Question

1.0 kg of water at 35°C is mixed with 1.0 kg of water at 45°C in a well-insulated container. Estimate the net change in entropy of the system.

Short answer

The net increase of entropy is \(1.1\;{\rm{J/K}}\).

Step-by-step solution

Concepts

The heat absorbed by the hot water equals the heat released by hot water.

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

Given data

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

The initial temperature of the hot water is \({T_1} = {45 \circ }{\rm{C}}\).

The initial temperature of the cold water is \({T_2} = {35 \circ }{\rm{C}}\)

Let T be the final temperature of the water mixture.

Calculation

The amount of heat absorbed by the cold water equals the heat released by the hot water.

Now the amount of heat is \(Q = mc\left( {{T_1} - T} \right)\).

As both types of water have the same mass, then the final temperature is,

\(\begin{array}{c}T = \frac{{{T_1} + {T_2}}}{2}\\ = \frac{{{{45} \circ }{\rm{C}} + {{35} \circ }{\rm{C}}}}{2}\\ = {40 \circ }{\rm{C}}\end{array}\)

Now the average temperature of the hot water is,

\(\begin{array}{c}{{T'}_1} = \frac{{{T_1} + T}}{2}\\ = \frac{{{{45} \circ }{\rm{C}} + {{40} \circ }{\rm{C}}}}{2}\\ = {42.5 \circ }{\rm{C}}\\ = 315.5\;{\rm{K}}\end{array}\)

Now the average temperature of the cold water is,

\(\begin{array}{c}{{T'}_2} = \frac{{{T_2} + T}}{2}\\ = \frac{{{{35} \circ }{\rm{C}} + {{40} \circ }{\rm{C}}}}{2}\\ = {37.5 \circ }{\rm{C}}\\ = 310.5\;{\rm{K}}\end{array}\)

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,

\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = - \frac{Q}{{{{T'}_1}}} + \frac{Q}{{{T_2}}}\\ = mc\left( {{T_1} - T} \right)\left( {\frac{1}{{{{T'}_2}}} - \frac{1}{{{{T'}_1}}}} \right)\end{array}\)

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

\(\begin{array}{c}\Delta S = mc\left( {{T_1} - T} \right)\left( {\frac{1}{{{{T'}_2}}} - \frac{1}{{{{T'}_1}}}} \right)\\ = \left( {1.0\;{\rm{kg}}} \right) \times \left( {4186\;{\rm{J/kg}}{ \cdot {\rm{o}}}{\rm{C}}} \right)\left( {{{45} \circ }{\rm{C}} - {{40} \circ }{\rm{C}}} \right) \times \left( {\frac{1}{{310.5\;{\rm{K}}}} - \frac{1}{{315.5\;{\rm{K}}}}} \right)\\ = 1.1\;{\rm{J/K}}\end{array}\)

Hence, the net increase of entropy is \(1.1\;{\rm{J/K}}\).