Calculate the total entropy change for converting 1text kg of ice at 273text K into water vapor at 383text K . (Given: Specific heat of liquid water = 4.2text kJ K ^ -1 text kg ^ -1 , Specific heat of steam = 2.0text kJ K ^ -1 text kg ^ -1 , Enthalpy of fusion = 334text kJ kg ^ -1 , Enthalpy of vaporization = 2491text kJ kg ^ -1 )

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EEJ MAIN Chemistry QUESTION #7133

Question 1

Calculate the total entropy change for converting $1\text{ kg}$ of ice at $273\text{ K}$ into water vapor at $383\text{ K}$. (Given: Specific heat of liquid water $= 4.2\text{ kJ K}^{-1}\text{ kg}^{-1}$, Specific heat of steam $= 2.0\text{ kJ K}^{-1}\text{ kg}^{-1}$, Enthalpy of fusion $= 334\text{ kJ kg}^{-1}$, Enthalpy of vaporization $= 2491\text{ kJ kg}^{-1}$)

$9.26\text{ kJ K}^{-1}$✔️
$8.55\text{ kJ K}^{-1}$
$7.12\text{ kJ K}^{-1}$
$10.44\text{ kJ K}^{-1}$

Correct Answer Explanation

The total entropy change ($\Delta S_{total}$) is the sum of four steps:

1. Melting ice: $\Delta S_1 = \frac{334}{273} = 1.223$
2. Heating water (273K to 373K): $\Delta S_2 = 4.2 \ln(\frac{373}{273}) = 1.31$
3. Vaporizing water: $\Delta S_3 = \frac{2491}{373} = 6.678$
4. Heating steam (373K to 383K): $\Delta S_4 = 2.0 \ln(\frac{383}{373}) = 0.053$

Summing these gives approximately $9.26\text{ kJ K}^{-1}$.

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