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Topic: calculting heat capacity  (Read 3558 times)

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sportystar

  • Guest
calculting heat capacity
« on: February 24, 2005, 09:23:50 PM »
Hi, I am doing an experiment and i have to calculate the heat gained or lost.

In the reaction i dissolved 200g solid NaOH into water and took the initial temp of the water and then after the NaOH was dissolved.  My initial temp was 20 degcelcius and my final temp was 22 degcelcius.  

I used the equation deltaH= m*c*deltaT
so i plug in DeltaH= 200g*0.00418kJ/gdegc*22degcel
                DeltaH= 18.392
since 5.5g NaOH   5.5/40=0.1375
18.392/0.1375= 133.76

What i am confused about is, what temp am i supposed to use, 20degcel, 22degcel or -2degcel??

In the next question it says to use deltaH=m*deltaT*Q to find the heat lost or gained, isnt this what i found in the above question? Whats the difference between the two??

Can someone please explain this to me cuz i am lost.

Thanks


savoy7

  • Guest
Re:calculting heat capacity
« Reply #1 on: February 24, 2005, 10:14:15 PM »
in science land delta means change

so delta T is change in temperature - not the final temperature


maybe this site might help you
http://dbhs.wvusd.k12.ca.us/webdocs/Thermochem/Thermochem.html

it's a high school site that's pretty good

It looks like you are a little confused between H and q

You should review enthalpy and heat energy (or even specific heat), try again and post what you find.
Good Luck!!!

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