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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

a. Calculate the calibration factor

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Answer

To find the calibration factor, we use the formula:

E=IimesVimestE = I imes V imes t

where:

  • EE is the energy in joules
  • II is the current in amperes (1.78 A)
  • VV is the potential difference in volts (5.65 V)
  • tt is the time in seconds (135 s)

Calculating the energy:

E=1.78extAimes5.65extVimes135exts=1,358extJE = 1.78 ext{ A} imes 5.65 ext{ V} imes 135 ext{ s} = 1,358 ext{ J}

Now, to find the calibration constant (in kJ °C⁻¹), we divide by the temperature rise:

Calibration Constant=ETemperature Rise=1,358extJ1.15°C=1,182.6extJ°C−1=1.18extkJ°C−1Calibration\, Constant = \frac{E}{Temperature\, Rise} = \frac{1,358 ext{ J}}{1.15 °C} = 1,182.6 ext{ J °C}^{-1} = 1.18 ext{ kJ °C}^{-1}

Step 2

b. Calculate the molar heat of combustion of glucose

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Given the temperature rise:

  • Initial temperature = 18.23 °C
  • Final temperature = 35.55 °C
  • Temperature change, ΔT = 35.55 °C - 18.23 °C = 17.32 °C.

To find the energy released: Q=Calibration×ΔT=1.18extkJ°C−1×17.32°C=20.42extkJQ = Calibration \times \Delta T = 1.18 ext{ kJ °C}^{-1} \times 17.32 °C = 20.42 ext{ kJ}

Now, with the mass of glucose burned (1.324 g), converting to moles (molar mass of glucose = 180.18 g/mol):

n=1.324extg180.18extg/mol=0.00734extmoln = \frac{1.324 ext{ g}}{180.18 ext{ g/mol}} = 0.00734 ext{ mol}

Finally, we calculate the molar heat of combustion:

Molar Heat of Combustion=Qn=20.42extkJ0.00734extmol=2,785.1extkJmol−1Molar\, Heat\, of\, Combustion = \frac{Q}{n} = \frac{20.42 ext{ kJ}}{0.00734 ext{ mol}} = 2,785.1 ext{ kJ mol}^{-1}

Step 3

c. Predict the approximate value of the ratio

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Answer

To estimate the ratio of molar heats of combustion of sucrose to glucose:

The molar heat of combustion of sucrose is generally known to be around double that of glucose because sucrose is a disaccharide made of two glucose molecules. Hence, the ratio can be approximated as:

Molar Heat of Combustion SucroseMolar Heat of Combustion Glucose≈2\frac{Molar\, Heat\, of\, Combustion\, Sucrose}{Molar\, Heat\, of\, Combustion\, Glucose} \approx 2

This estimate arises from the molecular composition of sucrose, which consists of two glucoses, effectively doubling the energy content.

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