Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)

Respuesta :

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

The change in entropy of the gas is 374.39 J/K.

The given parameters:

  • Mass of the ice, m = 0.3071 kg
  • Temperature of the gas, T = 273.15 K
  • Latent heat of fusion of water, L = 333,000 J/kg

The heat of fusion of the ice is calculated as follows;

[tex]\Delta H = mL\\\\\Delta H = 0.3071 \times 333,000\\\\\Delta H = 102,264.3 \ J[/tex]

The change in entropy of the gas is calculated as follows;

[tex]\Delta S = \frac{\Delta H}{T} \\\\\Delta S = \frac{102,264.3}{273.15} \\\\\Delta S = 374.39 \ J/K[/tex]

Thus, the change in entropy of the gas is 374.39 J/K.

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