Question

A researcher studying the nutritional value of a new candy places a 3.80 g3.80 g sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.92 ∘C.2.92 ∘C. If the heat capacity of the calorimeter is 41.70 kJ⋅K−1,41.70 kJ⋅K−1, how many nutritional Calories are there per gram of the candy?

Answer 1

Explanation:

The given data is as follows.

         Heat capacity = 41.70 kJ/K,        

Temperature change = $2.92^{o}C$

Hence, calculate the quantity of heat as follows.

      Quantity of heat = heat capacity × temp. change

                                  = 41.70 kJ/K × 2.92 K

                                  = 121.764 kJ

This heat energy is related to burning of 3.80 g of given sample.

Therefore, calculate the heat per gram as follows.

                  Heat = $\frac{121.764 kJ}{3.80 g}$

                           = 32.04 kJ/g

As, 1 nutritional calorie = 1000 calorie

      1 calorie = 4.2 J

Therefore, calculate the heat in terms of calorie as follows.

            Heat = $\frac{32.04 kJ/g \times 1000 J/kJ}{1 g} \times \frac{1 cal}{4.2 cal} \times \frac{1 cal}{1000 cal}$

                     = 7.62 calorie

Thus, we can conclude that there are 7.62 calories per gram of the candy.