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Physical Chemistry : Calorimetry

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Example Question #21 : Thermochemistry And Thermodynamics

The heat capacity of a bomb calorimeter assembly is $3.90\ \frac{\textup{kJ}}{^\circ\textup{C}}$ $\displaystyle 3.90\ \frac{\textup{kJ}}{^\circ\textup{C}}$ . What is the heat of combustion of $1.1\textup{ g}$ $\displaystyle 1.1\textup{ g}$ of sucrose in a bomb calorimeter that results in the temperature rising from $22.1\ ^\circ \textup{C}$ $\displaystyle 22.1\ ^\circ \textup{C}$ to $27.8\ ^\circ\textup{C}$ $\displaystyle 27.8\ ^\circ\textup{C}$?

Possible Answers:

$41.5\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle 41.5\ \frac{\textup{kJ}}{\textup{g}}$

$-20.2\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -20.2\ \frac{\textup{kJ}}{\textup{g}}$

$-15.1\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -15.1\ \frac{\textup{kJ}}{\textup{g}}$

$-62.9\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -62.9\ \frac{\textup{kJ}}{\textup{g}}$

$3.0\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle 3.0\ \frac{\textup{kJ}}{\textup{g}}$

Correct answer:

$-20.2\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -20.2\ \frac{\textup{kJ}}{\textup{g}}$

Explanation:

A bomb calorimeter is a device used to measure the quantity of heat change for a process. The heat of a reaction which is denoted as $\displaystyle q$, is the negative of the thermal energy gained by the calorimeter:

$q_{rxn}=-q_{calorimeter}$ $\displaystyle q_{rxn}=-q_{calorimeter}$

The heat capacity of a calorimeter is:

$q_{calorimeter}=heat \ capacity \ of\ calorimeter\times \bigtriangleup T$ $\displaystyle q_{calorimeter}=heat \ capacity \ of\ calorimeter\times \bigtriangleup T$

Plugging the values given into the equation gives:

$q_{calorimeter}=3.90\frac{kJ}{^{o}C}\times (27.8-22.1)^{o}C=22.3kJ$ $\displaystyle q_{calorimeter}=3.90\frac{kJ}{^{o}C}\times (27.8-22.1)^{o}C=22.3kJ$

Using the relation provided earlier:

$q_{rxn}=-q_{calorimeter}=-22.3kJ$ $\displaystyle q_{rxn}=-q_{calorimeter}=-22.3kJ$

Because we are dealing with 1.1 grams of sucrose, the heat of combustion of sucrose is:

$q_{rxn}=\frac{-22.3kJ}{1.1g}=-20.2\frac{kJ}{g}$ $\displaystyle q_{rxn}=\frac{-22.3kJ}{1.1g}=-20.2\frac{kJ}{g}$

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Example Question #21 : Thermochemistry And Thermodynamics

The heat capacity of a bomb calorimeter assembly is $2.60\ \frac{\textup{kJ}}{^\circ \textup{C}}$ $\displaystyle 2.60\ \frac{\textup{kJ}}{^\circ \textup{C}}$ . What is the heat of combustion of $1.65 \textup{ g}$ $\displaystyle 1.65 \textup{ g}$ of caffeine in a bomb calorimeter that results in the temperature rising from $23.3\ ^\circ \textup{C}$ $\displaystyle 23.3\ ^\circ \textup{C}$ to $31.8\ ^\circ\textup{C}$ $\displaystyle 31.8\ ^\circ\textup{C}$?

Possible Answers:

$-1.48\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -1.48\ \frac{\textup{kJ}}{\textup{g}}$

$50.0\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle 50.0\ \frac{\textup{kJ}}{\textup{g}}$

$13.0\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle 13.0\ \frac{\textup{kJ}}{\textup{g}}$

$-13.39\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -13.39\ \frac{\textup{kJ}}{\textup{g}}$

$3.01\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle 3.01\ \frac{\textup{kJ}}{\textup{g}}$

Correct answer:

$-13.39\ \frac{\textup{kJ}}{\textup{g}}$ $\displaystyle -13.39\ \frac{\textup{kJ}}{\textup{g}}$

Explanation:

A bomb calorimeter is a device used to measure the quantity of heat change for a process. The heat of a reaction which is denoted as q, is the negative of the thermal energy gained by the calorimeter:

$q_{rxn}=-q_{calorimeter}$ $\displaystyle q_{rxn}=-q_{calorimeter}$

The heat capacity of a calorimeter is:

$q_{calorimeter}=heat \ capacity \ of\ calorimeter\times \bigtriangleup T$ $\displaystyle q_{calorimeter}=heat \ capacity \ of\ calorimeter\times \bigtriangleup T$

Plugging the values given into the equation gives:

$q_{calorimeter}=2.60\ \frac{kJ}{^{o}C}\times (31.8-23.3)^{o}C=22.1kJ$ $\displaystyle q_{calorimeter}=2.60\ \frac{kJ}{^{o}C}\times (31.8-23.3)^{o}C=22.1kJ$

Using the relation provided earlier:

$q_{rxn}=-q_{calorimeter}=-22.1kJ$ $\displaystyle q_{rxn}=-q_{calorimeter}=-22.1kJ$

Because we are dealing with 1.65 grams of sucrose, the heat of combustion of sucrose is:

$q_{rxn}=\frac{-22.1kJ}{1.65g}=-13.39\frac{kJ}{g}$ $\displaystyle q_{rxn}=\frac{-22.1kJ}{1.65g}=-13.39\frac{kJ}{g}$

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Physical Chemistry : Calorimetry

Study concepts, example questions & explanations for Physical Chemistry

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Example Question #21 : Thermochemistry And Thermodynamics

Example Question #21 : Thermochemistry And Thermodynamics

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