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Q.

From the following data, the enthalpy change for the sublimation of ice at 223 K will be [mean heat capacity of ice =2JK1g1 , mean heat capacity of H2O(l)=4.2JK1g1 , mean heat capacity of H2O(v) =1.85JK1g1 enthalpy of fusion of ice at 0C=334 Jg1, enthalpy of evaporation of water at 100C=2255Jg1 

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a

3000 Jg1

b

3109 Jg1

c

3827 Jg1

d

4000 Jg1

answer is B.

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Detailed Solution

 Ice (223K) Ice (273K) Water (273K) Water (373K) Steam (373K)

For 1 g of lce

ΔH=2×50+334+4.2×100+2255=3109

 

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From the following data, the enthalpy change for the sublimation of ice at 223 K will be [mean heat capacity of ice =2JK−1g−1 , mean heat capacity of H2O(l)=4.2JK−1g−1 , mean heat capacity of H2O(v) =1.85JK−1g−1 enthalpy of fusion of ice at 0∘C=334 Jg−1, enthalpy of evaporation of water at 100∘C=2255Jg−1