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3 December, 14:39

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?

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  1. 3 December, 17:33
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    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
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