Question

A small pool of water of mass 3.3 kg at 0°C freezes to ice at 0°C. The energy from the water goes into the air in the room, which is at -12°C and does not significantly change temperature during the process. Calculate the change of entropy of the universe due to this process. Latent heat of fusion of water is 3.35 x 10⁵ J/kg.

Answer 1

The change in entropy of the universe due to the freezing of a small pool of water can be calculated using the latent heat of fusion and the melting temperature of ice. In this case, the change in entropy is 4.04 x 10^3 J/K.

Step-by-step explanation:

The change in entropy of the universe can be calculated by considering the melting of a small pool of water into ice. The heat transfer necessary to melt 1.00 kg of ice is given by Q = mLf, where m is the mass and Lf is the latent heat of fusion. In this case, the latent heat of fusion of water is 3.35 x 10^5 J/kg. Therefore, the energy required to melt the water is Q = (3.3 kg)(3.35 x 10^5 J/kg) = 1.10 x 10^6 J. The change in entropy is then given by AS = Q/T, where T is the melting temperature of ice (0°C or 273 K). Substituting the values, we get AS = (1.10 x 10^6 J)/(273 K) = 4036.36 J/K = 4.04 x 10^3 J/K.