Question

The entropy change associated with the conversion of 1 kg of ice at 273 k to water vapour at 383 k is : (specific heat of water liquid and water vapour are and heat of liquid fusion and vapourisation of water are and respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583) options

Answer 1

Answer:

According to the question, we can write the phase change path as follows:

H2O(s)  →  H2O(l)  →  H2O(l)  →  H2O(g) →  H2O(g)  

273K      ∆S1  273K   ∆S2   373K   ∆S3  373K   ∆S4   383K

1. ∆S1 = ∆Hfusion / T

Here, ∆Hfusion = heat of liquid fusion of water = 334 kJ/kg

T = transitional temperature = 273 K

∴ ∆S1 = 334 / 273 = 1.223 ….. (i)

2. ∆S2 = m s ln [Tf / Ti]

Here, m = mass = 1 kg

s = specific heat capacity of water liquid = 4.2 kJ K⁻¹kg⁻¹

Tf = final transitional temperature = 373 K

Ti = initial transitional temperature = 273 K

∴ ∆S2 = 1 * 4.2 ln [373 / 273] = 1.31 ……. (ii)

3. ∆S3 = ∆Hvap / T

Here, ∆Hvap = heat of liquid vapourisation of water = 2491 kJ/kg

T = transitional temperature = 373 K

∴ ∆S3 = 2491 / 373 = 6.67 ….. (iii)

4. ∆S4 = m s ln [Tf / Ti]

Here, s = specific heat capacity of water vapour = 2.0 kJ K^-1 kg^-1

Tf = final transitional temperature = 383 K

Ti = initial transitional temperature = 373 K

∴ ∆S4 = 1 * 2 ln [383 / 373] = 0.0529 ……. (iv)

5.Total entropy change associated is  

∆Stotal = ∆S1 +∆S2 + ∆S3 + ∆S4

Substituting the value from (i), (ii), (iii) & (iv), we get

∆Stotal = 1.223 + 1.31 + 6.67 + 0.0529 = 9.2559 ≈ 9.26 kJ K⁻¹kg⁻¹

Hence, The entropy change associated with the conversion of 1 kg of ice at 273 k of water vapour at 383 k is 9.26 kJ K⁻¹ kg⁻¹.

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