The density of ice at 0 degree centigrade is 0.915g/cc and that of liquid water at 0 degree centigrade is 0.99987 g/cc. The work done for melting 1 mole of ice at 1.00 bar ( assuming work is done only due to expansion) is approximately
a) 0.17 J
b) 1.7 × 1000 J
C)0.000017 J

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Solution

Work done for expansion = P*delta V
Volume of 18g(1mole) of ice = 18/0.915= 19.67 ml
Volume of 18g of water = 18/0.99987 = 18ml
So the volume change is 1.67ml
Work done = 1x10^5 pascal x 1.67x10^-6M^3= 0.167J
or 0.17J

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The density of ice at 0 degree centigrade is 0.915g/cc and that of liquid water at 0 degree centigrade is 0.99987 g/cc. The work done for melting 1 mole of ice at 1.00 bar ( assuming work is done only due to expansion) is approximately a) 0.17 J b) 1.7 × 1000 J C)0.000017 J

Work done for expansion = P*delta V Volume of 18g(1mole) of ice = 18/0.915= 19.67 ml Volume of 18g of water = 18/0.99987 = 18ml So the volume change is 1.67ml Work done = 1x10^5 pascal x 1.67x10^-6M^3= 0.167J or 0.17J

The density of ice at 0 degree centigrade is 0.915g/cc and that of liquid water at 0 degree centigrade is 0.99987 g/cc. The work done for melting 1 mole of ice at 1.00 bar ( assuming work is done only due to expansion) is approximately
a) 0.17 J
b) 1.7 × 1000 J
C)0.000017 J

Work done for expansion = P*delta V
Volume of 18g(1mole) of ice = 18/0.915= 19.67 ml
Volume of 18g of water = 18/0.99987 = 18ml
So the volume change is 1.67ml
Work done = 1x10^5 pascal x 1.67x10^-6M^3= 0.167J
or 0.17J