Question

A researcher studying the nutritional value of a new candy places a 5.40 g sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.47
c. If the heat capacity of the calorimeter is 41.00 kjk1, how many nutritional calories are there per gram of the candy?

Answer 1

Answer: Nutritional Calories = 4.48 Cal/g  

Explanation:

Conversion from Calorie to Kilo joules is given by the conversion factor,

1 Calorie = 1 Kilocalorie = 4.184 kJ

Given in the question are:

ΔT = Observed temperature increase  = 2.47

$C_{calorimeter}$  = Heat Capacity of the calorimeter = 41.00 kJ $K^{-1}$

m = mass of the sample = 5.40 g

We know the formula for Heat Capacity (Q)

Q = $C_{calorimeter}$  × ΔT    ....................... equation 1

We also know the heat capacity for the sample is given by the formula,

$C_{candy}$ = $\frac{Q}{m}$

Substituting value of Q from equation 1

= $C_{calorimeter}$ × ΔT/m

= 41.00 kJ $K^{-1}$ × $\frac{2.47 K}{5.40 g}$

= 18.75 kJ/g

= 4.48 KCal/g

= 4.48 Cal/g

Nutritional calories per gram of the candy is 4.48 Calories per gram.

Answer 2

The amount of nutritional calories is C candy = 4.48 KCal / g = 4.48 Cal/g

Explanation:

Given data:

• Mass of candy = 5.40 g
• 1 Cal = 1 kcal = 4.1868 kJ
• Increase in temperature = 2.47
• Heat capacity of the calorimeter = 41.00 KJK^-1
• To find: Nutritional calories = ?

Solution:

Q = C calorimeter x ΔT

C candy = Q / m = C calorimeter  x ΔT / m

C candy = 41.00 x 2.47 k / 5.40 g

C candy = 18.75 KJ / g

C candy = 4.48 K.Cal / g

Thus the amount of nutritional calories is C candy = 4.48 K.Cal / g = 4.48 Cal/g