Question

(II)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

Short answer

The estimated caloric content of candy is \(326.7\;{\rm{kcal}}\).

Step-by-step solution

Given data

The mass of the candy is\({m_{\rm{c}}} = 65\;{\rm{g}}\).

The mass of the sample of candy is\({m_{{\rm{sc}}}} = 15\;{\rm{g}}\).

The mass of the aluminum container is\({m_{\rm{a}}} = 0.325\;{\rm{kg}}\).

The mass of water is\({m_{\rm{w}}} = 1.75\;{\rm{kg}}\).

The mass of aluminum cup is\({m_{{\rm{ac}}}} = 0.624\;{\rm{kg}}\).

The initial temperature is\({T_{\rm{i}}} = 15\circ {\rm{C}}\).

The final temperature is \({T_{\rm{f}}} = 53.5\circ {\rm{C}}\).

Heat released by the candy

In this problem, the heat released by the candy while burning will be equivalent to the heat absorbed by the water and aluminum calorimeter cup. Utilize this relation to find the calorie content of candy.

Calculation of the heat gained by water and aluminum calorimeter cup

The relation to find the heat gained by aluminum is given by:

\(\begin{array}{c}{Q_{\rm{a}}} = \left( {{m_{\rm{a}}} + {m_{{\rm{ac}}}}} \right){c_{\rm{a}}}\left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)\\{Q_{\rm{a}}} = \left[ {\left( {0.325\;{\rm{kg}} + 0.624\;{\rm{kg}}} \right)\left( {0.22\;{\rm{kcal/kg}} \cdot \circ {\rm{C}}} \right)\left( {53.5\circ {\rm{C}} - 15\circ {\rm{C}}} \right)} \right]\\{Q_{\rm{a}}} = 8.03\;{\rm{kcal}}\end{array}\)

Here,\({c_{\rm{a}}}\)is the specific heat of aluminum.

The relation to find the heat gained by water is given by:

\(\begin{array}{c}{Q_{\rm{w}}} = {m_{\rm{w}}}{c_{\rm{w}}}\left( {{T_{\rm{f}}} - {T_{\rm{g}}}} \right)\\{Q_{\rm{w}}} = \left[ {\left( {1.75\,{\rm{kg}}} \right)\left( {1\,{\rm{kcal/kg}} \cdot \circ {\rm{C}}} \right)\left( {53.5\circ {\rm{C}} - 15\circ {\rm{C}}} \right)} \right]\\{Q_{\rm{w}}} = 67.375\;{\rm{kcal}}\end{array}\)

Here, \({c_{\rm{w}}}\) is the specific heat of water.

Calculation of the net heat transfer and calorie content of candy

The relation to find the net heat transfer is given by:

\(\begin{array}{l}{Q_{\rm{n}}} = \left( {8.03\;{\rm{kcal}}} \right) + \left( {67.375\;{\rm{kcal}}} \right)\\{Q_{\rm{n}}} = 75.4\;{\rm{kcal}}\end{array}\)

The relation to find the caloric content of candy is given by:

\(\begin{array}{l}{Q_{{\rm{65g}}}} = {Q_{\rm{n}}}\left( {\frac{{65\;{\rm{g}}}}{{15\;{\rm{g}}}}} \right)\\{Q_{{\rm{65g}}}} = \left( {75.4\;{\rm{kcal}}} \right)\left( {\frac{{65\;{\rm{g}}}}{{15\;{\rm{g}}}}} \right)\\{Q_{{\rm{65g}}}} = 326.7\;{\rm{kcal}}\end{array}\)

Thus, \({Q_{{\rm{65g}}}} = 326.7\;{\rm{kcal}}\) is the required caloric content.