Question

What is the entropy change when 1.0 kg of water at \( 100^\circ C \) is converted to steam at the same temperature? The latent heat of vaporization of water is \( 2.25 \times 10^6 \, \text{J/kg} \).

• A. \( 2.25 \times 10^3 \, \text{J/K} \)
• B. \( 2.25 \times 10^6 \, \text{J/K} \)
• C. \( 2.25 \times 10^9 \, \text{J/K} \)
• D. \( 2.25 \times 10^7 \, \text{J/K} \)

Hint:

The entropy change during a phase transition is given by \( \Delta S = \frac{Q}{T} \). For vaporization, \( Q = m L \), where \( m \) is the mass of the substance and \( L \) is the latent heat of vaporization.

Answer 1

The Correct Option is A

The entropy change for a phase transition (such as vaporization) is given by the formula: \[ \Delta S = \frac{Q}{T} \] Where: - \( Q \) is the heat absorbed during the phase transition, - \( T \) is the temperature at which the transition occurs (in Kelvin). For the vaporization of water, the heat absorbed is: \[ Q = m L \] Where: - \( m = 1.0 \, \text{kg} \) (mass of water), - \( L = 2.25 \times 10^6 \, \text{J/kg} \) (latent heat of vaporization). Substitute the values: \[ Q = 1.0 \times 2.25 \times 10^6 = 2.25 \times 10^6 \, \text{J} \] Now, convert the temperature from Celsius to Kelvin: \[ T = 100^\circ C = 100 + 273.15 = 373.15 \, \text{K} \] Now calculate the entropy change: \[ \Delta S = \frac{2.25 \times 10^6}{373.15} \approx 2.25 \times 10^3 \, \text{J/K} \] Thus, the entropy change is \( 2.25 \times 10^3 \, \text{J/K} \).