Answer to Question #99580 in General Chemistry for Beverlie

Question

The complete combustion of 0.591 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.

Answer 1

According to the law of calorimetry;

Heat given = Heat supplied

$\implies C_{cal}\Delta T=m_{bar}*(food value)$

$\implies C_{cal}=m_{bar}s_{bar}$

$\implies 6.15*10^3 *1.29=(0.591g)*(food value)$

$\implies food value=13.424 kJg^{-1}$

$=3208.38calg^{-1}$ (Answer)

(since, $1cal=4.184J$ )

Thus food value of the snack bar is $3208.38calg^{-1}$ .