Question

A 2.8-kg piece of aluminum at 28.5°C is placed in 1.0 kg of water in a Styrofoam container at room temperature (20.0°C). Estimate the net change in entropy of the system.

Short answer

The net increase of entropy is \(0.64\;{\rm{J/K}}\).

Step-by-step solution

Concepts

The lost heat by aluminum is equal to the absorbed heat by water.

The change in entropy is\(\Delta S = \frac{Q}{T}\)

Given data

The mass of water is \({m_1} = 1.0\;{\rm{kg}}\).

The mass of aluminum is \({m_2} = 2.8\;{\rm{kg}}\).

The initial temperature of the aluminum is \({T_2} = {28.5 \circ }{\rm{C}}\).

The initial temperature of the water is \({T_1} = {20.0 \circ }{\rm{C}}\)

Let T be the final temperature of the system.

Calculation

The specific heat of water is \({c_1} = 4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}\) , and the specific heat of aluminum is \({c_2} = 900\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}\).

Now for thermal equilibrium you get,

\(\begin{array}{c}{m_1}{c_1}\left( {T - {T_1}} \right) = {m_2}{c_2}\left( {{T_2} - T} \right)\\{m_1}{c_1}T - {m_1}{c_1}{T_1} = {m_2}{c_2}{T_2} - {m_2}{c_2}T\\\left( {{m_1}{c_1} + {m_2}{c_2}} \right)T = {m_2}{c_2}{T_2} + {m_1}{c_1}{T_1}\\T = \frac{{{m_2}{c_2}{T_2} + {m_1}{c_1}{T_1}}}{{{m_1}{c_1} + {m_2}{c_2}}}\end{array}\)

Now substituting the values in the above equation, you get,

\(\begin{array}{c}T = \frac{{\left[ {\left( {2.8\;{\rm{kg}}} \right)\left( {900\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)\left( {{{28.5} \circ }{\rm{C}}} \right)} \right] + \left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)\left( {{{20} \circ }{\rm{C}}} \right)} \right]}}{{\left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)} \right] + \left[ {\left( {2.8\;{\rm{kg}}} \right)\left( {900\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)} \right]}}\\ = {23.19 \circ }{\rm{C}}\end{array}\)

The amount of heat absorbed by the water equals the heat released by the aluminum.

The heat lost by aluminum is,

\(\begin{array}{c}Q = {m_1}{c_1}\left( {T - {T_1}} \right)\\ = \left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)\left( {{{23.19} \circ }{\rm{C}} - {{20} \circ }{\rm{C}}} \right)\\ = 13353\;{\rm{J}}\end{array}\)

Now the average temperature of the water is,

\(\begin{array}{c}{{T'}_1} = \frac{{{T_1} + T}}{2}\\ = \frac{{{{20} \circ }{\rm{C}} + {{23.19} \circ }{\rm{C}}}}{2}\\ = {21.60 \circ }{\rm{C}}\\ = 294.60\;{\rm{K}}\end{array}\)

Now the average temperature of the aluminum is,

\(\begin{array}{c}{{T'}_2} = \frac{{{T_2} + T}}{2}\\ = \frac{{{{28.5} \circ }{\rm{C}} + {{23.19} \circ }{\rm{C}}}}{2}\\ = {25.85 \circ }{\rm{C}}\\ = 298.85\;{\rm{K}}\end{array}\)

Now the rate of change in entropy for the hot water is \(\Delta {S_1} = - \frac{Q}{{{{T'}_1}}}\).

The rate of change in entropy for the cold water is \(\Delta {S_2} = \frac{Q}{{{{T'}_2}}}\).

Therefore the rate of change of net entropy is,

\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = + \frac{Q}{{{{T'}_1}}} - \frac{Q}{{{T_2}}}\\ = Q\left( {\frac{1}{{{{T'}_1}}} - \frac{1}{{{{T'}_2}}}} \right)\end{array}\)

Now substituting the values in the above equation, you get,

\(\begin{array}{c}\Delta S = Q\left( {\frac{1}{{{{T'}_2}}} - \frac{1}{{{{T'}_1}}}} \right)\\ = 13353\;{\rm{J}} \times \left( {\frac{1}{{294.60\;{\rm{K}}}} - \frac{1}{{298.85\;{\rm{K}}}}} \right)\\ = 0.64\;{\rm{J/K}}\end{array}\)

Hence, the net increase of entropy is \(0.64\;{\rm{J/K}}\).