1.2 kilogram 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)

Question

Quizplus · Accepted Answer

We can use the formula for entropy change during a phase transition:

ΔS = L/T

where L is the latent heat of vaporization and T is the boiling point temperature.

Using the given value for latent heat of vaporization and temperature, we get:

ΔS = (2.26 × 10^6 J/kg) / 373 K

ΔS = 6,057 J/K

However, this value is for 1 kg of water, and the problem asks for the entropy change for 1.2 kg. So we need to multiply by 1.2:

ΔS = 6,057 J/K * 1.2

ΔS = 7,268.4 J/K

which is closest to answer choice B: 7,271 J/K.

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