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Author Topic: kren needs homework help =O  (Read 792 times)

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kren needs homework help =O
« on: January 18, 2007, 11:26:07 pm »
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well, in chemistry class we are seeing Physics stuff, eventhough it's odd, i need to know how to get c m and ΔT from the next forumla pleassee!!

Q = m c ΔT

Q =heat energy
m = mass
c = specific heat
ΔT = temperature

It is the formula for specific heat. so please help Kren :o!!
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Re: kren needs homework help =O
« Reply #1 on: January 18, 2007, 11:50:51 pm »
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Well, I did this last year.  First you figure out which has the greatest temperature, for this is the one that loses heat.  This can be denoted as +Q (loss of heat), and the one that has a heat increase is denoted as -Q.  So, the loss of heat is equal to the heat gain (+Q = -Q).  You may be thinking, what the !@#$%, they cancel out, but you would be wrong, sort of.  Set up the formula like this:

+Q = -Q   //Heat lost equals heat gained
mC(T0-TF) = mC(T0-TF)//The right side of the equation is negative because you take the inital temperature minus the final temperature and get a little number minus a big number (which equals a negative number).

You end up solving for the TF or anything you are looking for I guess.
« Last Edit: January 18, 2007, 11:58:06 pm by 4Sword »
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Re: kren needs homework help =O
« Reply #2 on: January 18, 2007, 11:57:19 pm »
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hmm I understanded all until this part :
Quote
mC(T0-TF) = mC(T0-TF)//The left side of the equation is negative because you take the inital temperature minus the final temperature
i see no difference between the left and right <-< well, what i need to know is how to get c m and T
like..  M = C T Q (it's just an example so im sure it is wrong) and such on i know this stuff is really easy, just to pass one to the other side, but im confused for various reasons ...
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Re: kren needs homework help =O
« Reply #3 on: January 19, 2007, 12:02:42 am »
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The difference is that you have two substances.  For example, if we had ice and water (I think their caloric values are different), by putting the ice in the water, the ice gains heat and the water loses heat.

mCΔT =  mCΔT   //  Let us say that the left side is water and the right is ice, now let us look at the change in temperature values alone.
(50-40) and (40-50)

When trying to do the equation, it helps just to leave it in mCΔT form.  Leave both parts on their respective sides until the end of your problem.
« Last Edit: January 19, 2007, 12:05:01 am by 4Sword »
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Re: kren needs homework help =O
« Reply #4 on: January 19, 2007, 12:18:59 am »
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hmm good theory.. let me try it here..
3) If 50 J of heat are applyed to 10 g of Fe. How much will the temperature rise?
Q= 50 J
g= 10 g
ce = .449 J/G C

mCΔT =  mCΔT 
(10 g)(.449) ΔT = (10 g)(.449) ΔT
4.49 ΔT = 4.49 ΔT
then? what <.< erm <-< really hard, well the main reason i don't know this is because that day i can't see thanks to the stupid liquid in my contact lense, so i was just hearing >-<
4 swords I really appreciate your help, i didn't know that your technique existed :P
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Re: kren needs homework help =O
« Reply #5 on: January 19, 2007, 12:27:21 am »
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Oh, I am kind of dumb.  Sorry, I was thinking about how to to have one substance added to another.  Your problem is a lot easier.  All you are looking for then is ΔT which is the change in temperature itself.  So:

Q=mCΔT
50=10(.449)ΔT
50=4.49ΔT
ΔT~11.13585746102449888641425389755
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Re: kren needs homework help =O
« Reply #6 on: January 19, 2007, 12:34:49 am »
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so the formula is
q/g* c = T
right ?!
yay thanks alot dude! you really helped me a bunch!
thats what I needed to know :D:D:D:D!! buy I owe you a something if you ever need a sprite just tell me :d!!
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Re: kren needs homework help =O
« Reply #7 on: January 19, 2007, 12:44:54 am »
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It is fine, I do these things for free. 

But, if I were you, I would just leave the formula as Q=mCΔT.  That way you do not have to try to remember formulas for specific things, but rather you can just use one formula and "plug" things in.  Your equation would be fine, the only thing wrong is that "T" should be "ΔT" because it is a change in temperature and not the actual temperatures themselves.  So, Q/(m*C)=ΔT; make sure to have your parathesis in if you are using a graphing calculator.  Since that sample question you were asking was only asking for the "change in temperature", you are solving for ΔT alone.  Just remember that ΔT itself is actually (T0 - TF). 
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