A1.00-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

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer