Question

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?

Answer 1

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.

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

Answer 2

The entropy change of the sample of steam is equal to 6,058.98 J/Kmol.

Given the following data:

• Mass = 1.00 kg

• Temperature = 100°C to Kelvin = $273+100=373\;K$

• Latent heat of vaporization of water = $2.26 \times 10^6\;J/kg$

To determine the entropy change of the sample of steam:

Mathematically, entropy change is given by the formula:

$\Delta S = \frac{\Delta H}{T}$

Where:

• $\Delta H$ is the enthalpy change.

• T is the absolute temperature.

Substituting the given parameters into the formula, we have;

$\Delta S = \frac{2.26 \times 10^6\times 1}{373}$

Entropy change = 6,058.98 J/Kmol.

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