Thermochemistry Q

Question

A 1.50 g sample of pure sucrose is burned in a bomb calorimeter that has a heat capacity of 8.930 kJ/ºC. The temperature of the water jacket rises from 24.00 ºC to 26.77 ºC.

What is ΔE for the combustion of 1 mole of sucrose, C₁₂H₂₂O₁₁?

How many dietary Calories per gram does sucrose provide when metabolized?

Answer 1

The sample size is 1.50 g sucrose. This is equivalent to 1.50 g x 1 mol sucrose/342 g = 0.004386 moles

∆E = Ccal x ∆T

Ccal = heat capacity of calorimeter = 8.930 kJ/ºC

∆T = change in temperature = 26.77º- 24.00º = 2.77ºC

∆E = (8.930 kJ/ºC)(2.77ºC) = 24.74 kJ. This is the heat of combustion for 0.004386 moles

molar ∆E = 24.74 kJ/0.004386 moles = 5641 kJ/mole

Dietary Calories (with a capital C) is the same as 1000 calories (small c) or the same as 1 kcal.

24.74 kJ was from 1.50 g of sucrose. Converting this to Calories requires first changing units from Joules to calories. There are 4.184 J / calorie or 4.184 kJ/kcal.

24.74 kJ/1.5 g = 16.49 kJ/g

16.49 kJ/g x 1 kcal/4.184 kJ = 3.94 kcal/g = 3.94 Calories/gram