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sashaice [31]
3 years ago
9

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 hea

t of vaporization of water is 2.26 x 10⁶ J/kg?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

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.

<em>Given:  If 1 mole of water = 0.0018 kg,</em>

<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>

<em>T = 100 °C = (100+273)  K = 373 K.</em>

<em>Substituting these values into equation 1,</em>

<em>ΔS =2.26x 10⁶/373</em>

ΔS = 6.059 x 10³ J/K.mol

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

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