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.

Question

Quizplus · Accepted Answer

To solve the problem, we can use the formula Q = mcΔT, where Q is the thermal energy produced, m is the mass of the water (which we can assume to be 524 g or mL, since water has a density of 1 g/mL), c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature (which is 5.0°C).

First, we need to calculate the heat absorbed by the water and the calorimeter. The metal calorimeter has a heat capacity of 725 J/°C, so its heat absorbed can be calculated as:
Q_calorimeter = C_calorimeter ΔT = (725 J/°C)(5.0°C) = 3625 J

The heat absorbed by the water can be calculated as:
Q_water = m_water c_water ΔT = (524 g)(4.184 J/g°C)(5.0°C) = 10996 J or 10.996 kJ

The total heat produced by the food sample can then be calculated as:
Q_food = Q_total - Q_calorimeter - Q_water

where Q_total is the total heat absorbed by the system (water, calorimeter, and food).

Since the problem only asks for the thermal energy produced, we can ignore the negative sign in the equation above and take its absolute value. Thus,
Q_food = |Q_total - Q_calorimeter - Q_water| = |Q_total - 3625 J - 10.996 kJ|

We can now solve for Q_total by rearranging the equation and substituting the given values:
Q_total = Q_food + Q_calorimeter + Q_water
         = |14.6 kJ - 3625 J - 10.996 kJ|
         = 0.998 kJ or 998 J

However, the question asks for the thermal energy produced in kilojoules, so we need to convert the answer to that unit:
Q_total = 998 J = 0.998 kJ

Therefore, the best choice is B) 14.6 kJ.

A Food Sample Was Burned in a Bomb Calorimeter Containing