What is the specific heat capacity of the petroleum?

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SUMMARY

The specific heat capacity of petroleum was calculated using a calorimetry experiment involving a copper calorimeter, petroleum, and a copper weight. The heat capacity of the calorimeter was given as 75 J/K, and the specific heat capacity of copper was noted as 387 J/(kg·K). The experiment determined that the calculated specific heat capacity of petroleum was 20.4 J/(kg·K), which is significantly lower than the accepted value of approximately 1900 J/(kg·K). This discrepancy indicates a potential error in the calculation or assumptions made during the experiment.

PREREQUISITES
  • Understanding of calorimetry principles
  • Knowledge of specific heat capacity calculations
  • Familiarity with the heat transfer equation: C_pet*m_pet(t2-ts)+C_cop*m_cop(t2-t1)+C_cal*m_cal(t2-t1)=0
  • Basic knowledge of unit conversions (Joules, kilograms, Kelvin)
NEXT STEPS
  • Review the principles of calorimetry and heat transfer
  • Learn how to correctly isolate variables in heat capacity equations
  • Study common specific heat capacities of various substances, including petroleum
  • Practice solving calorimetry problems with varying initial conditions and materials
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, chemistry, and physics, particularly those focusing on calorimetry and heat transfer calculations.

Hoejer
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Homework Statement


A copper calorimeter with the heatcapacity of 75 J/K has 300 g of petrolium in it. The starttemp. is 17.8 C. A kobberweight of 100 g. is 100Celcius. It is being put in the petroleum, where the temperature ends up being 22 C. The specifik heatcapacity of coppe ris 387J(kg/K)
What is petroleums specifik heat capacity?


Homework Equations



C_pet*m_pet(t2-ts)+C_cop*m_cop(t2-t1)+C_cal*m_cal(t2-t1)=0

The Attempt at a Solution


I think I have overlooked something - so I just need hints on the way.

I don't have the mass of the calorimeter, but I now that it's heatcapacity is 75 J/K and I know that it's made of coper and the specifik heat capacity of that is 387 J(kg*K)

If that is the way to go, then I have all the variables except the specifik heat capacity of petroleum, and then I just isolate C-Pet and it's piece of cake!
 
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the the mass of the calorimeter must be 75/387 = 0.193 kg = 193 g

Which then means C_pet*300(22-100)+387*100(22-17.8)+387*193(22-17.8)=0

C_pet = 20.4 J(kg/K)
 
Which isn't true as looking in table it is around 1900 J/(kg*K)
 
No help?
 

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