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A quantity of 1.922 g of methanol (CH3OH) was burned in
Chapter 6, Problem 6PE(choose chapter or problem)
A quantity of 1.922 g of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose by \(4.20^{\circ} \mathrm{C}\). If the heat capacity of the bomb plus water was \(10.4 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), Calculate the molar heat of combustion of methanol.
Questions & Answers
QUESTION:
A quantity of 1.922 g of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose by \(4.20^{\circ} \mathrm{C}\). If the heat capacity of the bomb plus water was \(10.4 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), Calculate the molar heat of combustion of methanol.
ANSWER:Step 1 of 3
The goal of the problem is to calculate the molar heat of combustion of methanol.
Given:
Mass of methanol \(\left(\mathrm{CH}_3\mathrm{OH}\right)=1.922\mathrm{\ g}\)
Change in Temperature \((\Delta t)=4.20^{\circ} \mathrm{C}\)
Heat capacity of the bomb plus water \(=10.4\mathrm{\ KJ}/^{\circ}\mathrm{C}\)
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
\(q_{c a l}=C_{c a l} \Delta t\)
Where, \(q_{c a l}=\) heat of reaction
\(C_{c a l}=\) heat capacity of calorimeter
\(\Delta \mathrm{t}=\) change in temperature of the sample