Question

One tablespoon of peanut butter has a mass of 16.0 g. When it is combusted in a bomb calorimeter with a heat capacity of 103.3 kJ/ ^$\degree$ C, submerged in 4.0 L of water. The temperature of the water rises from 22.2 ^$\degree$ C to 25.4 ^$\degree$ C. Find the food caloric content of peanut butter in Cal/g. Assume density of water is 1.0 g/mL. The specific heat of water is 4.184 J/g ^$\degree$ C

Answer 1

Mass of water, m = 4.0 LX 1000 mL 1g - 40 1 mL Specificheat of water s = 4.184 J/g.°C Change in temperature AT = 25.4°C-22.2°C = 3.2°C Amount of heat absorbed by water, Gyvater = mSAT = 4000 gx4.184*(3.2°C) = 53555 J = 53.555 kJ Change in temperature produced, AT = 3.2°C Heat capacity of calorimeter,s = 103.3 kJ/°C Amount of heat absorbed by the calorimeter, cal = CAT = (103. 4. (3.2°C) = 330.56 kJ Amount of heat relased in the reaction is equal to the heat absorbed by system 4system = -( 9ca+waxy) =-(330.56 kJ + 53.555 kJ ) =-384.115 kJ x 11 1 Cal * 4.184 kJ = -91.80 Cal Amount of peanut butter = 16.0 g - 91.80 Cal Food caloric content of peanut butter, AH - = 16.0 g = -5.74 Calg