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Physics Help

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Igakusei
Profile Blog Joined October 2002
United States610 Posts
Last Edited: 2010-05-17 15:45:45
May 16 2010 19:43 GMT
#1
Edit: new question, see below

I'm studying for a general physics (non-calc based) test on Wednesday. I have possibly the worst physics teacher in the universe.

Adiabatic vs Isothermic

Isothermic processes remain the same temperature
Adiabatic processes can change temperature, but there is no heat transfer.

How can you have temperature change with no heat transfer? I'm so totally not understanding this at all, and all the references I can find online are way too advanced for the bonehead version of the class I'm taking. Guy didn't explain any of this at all in lecture, but then it's all over the homework.



Loser777
Profile Blog Joined January 2008
1931 Posts
Last Edited: 2010-05-16 19:45:31
May 16 2010 19:44 GMT
#2
Remember, temperature is a measure of the INTERNAL ENERGY of the gas, not heat... it's a common misconception.
You can raise the temperature, aka increase the INTERNAL ENERGY if you do work on the gas instead of heating it.
Just remember Change in Internal Energy = Heat added + Work done on the gas. So you can raise internal energy by just adding heat or just by doing work on the gas.
6581
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
May 16 2010 19:47 GMT
#3
http://www.physicsforums.com/forumdisplay.php?f=153
seppolevne
Profile Blog Joined February 2009
Canada1681 Posts
May 16 2010 19:59 GMT
#4
If the pressure (constant V) or volume (constant P) change then the temperature has to change, as PV = nRT and n and R do not. There is no transfer of heat, but work is done.
J- Pirate Udyr WW T- Pirate Riven Galio M- Galio Annie S- Sona Lux -- Always farm, never carry.
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
May 16 2010 20:22 GMT
#5
On May 17 2010 04:59 seppolevne wrote:
If the pressure (constant V) or volume (constant P) change then the temperature has to change, as PV = nRT and n and R do not. There is no transfer of heat, but work is done.

In an adiabatic process neither V nor P is constant... There is however a function f(P,V) that is constant so they are still not independent variables so you have a point.

Anyhow, to answer the OP, when you push the gas together you are actively accelerating the molecules inside. Think of it as a baseball and a baseball bat, if the batter is swinging the baseball bat the ball will go faster than if it stood still, it is the same thing when you push the walls together. Faster mpolecules means that you have a higher temperature, and no heat transger was done you just directly accelerated the molecules.
love1another
Profile Blog Joined December 2009
United States1844 Posts
Last Edited: 2010-05-16 21:11:03
May 16 2010 21:09 GMT
#6
How can you have temperature change with no heat transfer? Good question.

How can Flash own my ass at Starcraft without us ever having played?
"I'm learning more and more that TL isn't the place to go for advice outside of anything you need in college. It's like you guys just make up your own fantasy world shit and post it as if you've done it." - Chill
cgrinker
Profile Blog Joined December 2007
United States3824 Posts
May 16 2010 21:10 GMT
#7
Is it possible that Physics might also be a difficult subject or is it just the teacher's fault?
love1another
Profile Blog Joined December 2009
United States1844 Posts
May 16 2010 21:11 GMT
#8
Where are the snowdens of yesteryear?
"I'm learning more and more that TL isn't the place to go for advice outside of anything you need in college. It's like you guys just make up your own fantasy world shit and post it as if you've done it." - Chill
SagaZ
Profile Blog Joined May 2009
France3460 Posts
May 16 2010 21:24 GMT
#9
I recomend you to check out academicearth's lessons about termodinamics, the clases are very well done.
http://academicearth.org/subjects/chemistry
Adiabatic and isometric are comonly covered in termodinamics. Also there are other lessons in case your teacher is really bad and you need external help.
gl
Be nice, buy wards and don't feed double buff.
Igakusei
Profile Blog Joined October 2002
United States610 Posts
Last Edited: 2010-05-16 21:33:34
May 16 2010 21:32 GMT
#10
On May 17 2010 06:10 cgrinker wrote:
Is it possible that Physics might also be a difficult subject or is it just the teacher's fault?


It's the teacher's fault. His class averages are below 40% and he's teaching freshmen. I learn more from wikipedia than I do from his lectures, and he routinely tests on things he never mentioned in class. He hasn't solved a single problem in class all quarter, and spends somewhere between 2/3rds and 3/4ths of his lectures on unrelated tangents. I recognize that most people find physics hard, but it's hard to learn something when it's never even explained to you in the first place.

Thanks for all the help, everyone!

