One kilogram of water at 1.00 atm at the boiling point of 100 $\degree$C is heated until all the water vaporizes.What is its change in entropy? (For water,L_v = 2.26 x 10⁶ J/kg)

Question

Quizplus · Accepted Answer

The change in entropy of the water can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature at which the heat is added.

In this case, all of the water is vaporized, meaning that the heat added to the system is equal to the heat of vaporization of water, which is 40.7 kJ/mol or 2.26 x 10^6 J/kg.

We can calculate the initial and final temperatures of the water using the conversion formula from Fahrenheit to Celsius:

100°F = 37.8°C
100°C = 373.15K

Using these values, we can calculate the change in temperature of the water:

ΔT = Tfinal - Tinitial = 373.15 K - 310.15 K = 63 K

Therefore, the change in entropy is:

ΔS = Q/T = 2.26 x 10^6 J/kg / 373.15 K = 6,059.6 J/K

Rounding to three significant figures, the answer is B) 6 060 J/K.

One Kilogram of Water at 1 °\degree° C Is Heated Until All the Water Vaporizes

One Kilogram of Water at 1 $\degree$ C Is Heated Until All the Water Vaporizes