When 0.18 g of glucose was burnt in a bomb calorimeter, the temperature rose by 4 K. The heat capaci — Thermodynamics and Thermochemistry Chemistry Question
Question
When 0.18 g of glucose was burnt in a bomb calorimeter, the temperature rose by 4 K. The heat capacity of the calorimeter system is ($\Delta_c H^\circ = -2.8 \times 10^6\text{ J/mol glucose}$)
Answer: A
💡 Solution & Explanation
Molar mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) = $180\text{ g/mol}$. Moles burnt = $0.18 / 180 = 0.001\text{ mol}$. Total heat released $Q = 0.001 \times 2.8 \times 10^6\text{ J} = 2800\text{ J}$. Heat capacity $C = Q / \Delta T = 2800 / 4 = 700\text{ J/K}$.
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