Find the ΔE of the combustion of octane in a bomb calorimeter

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SUMMARY

The combustion of octane in a bomb calorimeter was analyzed, revealing a calculated ΔE of -5494.25 kJ/mol, which deviates from the expected value of -5506.3 kJ/mol. The experiment involved burning 8.10 mL of octane in 10.00 kg of water, resulting in a temperature increase of 5.66°C. The heat capacity of the bomb calorimeter was determined to be 6.656 kJ/°C, and the specific heat of water was 4.184 J/g°C. The discrepancy in the ΔE calculation was attributed to the method of combining heat values from both the calorimeter and water.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with bomb calorimeter operations and calculations.
  • Knowledge of specific heat capacity and its application in calorimetry.
  • Ability to perform mole calculations from mass and density of substances.
NEXT STEPS
  • Review the principles of bomb calorimetry and its applications in thermodynamic calculations.
  • Learn about the significance of heat capacities in calorimetric experiments.
  • Study the calculation of ΔE and its relationship to reaction enthalpy.
  • Explore common sources of error in calorimetry experiments and how to mitigate them.
USEFUL FOR

Chemistry students, educators, and professionals involved in thermodynamics and calorimetry, particularly those studying combustion reactions and energy calculations.

MarcL
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Homework Statement



Consider the combustion of octane that is burned in a bomb calorimter in the presence of 10.00kg of water. When burning 8.10mL of octane (.703g/ml=d) the temp. of the surrounding water increase by 5.66°C. The heat capacity of the bomb calorimter is 6.656Kj/ °C and the water is 4.184 j/g x °C

Homework Equations



q=mcΔT
qcal=Ccal x ΔT
ΔErxn= qrxn / moles of octane

The Attempt at a Solution



find Q of water which gave me 236.8 KJ by q=mcΔT ( m = 10000KG )
and find Q cal which ended up to 37.13 KJ by q=Ccal ΔT
then I added both Heat to get 273.9 Kj and then I divided by the number of mole found which was .049852045 moles.

However I got as an answer -5494.25 and the answer is -5506.3 Kj/mol

P.s: why does my textbook says to just use the Q of the calorimter and not add it up to the one of water in order to find the ΔE?
 
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