Chapter 14, Problem 21P

Estimate the Calorie content of 65 g of candy from the following measurements. A 15-g sample of the candy is placed in a small aluminum container of mass 0.325 kg filled with oxygen. This container is placed in 1.75 kg of water in an aluminum calorimeter cup of mass 0.624 kg at an initial temperature of 15.0°C. The oxygen-candy mixture in the small container (a “bomb calorimeter”) is ignited, and the final temperature of the whole system is 53.5°C.

To determine

Section 1:

The heat gained.

Answer to Problem 21P

Solution:

The heat gained is $75.41\text{ kcal}$.

Explanation of Solution

Given info:

The mass of candy is $\text{15 g}$, the mass of aluminum container is $0.325\text{ kg}$, the mass of water is $1.75\text{ kg}$, the mass of aluminum calorimeter cup is $\text{0}\text{.624 kg}$, the initial temperature is $15°\text{C}$ and the final temperature is $53.5°\text{C}$.

The principle of calorimetric is written as,

$\text{Heat lost}=\text{Heat gained}$

The heat gained is calculated as,

$\begin{array}{c}\text{Heat gained}=+\left(m_{\text{Al}}{c}_{\text{Al}}\left(T_{\text{f}}-T_{\text{i}}\right)+m_{\text{w}}{c}_{\text{w}}\left(T_{\text{f}}-T_{\text{i}}\right)+m_{\text{C}}{c}_{\text{Al}}\left(T_{\text{f}}-T_{\text{i}}\right)\right)\\ =\left(T_{\text{f}}-T_{\text{i}}\right)\left(m_{\text{Al}}{c}_{\text{Al}}+m_{\text{w}}{c}_{\text{w}}+m_{\text{C}}{c}_{\text{Al}}\right)\end{array}$

• $m_{\text{Al}}$ is the mass of aluminum container.

• $m_{\text{w}}$ is the mass of water.

• $m_{\text{C}}$ is the mass of calorimeter.

• $c_{\text{Al}}$ is the specific heat capacity of aluminum.

• $c_{\text{w}}$ is the specific heat capacity of water.

• $T_{\text{f}}$ is the final temperature.

• $T_{\text{i}}$ is the initial temperature.

Substitute $\text{0}\text{.325 kg}$ for $m_{\text{Al}}$, $1.75\text{ kg}$ for $m_{\text{w}}$, $0.624\text{kg}$ for $m_{\text{C}}$, $0.22\text{kcal}/\text{kg }°\text{C}$ for $c_{\text{Al}}$, $\text{1 }\text{kcal}/\text{kg }°\text{C}$ for $c_{\text{w}}$, $15°\text{C}$ for $T_{\text{i}}$ and $53.5°\text{C}$ for $T_{\text{f}}$ in above equation to find heat gained.

$\begin{array}{c}\text{Heat gained}=\left(53.5°\text{C}-15°\text{C}\right)\left\{\begin{array}{l}\left(\text{0}\text{.325 kg}\right)\left(0.22\text{kcal}/\text{kg }°\text{C}\right)\\ +\left(1.75\text{ kg}\right)\left(\text{1 }\text{kcal}/\text{kg }°\text{C}\right)\\ +\left(0.624\text{kg}\right)\left(0.22\text{kcal}/\text{kg }°\text{C}\right)\end{array}\right\}\\ =75.41\text{ kcal}\end{array}$

To determine

Section 2:

The energy liberated from calorie content of $65\text{ g}$ candy.

Answer to Problem 21P

Solution:

The energy liberated from calorie content of $65\text{ g}$ candy is $326.78\text{ kcal}$.

Explanation of Solution

Given info:

The mass of candy is $\text{15 g}$, the mass of aluminum container is $0.325\text{ kg}$, the mass of water is $1.75\text{ kg}$, the mass of aluminum calorimeter cup is $\text{0}\text{.624 kg}$, the initial temperature is $15°\text{C}$ and the final temperature is $53.5°\text{C}$.

The heat gained is nothing but the amount of energy that was liberated from the $\text{15 g}$ sample.

The energy liberated from the $65\text{ g}$ is calculated as,

$\begin{array}{c}E=\text{Heat gained}\times \frac{65\text{ g}}{15\text{ g}}\\ =75.41\text{ kcal}\times \frac{65\text{ g}}{15\text{ g}}\\ =326.78\text{ kcal}\end{array}$