# Homework Help: Measuring heat of combustion

1. Oct 7, 2012

### doctordiddy

1. The problem statement, all variables and given/known data

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
(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?

2. Relevant equations

dU=dQ-dW
dW=nRT ???

3. 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 im 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

Last edited: Oct 7, 2012
2. Oct 7, 2012

### Staff: Mentor

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

3. Oct 7, 2012

### doctordiddy

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.

Last edited: Oct 7, 2012
4. Oct 8, 2012

### Staff: Mentor

That's the right idea, though you'd better check that 0.2 gram figure.
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.

5. Oct 8, 2012

### doctordiddy

wouldn't 200ml equal 200g? And since im using kg it would become 0.2kg?

6. Oct 8, 2012

### Staff: Mentor

If you use kg, then you must adjust 4.2Jg^-1K^-1 accordingly. http://imageshack.us/a/img593/2293/starwarssmiley010.gif [Broken]

Last edited by a moderator: May 6, 2017