Aresearcher studying the nutritional value of a new candy places a 5.90 g sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. the observed temperature increase is 2.30°c. if the heat capacity of the calorimeter is 45.80 kj ⋅ k⁻¹, how many nutritional calories are there per gram of the candy?

4.267 calories per gram ≈ 4.27 Cal/g

Explanation:

Calorimeter

A calorimeter is a device used to measure the heat flow of a chemical reaction or physical change. The basic calorimeter consists of a metal container of water above a combustion chamber, in which a thermometer is used to measure the change in water temperature.

Nutritional Value of Food

The nutritional value of food is the content of the different classes of food it contains. It has to with the amount or the number of calories present in them, and other classes of food. When food substances are burned in the body heat is released. In biological systems, this process is known as respiration. To know the nutritional value of a food substance, they are burned in a calorimeter of known heat capacity. The rise in temperature, measured, provides a useful value for the nutritional content of the foodstuff.

The heat given off by the   (Q)

Q = cΔT(1).

where c = heat capacity of the calorimeter, in kJ/K, and ΔT = Temperature rise ( change in temperature).

c=45.80 kJ/K, ΔT= 2.30K

∴ The heat given off by the combustion of the candy

Q = 45.80 × 2.30

 Q = 105.34 kJ.

  If  4.184 kJ = 1 calories

∴ 105.34 kJ = (105.34 kJ × 1 calories)/4.184 kJ

= 25.177 calories.

≈ 25.18 Cal.

The nutritional Calories per gram (q) = Q(calories)/M(g)

where q= nutritional calories per gram in C/g, Q= Heat or Energy in Calories and M= Mass in gram.

Q= 25.18 Calories, M = 5.9 g

∴ q = 25.18/5.9

q = 4.267 calories per gram ≈ 4.27 Cal/g

e

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%