91影视

It works on the principle of the conservation of energy. In this question, the heat lost or rejected by the water equals the heat absorbed by the aluminum cup.

Equate the heat energy equations. Then, obtain the equilibrium temperature. The change in the entropy is the sum of the individual entropies of water and the aluminum cup.

The given data can be listed as

• The mass of an insulated aluminum cup is\({m_{{\rm{Al}}}} = 110{\rm{ g}}\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right) = 0.11{\rm{ kg}}\).
• The mass of the water is\({m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = 150{\rm{ g}}\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right) = 0.150{\rm{ kg}}\).
• The initial temperature of the aluminum cup is\({T_{{\rm{A}}{{\rm{l}}_{\rm{i}}}}} = 35\circ {\rm{C}} = \left( {35\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}} = 308{\rm{ K}}\).
• The initial temperature of the water is\({T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} = 45\circ {\rm{C}} = \left( {45\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}} = 318{\rm{ K}}\).
• The specific heat capacity of aluminum is\({c_{{\rm{Al}}}} = 900{\rm{ J/kg}} \cdot \circ {\rm{C}}\).
• The specific heat capacity of water is \({c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = 4186{\rm{ J/kg}} \cdot \circ {\rm{C}}\).

From calorimetry, the heat energy鈥檚 equation can be expressed as

鈥楢t equilibrium, equate the heat lost by water and the heat gained by the aluminum cup to obtain the final temperature.鈥�

\({Q_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = {Q_{{\rm{Al}}}}\)

Here,\({Q_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\)is the heat energy lost by the water, and\({Q_{{\rm{Al}}}}\)is the heat energy gained by the aluminum cup.

\(\begin{array}{c}{m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\left( {{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} - {T_f}} \right) = {m_{{\rm{Al}}}}{c_{{\rm{Al}}}}\left( {{T_f} - {T_{{\rm{A}}{{\rm{l}}_{\rm{i}}}}}} \right)\\{m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} - {m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{T_f} = {m_{{\rm{Al}}}}{c_{{\rm{Al}}}}{T_f} - {m_{{\rm{Al}}}}{c_{{\rm{Al}}}}{T_{{\rm{A}}{{\rm{l}}_{\rm{i}}}}}\\{m_{{\rm{Al}}}}{c_{{\rm{Al}}}}{T_f} + {m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{T_f} = {m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} + {m_{{\rm{Al}}}}{c_{{\rm{Al}}}}{T_{{\rm{A}}{{\rm{l}}_{\rm{i}}}}}\\{T_f} = \frac{{\left( {{m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} + {m_{{\rm{Al}}}}{c_{{\rm{Al}}}}{T_{{\rm{A}}{{\rm{l}}_{\rm{i}}}}}} \right)}}{{\left( {{m_{{\rm{Al}}}}{c_{{\rm{Al}}}} + {m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}} \right)}}\end{array}\).

Here,\({T_f}\)is the final temperature or equilibrium temperature.

Substituting the values in the above equation,

\(\begin{array}{c}{T_f} = \frac{{\left( {0.150{\rm{ kg}} \times 4186{\rm{ J/kg}} \cdot \circ {\rm{C}} \times 45\circ {\rm{C}} + 0.11{\rm{ kg}} \times 900{\rm{ J/kg}} \cdot \circ {\rm{C}} \times 35\circ {\rm{C}}} \right)}}{{\left( {0.11{\rm{ kg}} \times 900{\rm{ J/kg}} \cdot \circ {\rm{C}} + 0.150{\rm{ kg}} \times 4186{\rm{ J/kg}} \cdot \circ {\rm{C}}} \right)}}\\ = \frac{{31720.5{\rm{ J}}}}{{726.9{\rm{ J/}}\circ {\rm{C}}}}\\ = 43.64\circ {\rm{C}}\end{array}\).

Thus, the final temperature is \(43.64\circ {\rm{C}}\).

The heat lost by the water is equal to the heat gained by the aluminum cup. The heat energy can be expressed as

\(\begin{array}{c}Q = {Q_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\\ = {Q_{{\rm{Al}}}}\\ = {m_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{c_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\left( {{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} - {T_f}} \right)\end{array}\).

Substituting the values in the above equation,

\(\begin{array}{c}Q = 0.150{\rm{ kg}} \times 4186{\rm{ J/kg}} \cdot \circ {\rm{C}} \times \left( {45\circ {\rm{C}} - 43.64\circ {\rm{C}}} \right)\\ = 853.94{\rm{ J}}\end{array}\)

There is a variation in the temperatures of the aluminum cup and water. So, evaluate the average temperature with the help of their temperatures. These temperatures can be used to obtain the change in the entropy.

The average temperature of the water can be calculated as

\(\begin{array}{c}{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{{\rm{avg}}}}}} = \frac{{\left( {{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{i}}}}} + {T_f}} \right)}}{2}\\ = \frac{{\left( {45\circ {\rm{C}} + 43.64\circ {\rm{C}}} \right)}}{2}\\ = \left( {44.32\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}}\\ = 317.32{\rm{ K}}\end{array}\).

The average temperature of the aluminum cup can be calculated as

\(\begin{array}{c}{T_{{\rm{A}}{{\rm{l}}_{{\rm{avg}}}}}} = \frac{{\left( {{T_f} + {T_{{\rm{A}}{{\rm{l}}_{\rm{i}}}}}} \right)}}{2}\\ = \frac{{\left( {43.64\circ {\rm{C}} + 35\circ {\rm{C}}} \right)}}{2}\\ = \left( {39.32\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}}\\ = 312.32{\rm{ K}}\end{array}\).

The total change in the entropy can be expressed as

\(\begin{array}{c}\Delta {S_{\rm{T}}} = \Delta {S_{{\rm{Al}}}} - \Delta {S_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\\\Delta {S_{\rm{T}}} = \frac{Q}{{{T_{{\rm{A}}{{\rm{l}}_{{\rm{avg}}}}}}}} - \frac{Q}{{{T_{{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{{\rm{avg}}}}}}}}\end{array}\).

Substituting the values in the above equation,

\(\begin{array}{c}\Delta {S_{\rm{T}}} = \frac{{853.94{\rm{ J}}}}{{312.32{\rm{ K}}}} - \frac{{853.94{\rm{ J}}}}{{317.32{\rm{ K}}}}\\ = 2.734{\rm{ J/K}} - 2.691{\rm{ J/K}}\\ = 0.043{\rm{ J/K}}\end{array}\).

Thus, the total change in the entropy is \(0.043{\rm{ J/K}}\).