JitnikoVi
Profile Joined May 2010
Russian Federation396 Posts
May 16 2010 22:22 GMT
#11
On May 17 2010 06:10 cgrinker wrote:
Is it possible that Physics might also be a difficult subject or is it just the teacher's fault?


could be both i think
In theory yes, but theoretically, no.
lowbright
Profile Blog Joined March 2010
308 Posts
May 16 2010 22:57 GMT
#12
my inorganic chemistry teacher was very similar to your teacher

but wouldn't this question be more chemistry than physics?
TeamLiquid CJ Entusman #49
Igakusei
Profile Blog Joined October 2002
United States610 Posts
May 16 2010 23:34 GMT
#13
On May 17 2010 07:57 lowbright wrote:
my inorganic chemistry teacher was very similar to your teacher

but wouldn't this question be more chemistry than physics?


Yeah. I don't doubt that I already understand all the concepts involved, it's just the vocabulary. Chemistry doesn't tend to use words like adiabatic.
Igakusei
Profile Blog Joined October 2002
United States610 Posts
May 17 2010 15:45 GMT
#14
New question:

1. The problem statement, all variables and given/known data
80g of ice initially at zero degrees C is allowed to melt in 200g of water initially at 20 degrees C.

Find the energy change of the water, energy change of ice
Find entropy change in water, ice, and universe.

2. Relevant equations
Delta S = Q/T
Q = (Specific heat)(mass)(Delta T)
Lf = heat of fusion
Cw = specific heat of water (yes I know it's 4.184ish J/grams degrees C, but he doesn't like us using numbers).

3. The attempt at a solution
All the instructor did in class was try to explain what entropy was in words. He didn't write any equations down and he didn't do any problems. So I'm just going to attempt to solve this using methods I remember from high school chemistry and I'm hoping maybe someone can tell me if I'm way off my rocker.

so Qice = (80)(Lf) + (80)(Cw)(Tf-0)
and Qwater = (200)(Cw)(Tf-20)

Since I know energy from the water is going into the ice:

(80)(Lf) + (80)(Cw)(Tf-0) = -(200)(Cw)(Tf-20)

Since the final temperature is the only real variable there, I can solve for that and come up with values for the Qs.

As far as entropy goes though, this isn't a constant temperature process. In the equation Delta S = Q/T, if indeed that is the equation I should be using, do I put the final temperature there below the Q change required to reach it? Would the entropy change of the universe be the sum of those two Delta Ss?

In case you're curious we don't turn in homework, so it's not graded. He gives us homework so we have some idea of the types of problems we might be expected to solve, but because he almost never solves any in class and doesn't provide solutions to the homework most people just totally fail the exams.
15vs1
Profile Joined November 2007
64 Posts
May 17 2010 18:13 GMT
#15
You are right, you need to find the final temperature first. Then it easy to find the energy changes. So as i understand the problem is enthropy.
Lets consider the ice (rigorously speaking the subsystem that were ice initialy). The enthropy change during the fusion will be (in your notations): m*Lf/T_f where T_f - is the temperature of fusion.
To find the enthropy change during the heating of resulting water we should write the first law of thermodynamics:
dE=TdS-pdV neglecting the change of volume
dS=dE/T=Cv*dT/T where Cv - specific heat during isochoric process (for liquids Cv almost equals Cp and we assume it does not depend on temperature )
integrating this equation and neglecting additional constant to enthropy (which is irrelevant and cannot be found in phenomenological theory anyway)
S=Cv*lnT
so if water is heated from 273K to Tf the change in enthropy is m*Cv*ln(Tf/273).

And then you should do the same for the second subsystem (200g of water). But there will not be phase transitions and the water will be just cooled. So the enthropy change will be m*Cv*ln(Tf/293).

The resulting change of enthropy (entropy change of the universe) is the sum of the enthropy changes of these two subsystems: "ice" and water. And it will be positive because the process was irreversible.
KissBlade
Profile Blog Joined October 2004
United States5718 Posts
May 18 2010 02:46 GMT
#16
Hey Nevvie, I haven't taken physics for a long long time but I had a decent knack for it. Looking through the material, temperature can be raised by pressure. This is related to what you refer to as "Adiabatic" processes. Since pressure obviously exudes a force, the force causes vibrations upon the atoms and that gets you the temperature change. I'm giving a very very basic gist of it and could be in error but that's what I understand about it.
Boblion
Profile Blog Joined May 2007
France8043 Posts
Last Edited: 2010-05-18 02:55:42
May 18 2010 02:55 GMT
#17
I feel your pain man.
Bad teachers are horrible but the worst thing ever is a bad Algebra teacher.
fuck all those elitists brb watching streams of elite players.
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