Determining Heat of Combustion for a Fuel and Oxygen Mixture

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Discussion Overview

The discussion revolves around a combustion experiment involving a mixture of fuel and oxygen, focusing on the heat of combustion and the associated thermodynamic principles. Participants explore concepts such as heat transfer, work done on or by the system, and the calculation of energy released during combustion, with specific attention to the temperature change in a water bath.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests that heat is transferred from the system since the surrounding water's temperature rises, questioning if this is a valid explanation.
  • Another participant proposes that no work is done on or by the system due to it being a constant volume system.
  • Concerns are raised about the sign of internal energy (U) and whether it would be negative, depending on the heat transfer.
  • Participants discuss the appropriate equations to use for calculating the heat of combustion, with one suggesting the use of dU=nCvdT, while expressing uncertainty about the relevance of the water volume.
  • One participant calculates the heat added to the water using Q=mcdT, questioning the accuracy of their mass figure and the significance of heat energy in combustion by-products.
  • There is a request for clarification on the correctness of various work-related equations, with a specific focus on the implications of constant volume on work done.
  • Another participant notes the need to justify ignoring heat energy from combustion by-products and discusses the distinction between energy transferred as heat and energy lost as work.
  • Clarification is sought regarding the conversion of water mass from volume, with a participant questioning the conversion from 200 mL to grams.

Areas of Agreement / Disagreement

Participants express differing views on the work done in a constant volume system and the interpretation of heat transfer. The discussion remains unresolved regarding the calculations and the implications of the various thermodynamic principles involved.

Contextual Notes

Participants have not reached consensus on the correct approach to calculating the heat of combustion, the significance of work done in a constant volume system, or the treatment of combustion by-products. There are also unresolved questions about the assumptions made in the calculations.

doctordiddy
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Homework Statement



n A student performs a combustion experiment by burn-
ing a mixture of fuel and oxygen in a constant-volume metal can surrounded by a water
bath. During the experiment, the temperature of the water is observed to rise. Regard the
mixture of fuel and oxygen as the system. Steps (a-c) help you think about the problem before
calculating the answer in (d).
(a) Has heat been transferred to/from the system? How can you tell?
(b) Has work been done by/on the system? How can you tell?
(c) What is the sign of U of the system? How can you tell?
(d) 1g of fuel is placed in the can for combustion. During combustion, the 200mL water bath
is observed to increase in temperature by 35.5C. What is the heat of combustion (energy
released per gram) of the fuel?

Homework Equations



dU=dQ-dW
dW=nRT ?

The Attempt at a Solution



a) heat is transferred from the system since the surrounding water rises in temperature?

is this correct and a valid explanation?

b)would this be 0 work done on/by the system as it's a constant volume system?

c)would dU be negative since in this case with no work dU=dQ and dQ is negative (assuming my answer in part a is correct?)

d)I'm not sure what to do here. I thought that maybe I could use W=nRT to find work, but then I realized there is no work. Do I perhaps use the formula dU=nCvdT? This seems too easy, and it doesn't account for the 200ml water (which I am not sure is important). Also, by heat of combustion, does it just mean the Q or U since there is 1g of fuel and it says heat of combustion is energy released per gram?

edit:nvm, Cv for this system is not even given
 
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doctordiddy, meet your classmate. https://www.physicsforums.com/showthread.php?t=641745
d)I'm not sure what to do here. I thought that maybe I could use W=nRT to find work, but then I realized there is no work. Do I perhaps use the formula dU=nCvdT? This seems too easy, and it doesn't account for the 200ml water (which I am not sure is important). Also, by heat of combustion, does it just mean the Q or U since there is 1g of fuel and it says heat of combustion is energy released per gram?

edit:nvm, Cv for this system is not even given
Can you determine how many Joules have been added to the water?
 
NascentOxygen said:
doctordiddy, meet your classmate. https://www.physicsforums.com/showthread.php?t=641745

Can you determine how many Joules have been added to the water?

I'm assuming you would use the equation Q=mcdT where cwater is given to be 4.2Jg^-1K^-1

so it would be

Q=0.2*4.2*35.5
Q=29.82
?

On a somewhat unrelated note, can you tell me which of these equations are correct and which are not?

W=nRT
W=nRdT
dW=nRdW
W=PV
W=PdV
dW=PdV

thanks.

Also I noted that in the other thread you mentioned that there is work being done on the surroundings for part b? Why is this the case if it is a constant volume system? How can there be work done and heat transferred into the water?

From my textbook: In any process in which the volume is constant, the system does no work because there is no displacement.
 
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doctordiddy said:
I'm assuming you would use the equation Q=mcdT where cwater is given to be 4.2Jg^-1K^-1

so it would be

Q=0.2*4.2*35.5
Q=29.82
?
That's the right idea, though you'd better check that 0.2 gram figure. :smile:
Note, are you able to justify ignoring the heat energy in the by-products of combustion?
Also I noted that in the other thread you mentioned that there is work being done on the surroundings for part b? Why is this the case if it is a constant volume system?
For the energy equation, I was drawing no distinction between energy transferred as heat, and energy lost as work. Both represent transformation of chemical energy, and it's that total energy we seek. No mechanical work is done here.
 
NascentOxygen said:
That's the right idea, though you'd better check that 0.2 gram figure. :smile:
Note, are you able to justify ignoring the heat energy in the by-products of combustion?

For the energy equation, I was drawing no distinction between energy transferred as heat, and energy lost as work. Both represent transformation of chemical energy, and it's that total energy we seek. No mechanical work is done here.

wouldn't 200ml equal 200g? And since I am using kg it would become 0.2kg?
 
If you use kg, then you must adjust 4.2Jg^-1K^-1[/color] accordingly. http://imageshack.us/a/img593/2293/starwarssmiley010.gif
 
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