91影视

The mass of the cookies is\({m_{\rm{c}}} = 100\;{\rm{g}}\).

The sample of cookies is\({m_{{\rm{sc}}}} = 10\;{\rm{g}}\).

The mass of the aluminum bomb is\({m_{\rm{a}}} = 0.615\;{\rm{kg}}\).

The mass of water is\({m_{\rm{w}}} = 2\;{\rm{kg}}\).

The mass of the aluminum cup is\({m_{{\rm{ac}}}} = 0.524\;{\rm{kg}}\).

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

The final temperature is \({T_{\rm{f}}} = 36\circ {\rm{C}}\).

In this problem, the heat released by the cookies is equivalent to the heat absorbed by the water and aluminum calorimeter cup.

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

\(\begin{array}{c}{Q_{\rm{a}}} = \left( {{m_{\rm{a}}} + {m_{{\rm{ac}}}}} \right){c_{\rm{a}}}\left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)\\{Q_{\rm{a}}} = \left[ {\left( {0.615\;{\rm{kg}} + 0.524\;{\rm{kg}}} \right)\left( {0.22\;{\rm{kcal/kg}} \cdot \circ {\rm{C}}} \right)\left( {36\circ {\rm{C}} - 15\circ {\rm{C}}} \right)} \right]\\{Q_{\rm{a}}} = 5.26\;{\rm{kcal}}\end{array}\)

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

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

\(\begin{array}{c}{Q_{\rm{w}}} = {m_{\rm{w}}}{c_{\rm{w}}}\left( {{T_{\rm{f}}} - {T_{\rm{g}}}} \right)\\{Q_{\rm{w}}} = \left[ {\left( {2\;{\rm{kg}}} \right)\left( {1\;{\rm{kcal/kg}} \cdot \circ {\rm{C}}} \right)\left( {36\circ {\rm{C}} - 15\circ {\rm{C}}} \right)} \right]\\{Q_{\rm{w}}} = 42\;{\rm{kcal}}\end{array}\)

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

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

\(\begin{array}{l}{Q_{\rm{n}}} = \left( {5.26\;{\rm{kcal}}} \right) + \left( {42\;{\rm{kcal}}} \right)\\{Q_{\rm{n}}} = 47.26\;{\rm{kcal}}\end{array}\)

The obtained solution\({Q_{\rm{n}}} = 47.26\;{\rm{kcal}}\)is the energy content of 10 grams; so for a content of 100 g, the value of energy will increase by 10 times, that is,\({Q_{\rm{n}}} = 472.6\;{\rm{kcal}}\).

Thus, \({Q_{\rm{n}}} = 472.6\;{\rm{kcal}}\) is the required caloric content.