# Homework Help: Series of Systems/Heat/Work Questions

1. Jul 11, 2014

### RJLiberator

I am trying to fully grasp the concepts behind the series of questions here.

"You're baking a batch of shepard's pie in your oven and fall asleep while it's baking. When you are awakened by the fire alarm, you quickly clear out the smoke and open the oven. Your shepard's pie is reduced to ash. Within the tragic scenario of destroying your pie, would the oven be considered part of the system or is it the surroundings? (Write system or surroundings)"

Why? Because the oven is not part of the molecular environment.

"Based upon your answer in question 28, is the ashing of the pie an exothermic or endothermic process?"
Why? Endothermic means that heat is being absorbed here. The outside surrounding (oven) with its heat is adding to the molecular "pie."

"Your kitchen is an anomaly in that it always maintains atmospheric pressure. If your raw shepard's pie occupied 2.55L of space and released 622L of gas as it burned, what amount of work was done under these conditions? (1 L*atm = 101.325J)."
This is where I am struggling big time. There seems to be many different equations here and I am not sure what equation to begin with. I see I need to find work done, but I don't understand how this translates to gas and the numbers given. Any guidance on where to begin? (equation, link, anything)

"If there was a total energy change of 3.32x10^5J during the burning of your pie, what amount of heat was released?

I think I will be able to answer this one after I get help with the previous question.

What was the enthalpy change in the reaction given your results? (remember the sign conventions when determining your answer)

With this one, I assume I need to find the balanced equation first, correct?

2. Jul 11, 2014

### Staff: Mentor

W=PΔV

(easy to derive if you remember W=force × displacement and pressure definition)

3. Jul 11, 2014

### RJLiberator

Thanks for replying, Borek. You seem to help out a lot here. It is appreciated. I hope to help out here someday as I get further advanced.

W=PChange in Volume

Okay. Well, we know that pressure is atmospheric pressure at 1atm
We know we need to find work.

The change in volume, hm. I wonder if I add or subtract the two numbers. Since the pie occupies initially 2.55L of space and RELEASED 655L of energy, I would imagine I subtract. So 655L-2.55L and then multiply by 101.325 to get the answer.

Is my reasoning correct?

Last edited: Jul 11, 2014
4. Jul 11, 2014

### RJLiberator

Okay, Update:

So with the atmospheric pressure being equal to 1atm we can negate that from the equation essentially (taking in account for units.)

W = 1atm*(622L-2.55L)
We then multiple the answer by 101.325 to get the answer in Joules.

:D?

5. Jul 11, 2014

### Staff: Mentor

Don't guess, check the units.

Yes, 622L-2.55L is the ΔV.

6. Jul 11, 2014

### RJLiberator

Thank YOU. That part of the equation has been bugging me all morning.

622-2.55 is correct volume change. So this means that we have

W = 1atm(622L-2.55L)
This would equal a number in L*atm Which we can convert to joules per the problems request by multiplying by 101.325.

Okay, most importantly, I can see how the wording of the equation translates into the equation. So it is not cumulative and needs to be subtracted.

Thank you.

My final answer was calculated to be 62765. With significant figures this comes out to be 62800J.

Last edited: Jul 11, 2014
7. Jul 11, 2014

### RJLiberator

For the next problem:
"If there was a total energy change of 3.32x10^5J during the burning of your pie, what amount of heat was released?

Changei n Energy = q+w where q is heat and w is work
(3.32*10^5) = (622-2.55)(101.325)+q
= 269,200J

Sig figs = 2.69*10^5J

Last edited: Jul 11, 2014
8. Jul 11, 2014

### RJLiberator

For the question:
What was the enthalpy change in the reaction given your results? (remember the sign conventions when determining your answer)

"Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined by the equation:

ΔH=q"

So with this equation, my answer in question 4 (269000J) is ALSO the answer for #5. For some reason, I think I need to put a negative in front of it and it has to do with the heat being absorbed or released. I just can't wrap my mind around whats happening here.

Last edited: Jul 11, 2014