The complete combustion of  0.521  g of a snack bar in a calorimeter (ccal = 6.15 kj/°c) raises the temperature of the calorimeter by  1.29  °c.  calculate the food value (in cal/g) for the snack bar.

The food raises the temperature of the calorimeter by 1.29 °C. The amount of calorie produced by the combustion would be: 1.29°C *  6.15 kJ/°C= 7.9335 kJ.

If the food weight is 0.521g, then the food value would be: 
7.9335 kJ/ 0.521g / (4.184kcal/kJ) =
(15.227 kJ/g) / (4.184kcal/kJ)= 3.64 kilo calorie/gram

3,640 cal/g

Explanation:

1) Data:

m = 0.521 g

C cal = 6.15 kJ/ °C

ΔT = 1.29°C

2) Principles and formulae:

The calorimeter is used to measure the heat exchanged between to objects.As the calorimeter is isolated, the heat released by the combustion of the snack bar equals the heat absorbed by the calorimeter.The food value of the snack bar (in Cal/g) equals the heat realeased during its combustion, so it equals the heat absorbed by the calorimeter.The heat absorbed by the calorimeter = C cal × ΔT

3) Solution

Heat released by the calorimeter = 6.15 kJ/°C × 1.29°C = 7.93 kJFood value =  7.93 kJ = 7,930 JConversion of units: 1.00 cal = 4.184 J

        ⇒ 7,930 J = 7,930 × 1.00 cal / 4.184 J  = 1,895cal

Cal/g = cal / m = 1,895 cal / 0.521 g = 3,637 cal/g = 3,640 cal/g

Only one reactant yields 2 or more products. meaning you will only have 1 product and it will break down into 2 or more products. eg. h2o -> h2 + o2