Question

A researcher studying the nutritional value of a new candy places a 4.70 g 4.70 g sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.41 ∘ C. 2.41 ∘C. If the heat capacity of the calorimeter is 43.90 kJ ⋅ K − 1 , 43.90 kJ⋅K−1, how many nutritional Calories are there per gram of the candy?

Answer 1

Answer : The nutritional Calories per gram of the candy are, 5.36 calories

Explanation :

First we have to calculate the amount of heat.

$q=c\times \Delta T$

where,

q = heat = ?

c = specific heat = $43.90kJ/K$

$\Delta T$ = change in temperature = $2.41^oC=2.41K$

Now put all the given values in the above formula, we get:

$q=43.90kJ/K\times 2.41K$

$q=105.799kJ$

Now we have to calculate the heat for per gram of sample.

Heat = $\frac{105.799kJ}{4.70g}=22.51kJ/g=22510J/g$

Now we have to calculate the heat in terms of calories.

As, 1 nutritional Calories = 1000 calories

and, 1 calories = 4.2 J

As, 4.2 J = 1 calories

So, 22510 J = $\frac{22510J}{4.2J}\times \frac{1}{1000}cal=5.36cal$

Therefore, the nutritional Calories per gram of the candy are, 5.36 calories