A food sample was burned in a bomb calorimeter containing 529 mL water.How much thermal energy was produced when the temperature of the water [c_v = 4.15 J/(g.°C)] 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.

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Quizplus · Accepted Answer

The correct answer is A.The thermal energy produced is equal to the heat absorbed by the water and the calorimeter. The heat absorbed by the water is:Q_water = m_water * c_water * ΔT_water= 529 g * 4.15 J/(g °C) * (25.0 °C - 20.0 °C)= 10.9 kJThe heat absorbed by the calorimeter is:Q_calorimeter = C_calorimeter * ΔT_calorimeter= 725 J/°C * (25.0 °C - 20.0 °C)= 3.63 kJThe total thermal energy produced is:Q_total = Q_water + Q_calorimeter= 10.9 kJ + 3.63 kJ= 14.6 kJ

A Food Sample Was Burned in a Bomb Calorimeter Containing