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

Step-by-step 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 =
`(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 =
`(32.04 kJ/g * 1000 J/kJ)/(1 g) * (1 cal)/(4.2 cal) * (1 cal)/(1000 cal)`

= 7.62 calorie

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