Question

1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.

a) Calculate entropy change of the system, surrounding and universe. (temperature of the
environment is -10 °C)
b) Make some comments on entropy changes from the obtained data.
Please use the following data for water :
Melting entalpy of ice (ΔHmelting) at 0°C and 1 bar is 6020 J mol-1
.
Cp (H2O (s)) = 37,7 J mol-1 K-1
Cp (H2O (l)) = 75,3 J mol-1 K-1

Answer 1

Step-by-step explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

`\Delta \ S = \int\limits^(T_2)_(T_1)(nC_p(s)dT)/(T)`

`\Delta \ S = {nC_p(s)In (T_2)/(T_1)`

`\Delta \ S = {(1*37.7)In (263)/(273)`

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.