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Specific heat capacity lab (DUE TOMORROW PLEASE HELP)

  • Thread starter dgoad
  • Start date
  • #1
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Homework Statement


I heated 50mL of water in a Dewar and measured the temp. change with time. however, this only gives me the heat capacity of the system. I again measured but with 100mL of water, in an attempt to correct for the unknown heat capacity of everything but the water. However, after hours of trying, i can't figure out how to correct for the heat capacity of everything else. This is due tomorrow and is for a University thermodynamics course, I could really use the help.

trial 1: dQ=258.5J, dT=0.981 K, m=0.050 kg
trial 2: dQ = 1180.2J, dT=2.230 K, m=0.100kg

I am trying to correct for the unknown heat capacities of everything but the water in an attempt to find the specific heat capacity of water. It seems like simple algebra but its not working out.


Homework Equations


c (specific heat capacity) = 1/M (dQ/dT)

C (heat capacity) = dQ/dT


The Attempt at a Solution


I tried Ctotal = Cwater + Cother for each and equated Cother to no luck... i tried everything i can think of
 

Answers and Replies

  • #2
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I'm not sure I like you using dQ... I think it should just be Q.

But let's start with getting your algebra started on the right track. You say you tried Ctotal = Cwater +Cother for each trial? Let's just double check... you should have two equations:
Ct1 = Cw1 + Co
Ct2 = Cw2 + Co

where do you go from there?
 

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