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Enthalpy Combustion of Alcohols, Lab Report Example

Pages: 1

Words: 363

Lab Report

Enthalpy Change during combustion (âˆ†H) = mcâˆ†T

Note: The mass of the spirit lamp with alcohol was averaged together to represent all three trials for each type of alcohol.

Methanol

Volume of water = 100 cm3

Mass of Water (m) = Volume x Density

= 100 cm3 x 1 g/cm3

= 100 g

Temperature Before Heating: 27 degrees Celsius

Temperature After Heating: 37 degrees Celsius

Total Change in Temperature: 37 degrees Celsius â€“ 27 degrees Celsius

= 10 degrees Celsius

Specific Heat Capacity of Water = 4.18 J g-1 K-1

Therefore:

Enthalpy Change = 100 cm3 x 4.18 J g-1 K-1 x 10 K

Enthalpy Change = 4180 J

Ethanol

Volume of water = 100 cm3

Mass of Water (m) = Volume x Density

= 100 cm3 x 1 g/cm3

= 100 g

Temperature Before Heating: 27 degrees Celsius

Temperature After Heating: 39 degrees Celsius

Total Change in Temperature: 39 degrees Celsius â€“ 27 degrees Celsius

= 12 degrees Celsius

Specific Heat Capacity of Water = 4.18 J g-1 K-1

Therefore:

Enthalpy Change = 100 cm3 x 4.18 J g-1 K-1 x 12 K

Enthalpy Change = 5016 J

Propanol

Volume of water = 100 cm3

Mass of Water (m) = Volume x Density

= 100 cm3 x 1 g/cm3

= 100 g

Temperature Before Heating: 27 degrees Celsius

Temperature After Heating: 40 degrees Celsius

Total Change in Temperature: 40 degrees Celsius â€“ 27 degrees Celsius

= 13 degrees Celsius

Specific Heat Capacity of Water = 4.18 J g-1 K-1

Therefore:

Enthalpy Change = 100 cm3 x 4.18 J g-1 K-1 x 13 K

Enthalpy Change = 5434J

Butanol

Volume of water = 100 cm3

Mass of Water (m) = Volume x Density

= 100 cm3 x 1 g/cm3

= 100 g

Temperature Before Heating: 28 degrees Celsius

Temperature After Heating: 39 degrees Celsius

Total Change in Temperature: 41 degrees Celsius â€“ 28 degrees Celsius

= 12 degrees Celsius

Specific Heat Capacity of Water = 4.18 J g-1 K-1

Therefore:

Enthalpy Change = 100 cm3 x 4.18 J g-1 K-1 x 13 K

Enthalpy Change = 5434 J

Pentanol

Volume of water = 100 cm3

Mass of Water (m) = Volume x Density

= 100 cm3 x 1 g/cm3

= 100 g

Temperature Before Heating: 27 degrees Celsius

Temperature After Heating: 39 degrees Celsius

Total Change in Temperature: 43 degrees Celsius â€“ 28 degrees Celsius

= 12 degrees Celsius

Specific Heat Capacity of Water = 4.18 J g-1 K-1

Therefore:

Enthalpy Change = 100 cm3 x 4.18 J g-1 K-1 x 15 K

Enthalpy Change = 6270 J

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Lab Report

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