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 latent heat of vaporization of water is 2.26 x 10⁶ J/kg?

The entropy change of the sample of water = 6.059 x 10³ J/K. mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S. I unit of Entropy is J/K. mol

Mathematically, entropy is expressed as

ΔS = ΔH/T ... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

Given: If 1 mole of water = 0.0018 kg,

ΔH = latent heat * mass = 2.26 x 10⁶ * 1 = 2.26x 10⁶ J.

T = 100 °C = (100+273) K = 373 K.

Substituting these values into equation 1,

ΔS = 2.26x 10⁶/373

ΔS = 6.059 x 10³ J/K. mol

Therefore the entropy change of the sample of water = 6.059 x 10³ J/K. mol