Answer to Question #105019 in General Chemistry for Natalie

Question

The complete combustion of 0.731 g of a snack bar in a calorimeter (Ccal = 6.15 kJ/°C) raises the temperature of the calorimeter by 1.93 °C. Calculate the food value (in Cal/g) for the snack bar.

Answer 1

Here, it is given that mass of snack bar as 0.731 g ,

Temp of calorimeter is given as $\Delta T= 1.93 ^\circ$

for calorimeter $C$ _cal= 6.15 $\frac{ kJ}{^oC}$

So as per question , Heat energy released by the calorimeter = C $\Delta T$

Q= 6.15 $\times$ 1.93 = 11.8695 kJ

heat energy is calorie= $\frac{11.8695 kJ}{4.184 \frac{kJ}{kcal}}$

Q= 2.8368 kcal

now as per question the food value for snack bar = $\frac{ Q}{mass of snack bar }$

food value = $\frac{2.8368 kcal }{0.731g} = \frac{2.8368 \times1000 cal}{0.731g}$

food value = 3880.8 $\frac{cal}{g}$

This is the required food value of snack bar