Two kilograms of water at 1.00 atm at the boiling point of 100°C is heated until all the water vaporizes.What is its change in entropy? (For water,L_v = 2.26 × 10⁶ J/kg) A) 12,100 J/K B) 6060 J/K C) 3030 J/K D) 1220 J/K

Question

Quizplus · Accepted Answer

The change in entropy of the water can be calculated using the formula ΔS = L/T, where L is the latent heat of vaporization and T is the temperature at which the water is boiling. 
The latent heat of vaporization for water is given as 2.26 × 106 J/kg. 
The temperature at which water is boiling is 100°C or 373 K. 
Therefore, the change in entropy of 2 kg of water is ΔS = (2.26 × 106 J/kg)/373 K = 6060 J/K. 
However, since all the water vaporizes, the entropy change will be doubled as the number of moles of H2O changes from 2 to 1. 
Therefore, the total change in entropy is ΔS = 2 × 6060 J/K = 12,120 J/K, which is closest to option A.

Two Kilograms of Water at 1