A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the heat of vaporization of water is 2.26 × 10⁶J/kg?

Question

Quizplus · Accepted Answer

The entropy change (ΔS) is calculated using ΔS = -Q/T, where Q is the heat removed (2.26 × 10^6 J) and T is the temperature in Kelvin (373 K). ΔS = -(2.26 × 10^6 J) / 373 K = -6.06 × 10^3 J/K.

A 100-Kg Sample of Steam at 100