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

( Specific heat of water liquid and water vapour are 4.2kJK^{-1}kg^{-1} and \: 2.0kJ K^{-1}kg^{-1} ; heat of liquid fusion and vapourisation of water are 334kJkg^{-1}\: and\: \: 2491kJkg^{-1} , respectively ). ( log 273 = 2.436 , log 373=2.572 , log 383 = 2.583)

  • Option 1)

    2.64kJkg^{-1}K^{-1}

    \: \:

     

  • Option 2)

    7.90kJkg^{-1}K^{-1}

  • Option 3)

    8.49 kJ kg^{-1}K^{-1}

  • Option 4)

    9.26\: \: kJkg^{-1}K^{-1}

Answers (1)
A admin

 

Entropy for phase transition at constant pressure -

\Delta S= \frac{\Delta H_{Transition}}{T}

- wherein

Transition \RightarrowFusion, Vaporisition, Sublimation

\Delta H\RightarrowEnthalpy

\Delta E\RightarrowInternal Energy

T\RightarrowTransitional temperature

 

 

Entropy for solid and liquid -

\Delta S= nC\, \ln \frac{T_{f}}{T_{i}}

or

\Delta S= ms\, \ln \frac{T_{f}}{T_{i}}
 

- wherein

n= no. of moles

C= molar heat capacity

m= mass

s= specific heat capacity

 

Phase change path 

H_{2}O(S)\underset{\Delta S_{1}}{\rightarrow}H_{2}O(l)\overset{\Delta S_{2}}{\rightarrow}H_{2}O(l)\overset{\Delta S_{3}}{\rightarrow}H_{2}O(g)\overset{\Delta S_{4}}{\rightarrow}H_{2}O(g)

273 K                    273 K                 373 K            373 K                   383 K 

\Delta S_{1}=\frac{\Delta H_{fusion}}{273}=\frac{334}{273}=1.22       \Delta S_{2}=4.2 ln (\frac{373}{273})=1.31

\Delta S_{3}=\frac{\Delta H_{vap}}{373}=\frac{2491}{373}=6.67          \Delta S_{4}=2.0 ln (\frac{383}{373})=0.05

\Delta S \: Total = \Delta S_{1}+\Delta S_{2}+\Delta S_{3}+\Delta S_{4}= 9.26 KJ \: Kg^{-1}K^{-1}


Option 1)

2.64kJkg^{-1}K^{-1}

\: \:

 

Option 2)

7.90kJkg^{-1}K^{-1}

Option 3)

8.49 kJ kg^{-1}K^{-1}

Option 4)

9.26\: \: kJkg^{-1}K^{-1}

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