Work and Heat

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coolchix321

coolchix321

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Could someone please explain to me the relationship between work and heat in regards to internal energy...
For example
When a piston is compressed... what happens? correct me if i am wrong
volume decreases, thus work is done on the system (+W), thus heat is absorbed/gained (+Q), thus the gas gains energy (+U)?
I am confused...
thanks
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coolchix321 said:
Could someone please explain to me the relationship between work and heat in regards to internal energy...
For example
When a piston is compressed... what happens? correct me if i am wrong
volume decreases, thus work is done on the system (+W), thus heat is absorbed/gained (+Q), thus the gas gains energy (+U)?
I am confused...
thanks
Click to expand...

you never stated if this is adiabatic or not.
 
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is my reasoning correct? What are the steps? Can you clarify please....

What if it is adiabatic? Then what
 
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Its just simple 7th grade algebra here.

U = W + Q

If is adiabatic, then Q = 0 (no heat transfer since the thing is insulated)

U = W.

If you add W, what do you think happens to U.



If Poop = Eat,

and you add Eat, what happens to Poop? It goes up doesnt it?
 
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coolchix321 said:
What if it is NOT adiabatic?
Click to expand...

U = W + Q


If you compress a gas, it becomes hot. If its not adiabatic, that means heat will transfer out. If the process is slow and the Q is allows to release and equilibrate, then U is constant.

Again, this is just algebra here.
 
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ok here is the equation

IE= W + Q

IE= Internal energy

W= work

Q= heat

If you compress a volume of gas. You are doing work ON the system. Whenever you do work On the SYSTEM the sign of work is positive.

If you expand a volume of gas. The gas is doing working on the SURROUNDINGS. Work is negative.

If you add heat to a system. Then the sign is positive.

If you remove heat on a system then the sign is negative.
 
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ezsanche said:
ok here is the equation

IE= W + Q

IE= Internal energy

W= work

Q= heat

If you compress a volume of gas. You are doing work ON the system. Whenever you do work On the SYSTEM the sign of work is positive.

If you expand a volume of gas. The gas is doing working on the SURROUNDINGS. Work is negative.

If you add heat to a system. Then the sign is positive.

If you remove heat on a system then the sign is negative.
Click to expand...


Thanks for your help
So,
If you compress a volume of gas. You are doing work ON the system... the sign of work is positive.... so is Q negative or positive? Why?
Thanks
 
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coolchix321 said:
Thanks for your help
So,
If you compress a volume of gas. You are doing work ON the system... the sign of work is positive.... so is Q negative or positive? Why?
Thanks
Click to expand...

you cant get all your answers in life on SDN. go read Kaplan, EK, or Giancoli. Thanks.
 
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I read the Kaplan book... many times
but it is inconsistent with what you are saying... you are explaining differently
for example, Kaplan says: U=Q-W
wheras here you say U=Q+W... so I want to learn from ONE consistent source
could you please answer this one question
thanks
 
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coolchix321 said:
I read the Kaplan book... many times
but it is inconsistent with what you are saying... you are explaining differently
for example, Kaplan says: U=Q-W
wheras here you say U=Q+W... so I want to learn from ONE consistent source
could you please answer this one question
thanks
Click to expand...
W = -PV

So, U = Q + W = Q + (-PV) = Q - PV = Q - W

It's the same thing. The V in the equation is the change in volume, btw; not the absolute volume.
 
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ezsanche said:
:D

You keep toying with this girl and now you have confounded her. Be nice! :D
Click to expand...

let me guess, shes a bio major.

coolchix321 said:
I read the Kaplan book... many times
but it is inconsistent with what you are saying... you are explaining differently
for example, Kaplan says: U=Q-W
wheras here you say U=Q+W... so I want to learn from ONE consistent source
could you please answer this one question
thanks
Click to expand...

forget the signs. use your intuition. the MCAT could care less about signs.

There is either

1. Adiabatic or Not
2. Fixed Volume or Not

If its Adiabatic, Q = 0
If its Fixed Volume, W=0

what is there to not understand?

1. you pressurize an insulated can, the can heats up, Q=0 and the U goes up = W
2. you relax an insulated can, the can cools down, Q=0 and the U goes down = W.
3. you heat a rigged can, no work is done because its rigid, so U = Q in
4. you cool a rigged can, no work is done, U = Q out

5. you pressurize a non insulated can, it heats up but Q escapes. W goes in, but Q comes out. U is constant.
6. you relax a non insulated can, it cools down, but Q comes in. W goes out, but Q goes in. U is constant.
7. you heat up a non rigid can, the pressure increases and does work. Q goes in, but W leaves by doing work. U = constant.
8. you cool down a non rigid can, the pressure decreases and receives work. Q goes out, but W enters by receiving work. U = constant.

Im not proof reading this s*** and if you still don't get it, quit.
 
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ChemEngSoonMD said:
let me guess, shes a bio major.



forget the signs. use your intuition. the MCAT could care less about signs.

There is either

1. Adiabatic or Not
2. Fixed Volume or Not

If its Adiabatic, Q = 0
If its Fixed Volume, W=0

what is there to not understand?

