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Topic: Calculating heat lost/gained in calorimeter experiment  (Read 16514 times)

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r1509

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Calculating heat lost/gained in calorimeter experiment
« on: November 25, 2005, 11:22:19 PM »
We performed an experiment where a piece of hot metal (lead) was dropped into a coffee-cup calorimeter.  We are supposed to be able to use the data from the experiment to calculate the specific heat of the metal, then check this experimental value against the actual known specific heat.  

I know that specific heat =             (heat transferred)
                                                   --------------------
                                      (g of substance) (change in temperature)

But I am having trouble calculating the heat transferred (in J).

Here is the data I have:

Mass of lead:  21.599 g
Initial temp of lead:  99.9 C
Initial temp of water: 21.6 C
Final temp (water + metal):  24.1
Change in temp of lead:  -75.9 C (gave off 75.9 degrees to water)
Change in temp of water:  2.5 C   (absorbed 2.5 degrees from lead)
Heat capacity of calorimeter:  456 J/C

We have been given many formulas, but I am not sure which one(s) to use.  This one looked helpful, but I have not been able to get it to come out right:

-C(metal) x delta T(metal) = C(cal) x delta T(cal)

I understand this to mean that the heat capacity of the metal, which is 0.0134 J/C, multiplied by the change in temp of the metal, will equal the heat capacity of the calorimeter multiplied by the change in temperature of the calorimeter.  However, setting it up, this is what I get:

(-0.0134 J/C)x(-75.9 C) = (456 J/C)x(2.5 C)

Can anyone tell me what I am doing wrong, or perhaps point me in the right direction of a better formula to use?  

Offline JZ_1

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Re:Calculating heat lost/gained in calorimeter experiment
« Reply #1 on: November 26, 2005, 01:16:04 PM »
I failed in this portion of my science class...but i think..
•These are the equations we were suppposed to use..and maybe more..
  Specific heat = Q /(m * ?T)
  Q= -CCal * ?T

Specific heat should be in J/G-K
Q= Quantity of heat transferred
m= mass (in grams)
?T= Change in temp. (Kelvin or Celcius) <--Doesn't matter

CCal = heat capacity (J/°C or KJ/°C)
  i don't know if there is supposed to be a negative there
---------------------------------------------------------
I think you are supposed to use the first equation to find Q and then use plug it into the 2nd find the heat capacity...

might as well diereguard this .
« Last Edit: November 26, 2005, 01:16:50 PM by JZ_1 »

r1509

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Re:Calculating heat lost/gained in calorimeter experiment
« Reply #2 on: November 26, 2005, 02:47:20 PM »
Thanks.  The second equation (Q= -CCal * ?T) looks helpful but I am not sure which ?T to use - the ?T of the metal, or of the water.  

Offline JZ_1

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Re: Calculating heat lost/gained in calorimeter experiment
« Reply #3 on: May 11, 2006, 10:12:07 PM »
You can't determine the ?T of the metal because you can't stick a thermometer in a metal teh ?T
is the difference of temperature of the water

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