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Question: The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K ...

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is :
[Given that : Specific heat of water liquid and water vapour are 4.2 kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}and 2.0kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}; latent heat of fusion and vaporisation of water are 344kJkg−1kJk{g^{ - 1}}and 2491kJkg−1kJk{g^{ - 1}}, respectively].
[Take : log 273 = 2.436, log 373 = 2.572, log 383 = 2.583]
a.) 7.90kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}
b.) 2.64kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}
c.) 8.49kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}
d.) 9.26kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}

Explanation

Solution

Entropy can be defined as the disorder in a system. When a substance changes its state, its entropy also changes. The gases have highest entropy while the solids possess lowest entropy. The entropy change in a system can be calculated from the two formulas given as-
ΔS\Delta S= ΔHTransitionT\dfrac{{\Delta {H_{Transition}}}}{T}
ΔS\Delta S= msln⁡TfTims\ln \dfrac{{{T_f}}}{{{T_i}}}
Where ΔS\Delta S is the entropy change
‘n’ is the number of moles
C is the molar heat capacity
Tf{T_f}is final temperature
Ti{T_i}is initial temperature
‘m’ is the mass
‘s’ is the specific heat capacity.

Complete step by step answer :
Let us start by writing what is given to us and what we need to find.
Thus, Given :
Mass of ice = 1 kg
Initial temperature = 273 K
Final temperature = 383 K
Further, we have been given that Specific heat of water liquid = 4.2 kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}
Specific heat of water vapour = 2.0kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}
Latent heat of fusion of water = 344kJkg−1kJk{g^{ - 1}}
Latent heat of vaporisation of water = 2491kJkg−1kJk{g^{ - 1}}
To find :
Entropy change
We have the formula for entropy change is as -
ΔS\Delta S= ΔHTransitionT\dfrac{{\Delta {H_{Transition}}}}{T}
ΔS\Delta S= nCln⁡TfTinC\ln \dfrac{{{T_f}}}{{{T_i}}}
ΔS\Delta S= msln⁡TfTims\ln \dfrac{{{T_f}}}{{{T_i}}}
Where ΔS\Delta S is the entropy change
‘n’ is the number of moles
C is the molar heat capacity
Tf{T_f}is final temperature
Ti{T_i}is initial temperature
‘m’ is the mass
‘s’ is the specific heat capacity.
We are converting the solid ice into water vapour which is a gas. The phase change for this can be as -
H2O(s)→ΔS1H2O(l)→ΔS2H2O(l)→ΔS3H2O(g)→ΔS4H2O(g){H_2}O(s)\xrightarrow{{\Delta {S_1}}}{H_2}O(l)\xrightarrow{{\Delta {S_2}}}{H_2}O(l)\xrightarrow{{\Delta {S_3}}}{H_2}O(g)\xrightarrow{{\Delta {S_4}}}{H_2}O(g)
ΔS1\Delta {S_1}= ΔHFUSIONT\dfrac{{\Delta {H_{FUSION}}}}{T}
ΔS1\Delta {S_1}= 334273\dfrac{{334}}{{273}}
ΔS1\Delta {S_1}= 1.22
ΔS2\Delta {S_2}= msln⁡TfTims\ln \dfrac{{{T_f}}}{{{T_i}}}
ΔS2\Delta {S_2}= 4.2ln⁡3832734.2\ln \dfrac{{383}}{{273}}
ΔS2\Delta {S_2}= 1.31
ΔS3\Delta {S_3}= ΔHvaporisationT\dfrac{{\Delta {H_{vaporisation}}}}{T}
ΔS3\Delta {S_3}= 2491373\dfrac{{2491}}{{373}}
ΔS3\Delta {S_3}= 6.67
ΔS4\Delta {S_4}= msln⁡TfTims\ln \dfrac{{{T_f}}}{{{T_i}}}
ΔS4\Delta {S_4}= 2.0ln⁡3832732.0\ln \dfrac{{383}}{{273}}
ΔS4\Delta {S_4}= 0.05
ΔSTOTAL\Delta {S_{TOTAL}}=ΔS1\Delta {S_1}+ΔS2\Delta {S_2}+ΔS3\Delta {S_3}+ΔS4\Delta {S_4}
ΔSTOTAL\Delta {S_{TOTAL}}= 1.22 + 1.31 + 6.67 + 0.05
ΔSTOTAL\Delta {S_{TOTAL}}= 9.25 kJK−1kg−1kJ{K^{ - 1}}k{g^{ - 1}}
This value is similar to in option d.).

So, the correct answer is option d.).

Note: The gases have least intermolecular forces. So, the molecules are not bonded to each other and thus are in random motion. So, the gases have highest entropy while solids have lowest entropy. When we are observing the phase transitions, firstly, the solid will convert to liquid and then there will be a rise in temperature of liquid by heating and when the sufficient temperature has reached then it will convert into gas. After this, there will be a temperature increase in gas.