Question

Water of mass 13.0 g is initially at a temperature of 20.5°C. The water is frozen to −13.5°C at a constant pressure of 1.013x10^5 Pa. Find the total entropy change of the system.

Answer 1

Here,    s_water = specific heat of water

            s_ice =   specific heat of ice

L_fusion   = Latent heat of fusion

total entropy change of the system = m * s_water * ln(Tf/Ti) + m * L_fusion/T + m * s_ice * ln(Tf/Ti)

                  = 0.013 * 4186 * ln(273.15/293.65) - 0.013 * 333.55 * 1000/273.15 + 0.013 * 2050 * ln(259.65/273.15)

                  = - 3.938 - 15.87 - 1.35

                   = - 21.158 J/K