Specific heat capacity lab (DUE TOMORROW PLEASE HELP)

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SUMMARY

The discussion focuses on calculating the specific heat capacity of water while accounting for the heat capacity of other materials in a thermodynamics lab setting. The user conducted two trials using 50mL and 100mL of water, measuring heat transfer (dQ) and temperature change (dT). The equations provided include c (specific heat capacity) = 1/M (dQ/dT) and C (heat capacity) = dQ/dT. The user struggles to isolate the unknown heat capacities of other materials, indicating a need for algebraic clarification in their approach.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Familiarity with heat capacity and specific heat capacity concepts
  • Basic algebra skills for solving equations
  • Knowledge of experimental measurement techniques in a laboratory setting
NEXT STEPS
  • Review the derivation of the heat capacity equations used in thermodynamics
  • Learn about the method of least squares for fitting experimental data
  • Explore the concept of heat transfer in multi-component systems
  • Investigate the impact of measurement errors on experimental results
USEFUL FOR

University students in thermodynamics courses, laboratory technicians conducting heat capacity experiments, and educators teaching principles of heat transfer and thermodynamics.

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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
 
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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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