You make tea with 0.50 kg of 85.0°C water and let it cool to room temperature 120.0°C2.

(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?

a) Entropy change dS = dQ/T

= mcdT / T

Integrating both sides

S₂ - S₁ = - mclnT₂ / T₁

= -.5 X 4200 ln (85+273) / (20 + 273)

.5 X 4200 ln 358 / 293

= - 417.6 J/K

Entropy change will be negative as heat is lost by the system.

b) Sine there is no change in the temperature of air, This heat will enter air at temperature (20 + 273) K = 293 K

Heat entering air

=.5 x 4200 x 65

= 136500 J

Change in entropy

136500 / 293 (room temperature is constant at 293k

= + 465.87 J/K

Entropy change will be positive as heat is gained by the system.

Total change in the entropy of the system (tea + air)

= + 465.87 - 417.6

= 48.27 J/K

Entropy change will be negative as heat is lost by the system.