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image [SOLVED] Quantity of Heat/Specific Heat Water+Iron Share It Thread Tools Search this Thread image
Old Feb12-08, 07:32 PM                  #1
Kaleb

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Posts: 38
[SOLVED] Quantity of Heat/Specific Heat Water+Iron

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

What will be the final temperature of 100g of 20 degrees C water when 100g of 40 degrees C iron nails are submerged in it. (The specific heat of iron is .12 cal/g C. Here you should equate the heat gained by the water to the heat lost by the nails.)


2. Relevant equations

Q = cm(change in)t

Q = quantity of heat

c = specific heat
(water = 4.184J/g or 1 cal/g)
(iron = .502J/g or .12 cal/g)

m = mass
100g + 100g = 200g total substance(water+iron)

t = temperature
(40(final)-20(initial))

3. The attempt at a solution

I feel I slightly understand what I am supposed to do but the part where it says I need to find the heat lost by iron and the heat gained by water im totally confused about. Heres my attempt: My assumption is that I need to do two different calculations and difference them out. Also if someone could point me in the right direction to be able to post real equations instead of what I have shown that would be so awesome.

Iron First:
Q = cm(Tf-Ti)

Q = .502J*100g(40-20)

Q = 1004J of Energy Lost

Water Second:

q = 4.184*100g(40-20)

Q = 8368J of Energy Gained

Answer:
8368J - 1004J = 7364J of Energy or 1760.04cal/g

I have tried numerous ways to find the temperature difference, lost and gained but I am clueless. Also I cant figure out where to transfer back into Celcius. By my calc's it would have gained like 1700 degrees haha. >.> I feel dumb... Anyways TIPS are always appreciated in advance. Constructive critism is appreciated as well, and well anything else that can make me a better poster is welcome too. ^^ thanks!
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Old Feb12-08, 07:44 PM                  #2
HallsofIvy

PF Mentor

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Posts: 25,722
Your only problem appears to be that you have not read the problem correctly and don't seem to know what is being asked! Your start is correct but why are you using 40 degrees and 20 degrees as "initial" and "final" temperatures for both water and iron?

You also say
"Answer:
8368J - 1004J = 7364J of Energy or 1760.04cal/g "
which you should realize makes no sense. The question was "What will be the final temperature" and that's not even a temperature.

20 degrees is the initial temperature for the water and 40 degrees is the initial temperature for the iron. The final temperature of both will be the same, of course, so just call that "T".

Your equations should be
Iron: Energy lost Q = .502J/g*100g(T- 40) and
Water: Energy gained q = 4.184J/g*100g(T-20) and they must be equal. Solve for T.
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Old Feb12-08, 08:00 PM       Last edited by Kaleb; Feb12-08 at 08:06 PM.. Reason: Got answer            #3
Kaleb

Kaleb is Offline:
Posts: 38
I see the fault, Thanks a bunch Ivy. You have saved my day and made me feel all the more dumb O.o. That was a joke btw.

Edit: I came out with 32 degrees C as the final for both. Any chance I can get confirmation on this? Ill wait awhile and if nobody posts Ill mark as solved. Thanks again Ivy, your my hero!
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