Molar Enthelpy question with Combustion

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The discussion focuses on calculating the molar enthalpy of 3-pentanol using combustion data. The initial and final masses of the alcohol, along with the temperature change and water volume, are provided for the calculation. The calculated heat (Q) is 8778 J, leading to a molar enthalpy of approximately -823 kJ/mol. A point of contention is whether the remaining alcohol also experienced a temperature increase, which could slightly adjust the final enthalpy value. Overall, the calculations appear correct, with a minor consideration for the heat absorbed by the remaining alcohol.
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Homework Statement


Using the data for the combustion of 3-pentanol calculate the molar enthalpy for the alcohol.

Initial mass of alcohol: 27.35g
Final mass of alcohol: 26.41g
Change in temperature: 21.0C
Volume of water: 100.0mL (100.0g)
Specific heat capacity of water: 4.18 J/gC

Homework Equations


Q=mcΔT

The Attempt at a Solution


3-pentanol: C5H12O (88.14818 g/mol)
27.35g-26.41g=0.94g
0.94g/88.14818g/mol=0.010663861mol

Q=mcΔT
=(100)(4.18)(21.0)
=8778J
ΔH=-8778J
ΔH/n=-8778J/0.0106638161mol
=-823154.0152J/mol
=-8.23x10^2kJ/mol ←is this the correct answer? (I don't have an answer key.)
 
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It looks good. The only complicating factor is to determine if the 3-pentanol underwent an increase in temperature as well. You used about a gram of the alcohol and about 26 grams are left. Did that remaining alcohol undergo an increase in temperature of 21oK?
 
I believe the remaining alcohol increased 21 Celcius.
I just want to know if I determined the molar enthalpy of alcohol in the end.
 
staka said:
I believe the remaining alcohol increased 21 Celcius.

If that is the case you will need to revise (very slightly) your answer to account for the heat required to heat 26.41 g of the alcohol through 21o. It's probably small enough to neglect.
 
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