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The energy content of food can be determined by completely burning a sample of the food in a bomb calorimeter and then calculating the energy released - VCE - SSCE Chemistry - Question 3 - 2003 - Paper 1

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The energy content of food can be determined by completely burning a sample of the food in a bomb calorimeter and then calculating the energy released. a. The calor... show full transcript

Worked Solution & Example Answer:The energy content of food can be determined by completely burning a sample of the food in a bomb calorimeter and then calculating the energy released - VCE - SSCE Chemistry - Question 3 - 2003 - Paper 1

Step 1

Calibration Factor Calculation

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Answer

To find the calibration factor of the calorimeter, we can use the formula:

E=IimesVimestE = I imes V imes t

Where:

  • EE is the energy in joules,
  • II is the current (1.78 A),
  • VV is the potential difference (5.65 V), and
  • tt is the time (135 s).

Substituting the values:

E=1.78imes5.65imes135=1358.1extJE = 1.78 imes 5.65 imes 135 = 1358.1 ext{ J}

To convert joules to kilojoules, we divide by 1000:

E = rac{1358.1}{1000} = 1.3581 ext{ kJ}

Next, we calculate the calibration factor in kJ °C⁻¹:

ext{Calibration Factor} = rac{E}{ ext{Temperature Rise}} = rac{1.3581 ext{ kJ}}{1.15} ext{ °C} ext{ } = 1.1792 ext{ kJ °C⁻¹}

Step 2

Molar Heat of Combustion Calculation

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The temperature increase is calculated as follows:

extTemperatureRise=35.55−18.23=17.32ext°C ext{Temperature Rise} = 35.55 - 18.23 = 17.32 ext{ °C}

Using the calibration factor obtained previously, we can calculate the energy released:

E=extCalibrationFactorimesextTemperatureRise=1.1752extkJ°C−1imes17.32ext°C=20.327extkJE = ext{Calibration Factor} imes ext{Temperature Rise} = 1.1752 ext{ kJ °C⁻¹} imes 17.32 ext{ °C} = 20.327 ext{ kJ}

To find the molar heat of combustion of glucose (1.324 g), we convert the mass to moles using the molar mass of glucose (C₆H₁₂O₆ ≈ 180.18 g/mol):

ext{Moles of Glucose} = rac{1.324 g}{180.18 g/mol} ext{ } ext{ } ext{ ≈ 0.00734 mol}

Now we calculate the molar heat of combustion:

ext{Molar Heat of Combustion} = rac{E}{ ext{Moles}} = rac{20.327 ext{ kJ}}{0.00734 ext{ mol}} = 2761.81 ext{ kJ mol⁻¹}

Step 3

Ratio Prediction of Molar Heats

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Answer

Sucrose is composed of two glucose molecules, therefore, the molar heat of combustion of sucrose can be approximated as double that of glucose. Hence:

ext{Approximate Ratio} = rac{ ext{Molar Heat of Combustion Sucrose}}{ ext{Molar Heat of Combustion Glucose}} ext{ } ext{ } ext{ ≈ 2 }

This means the molar heat of combustion for sucrose is around double that of glucose because of the additional carbon and hydrogen atoms contributing to the enthalpy of combustion.

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