Answer to Question #153142 in Chemistry for aseel

Question

When a 1.50 g sample of sucrose, C12H22O11 (s), is burned in a bomb calorimeter assembly, the temperature of the system rises from  21.40°C to 28.28°C. If the heat capacity of the calorimeter is 3.60 kJ/°C, a) Calculate the molar heat of combustion of sucrose, in kJ/mol.

b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 Calories (food Calories).

Answer 1

The heat released upon combustion of the sample of sucrose can be calculated from the heat capacity $c$ and the temperature change $∆T$ :

$Q = c∆T = 3.60·(28.28-21.40) = 24.768$ kJ.

The number of the moles of sucrose is its mass divided by its molar mass $M = 342.30$ g/mol:

$n = \frac{m}{M} = \frac{1.50}{342.30} = 0.00438$ mol.

The molar heat of combustion is the heat released divided by the number of the moles of the sucrose. The negative sign of the heat of combustion must be there because the heat is released:

$∆H_{comb} = -\frac{Q}{n} = -\frac{24.768}{0.00438} = -5652$ kJ/mol.

Therefore, one teaspoon of sugar contains:

$Q = -∆H_{comb}·\frac{m}{M} = 5652·\frac{4.8}{342.30} = 79$ kJ.

Finally, as one kJ corresponds to 1/4.2 Cal:

$Q = \frac{79}{4.2} = 18.9$ Cal.

Answer: a) the molar heat of combustion of sucrose is -5652 kJ/mol. b) yes, one teaspoon of sugar produces only 19 Cal (more exactly, 18.9 Cal).