Question

(a) What is the entropy change of 1.00 mol of \(\mathrm{H}_{2} \mathrm{O}\) when it changes from ice to water at \(0.0^{\circ} \mathrm{C} ?\) (b) If the ice is in contact with an environment at a temperature of \(10.0^{\circ} \mathrm{C}\) what is the entropy change of the universe when the ice melts?

Short answer

Answer: The entropy change of the universe when the ice melts is 0.8 J/K.

Step-by-step solution

(a) Calculate the entropy change of the system

First, we need to find the heat transferred during the phase change. The heat of fusion (q) for water is 6.01 kJ/mol. Since we have 1 mol of ice, we can compute the heat transfer as follows: q = (1.00 mol) × (6.01 kJ/mol) = 6.01 kJ Now, we should convert the temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15 = 0.0 + 273.15 = 273.15 K Next, we can calculate the entropy change (ΔS) using the formula ΔS = q/T: ΔS_system = (6.01 kJ) / (273.15 K) = 0.0220 kJ/K = 22.0 J/K (rounded to 2 decimal places)

(b) Calculate the entropy change of the universe

In this part, we have to find the entropy change of the environment when the ice melts. The heat transfer to the environment is the same as the heat absorbed by the ice: q_environment = -6.01 kJ (negative because heat is released to the environment) Since the temperature of the environment is 10.0°C, we should first convert it to Kelvin: T_environment(K) = T_environment(°C) + 273.15 = 10.0 + 273.15 = 283.15 K Now we can calculate the entropy change (ΔS) for the environment: ΔS_environment = (q_environment) / (T_environment) = (-6.01 kJ) / (283.15 K) = -0.0212 kJ/K = -21.2 J/K (rounded to 2 decimal places) Finally, we'll find the total entropy change of the universe, which is the sum of the entropy changes for the system and the environment: ΔS_universe = ΔS_system + ΔS_environment = 22.0 J/K + (-21.2 J/K) = 0.8 J/K (rounded to 2 decimal places) The entropy change of the universe when the ice melts is 0.8 J/K.