Question

Question:You make tea with 0.250 kg of 85.0°C water and let it cool to room temperature (20.0°C). (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all of the heat lost by the water goes into the air. What is the total entropy change of the system tea + air?

Short answer

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

Step-by-step solution

Identification of given data

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 \(m = 0.250\;{\rm{kg}}\)

Conceptual Explanation

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

Determination of change in entropy of water

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 \( - 0.21\;{\rm{kJ}}/{\rm{K}}\).