1. you pressurize an insulated can, the can heats up, Q=0 and the U goes up = W
2. you relax an insulated can, the can cools down, Q=0 and the U goes down = W.
3. you heat a rigged can, no work is done because its rigid, so U = Q in
4. you cool a rigged can, no work is done, U = Q out

5. you pressurize a non insulated can, it heats up but Q escapes. W goes in, but Q comes out. U is constant.
6. you relax a non insulated can, it cools down, but Q comes in. W goes out, but Q goes in. U is constant.
7. you heat up a non rigid can, the pressure increases and does work. Q goes in, but W leaves by doing work. U = constant.
8. you cool down a non rigid can, the pressure decreases and receives work. Q goes out, but W enters by receiving work. U = constant.

Im not proof reading this s*** and if you still don't get it, quit.
Click to expand...

ChemEng has a point. You dont need to worry about a sign you just need to use your intution and just be consistent with how you apply your signs.

FOR ME....

If I do work on a system by compressing it I consider the W to be +

If I heat a system I consider the Q to be +

If the system does work on the surrondings by expanding then I consider W to be -

If the system loses heat and gives it off to the surroundings thenI consider Q to be -

You can obviously have the signs backward. But it doesn't matter as long as you keep it consistent.
 
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ezsanche said:
o what did you do to get a probationary status?
Click to expand...

i have no idea. i was bantering with a friend in a post-bac thread, and apparently even in the absence of offensive language you can be put on probation if you go off topic or some such stupid crap.
 
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bleargh said:
i have no idea. i was bantering with a friend in a post-bac thread, and apparently even in the absence of offensive language you can be put on probation if you go off topic or some such stupid crap.
Click to expand...

the red looks pretty tight.
 
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Like someone above me said. I find this so amusing. But anyhow, coolchix needed help. This thread should have only lasted one comment, not 20....
 
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In a NON adiabatic system... when you compress the piston... you ADD work (+W), so you are thus adding heat (+Q)?​
 
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reading this tread has made me confused on a topic that i was really good at..

i learned it the way kaplan taught it.. at least i think thats what they taught, i get sleepy during the sessions lol..

anyways i use U=Q-W
if work is done ON the system that means there is -W (as this is the only way you can make internal energy aka U go up), if work is done BY the system work is +W (this will lower the internal energy).

U is a state function because its directly related to the temperature of the system while W and Q are both non-state functions.

adiabatic means no heat goes in or out, so many of you say Q=0 which means U=-W if work is done BY the system and U=-(-W) when work is done ON the system.

if you have an isochoric process no work is done W=0 so U=Q
isothermal means U=0 so q-w=0...

U=0 for isotherm
Q=0 for adiabatic
W=0 for isochoric (const. volume)
if Pressure=0 then W=PV
am i correct??
 
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coolchix321 said:
In a NON adiabatic system... when you compress the piston... you ADD work (+W), so you are thus adding heat (+Q)?​
Click to expand...

The W makes the temperature increase. If the temperature is hotter than the environment, the Q will leave the system. Temperature and Q are not the same thing.

14732916.JPG



Read a book.
 
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Why does Heat leave when you add Work in a NON Adiabatic system??
In a NON adiabatic system... when you compress the piston... you ADD work (+W), so you are thus removing heat (-Q)? Why?
Thanks​
 
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coolchix321 said:
Why does Heat leave when you add Work in a NON Adiabatic system??
In a NON adiabatic system... when you compress the piston... you ADD work (+W), so you are thus removing heat (-Q)? Why?

Thanks​
Click to expand...

If you compress the piston you ADD WORK.

That is all.
 
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coolchix321 said:
Why does Heat leave when you add Work in a NON Adiabatic system??
In a NON adiabatic system... when you compress the piston... you ADD work (+W), so you are thus removing heat (-Q)? Why?
Thanks​
Click to expand...

1. When you squeeze stuff, it gets hot.
2. Why do hot things cool? You tell me. You dont know why a hot cup of coffee cools? Why do balls in the air drop? Why do molecules form bonds at lower energy?

You aren't intentionally removing heat. It leaves because its hotter than the environment and its not insulated.
 
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Since this thread is so long, convoluted, and confusing
could someone please give me a summary of the answer to my question
thanks
 
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coolchix321 said:
Since this thread is so long, convoluted, and confusing
could someone please give me a summary of the answer to my question
thanks
Click to expand...

I can read a few posts from someone and gauge pretty accurately what their level is. All I have to say is, you have some major gapping holes missing in your background knowledge. If I were you, I would be worried about other things.

millionaire.jpg
 
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ChemEngSoonMD said:
I can read a few posts from someone and gauge pretty accurately what their level is. All I have to say is, you have some major gapping holes missing in your background knowledge. If I were you, I would be worried about other things.

millionaire.jpg
Click to expand...
Wow that dude is a dummy. Everyone knows that it's D, Buyer's remorse. final answer!
 
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I've always wondered why doing work (in an adiabatic case) only increased kinetic energy but not potential energy. Increasing W increases U (which is KE + PE), so isn't it possible that the temperature stays the same but potential energy increases?
 
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