A food sample was burned in a bomb calorimeter containing 524 mL water. How much thermal energy was produced when the temperature of the water and the calorimeter rose from 20.0°C to 25.0°C? The metal calorimeter had a heat capacity of 725 J/°C without the water. The specific heat capacity of water is 4.184 J/(g °C).

Question

Quizplus · Accepted Answer

First, we need to find the mass of water in the calorimeter:
density of water = 1 g/mL
mass of water = density x volume = 1 g/mL x 524 mL = 524 g

Next, we can use the formula Q = mCΔT, where Q is the thermal energy produced, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature. However, we need to make an adjustment for the heat capacity of the calorimeter itself:

Q = (m water + m calorimeter) (C water + C calorimeter) ΔT

m calorimeter = 725 J/°C (change in temperature)
C calorimeter = 1 J/g°C (since it's a metal)
ΔT = 5°C (change in temperature)

Q = (524 g + 725 J/4.184 J/g°C x 25°C/1°C) (4.184 J/g°C + 1 J/g°C) 5°C
Q = (524 g + 4360 g) (5.184 J/g°C) 5°C
Q = 2884.9 J/g x 524 g
Q = 1.514 MJ or 14.6 kJ (rounded to two significant figures)

A Food Sample Was Burned in a Bomb Calorimeter Containing