an increase of internal energy. But as you implied, according to the first law of thermodynamics, it cannot be. So what's going on?

8. Jan 26, 2010

### Jobrag

fuidistic
Heat isn't the only form of energy.

9. Jan 26, 2010

### fluidistic

Hmm I know (for instance work), but how does it matter here?
I don't want to hijack the thread, so if you want to reply me by PM, it's ok.

10. Jan 26, 2010

### Jobrag

fluidistic
If you accept that the thermal energy of the system has increased (the gas has got hotter) and that there has been no energy imported into the system, what else might have happened.
Work is heat and heat is work is a true statement but NOT excusivly true.

11. Jan 27, 2010

### fluidistic

I accept that the thermal energy (temperature) of the system has increased and that no energy has been imported into the system. It is explicitly said that
I also agree that there is no work done since the volume is constant. Hence one might say that the variation of internal energy of the system is 0, according to the first law.
My problem is looking at the definition of the internal energy of an ideal gas: it depends on temperature. So a hotter gas means a gas with a greater internal energy.
But how can that be? It seems to be in contradiction with the first law of thermodynamics here.
Do you understand where I'm misunderstanding?

12. Jan 27, 2010

### Jobrag

fluiodistic
What happens when hydrogen and oxygen are mixed and a spark applied?

13. Jan 27, 2010

### fluidistic

It forms water if I remember well. (from high school since I've no chemistry in my physics degree).
Hmm ok... But to be rigorous, don't you think we'd have to show that it is under the liquid form? Because if it would be a gas, it seems the first law of thermodynamics couldn't be applied. Is it sufficient to affirm that it is under the liquid form otherwise the first law wouldn't be right?

14. Jan 27, 2010

### Jobrag

It forms water but what is given off during the reaction?

15. Jan 27, 2010

### fluidistic

I don't remember. Heat I guess.

16. Jan 27, 2010

### mmmboh

Hydrogen and oxygen form water and release energy. Having said that I hope my second answer is right...

17. Jan 27, 2010

### Jobrag

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.

18. Jan 28, 2010

### fluidistic

Ok thank you very much. So as in this (https://www.physicsforums.com/showthread.php?p=2552092#post2552092) thread says, there is the chemical energy that Resnick-Halliday never mentioned.

19. Jan 28, 2010

### Andrew Mason

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: $\Delta Q = \Delta U + W$

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

20. Jan 28, 2010

### fluidistic

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 https://www.physicsforums.com/showthread.php?p=2552092#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.