91Ó°ÊÓ

The temperature of water is\({T_1} = 85\;^\circ {\rm{C}}\)

The temperature of room is\({T_2} = 20\;^\circ {\rm{C}}\)

The mass of water and tea is \(m = 0.250\;{\rm{kg}}\)

The change in entropy of water is found by molecular disorder between water temperature and room temperature.

The change in entropy of the water is given as:

\(\Delta {s_w} = mc\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right)\)

Here,\(c\)is the specific heat of water and its value is\(4.2\;{\rm{kJ}}/{\rm{kg}} \cdot {\rm{K}}\).

Substitute all the values in the above equation.

\(\begin{array}{l}\Delta {s_w} = \left( {0.250\;{\rm{kg}}} \right)\left( {4.2\;{\rm{kJ}}/{\rm{kg}} \cdot {\rm{K}}} \right)\ln \left( {\frac{{\left( {20\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{\left( {85\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}} \right)\\\Delta {s_w} = - 0.21\;{\rm{kJ}}/{\rm{K}}\end{array}\)

Therefore, the entropy change of water is \( - 0.21\;{\rm{kJ}}/{\rm{K}}\).