Question

calculate that change in entropy of 10kg of water at 100dgree when it changes to vapour​

Answer 1

Answer:

This question doesn’t specify the water’s final state (100 C vapor or liquid?) and assigns an incorrect value to liquid water’s specific heat ( “heat capacity”) - it’s 4180 J per kg, not mole

The liquid’s entropy change would be its heat capacity (Cp) times the natural log of the two absolute temperatures or 4.18 J/g degree K*18 g/mol *(ln*((273+100)/(273+90)= 2.0447 J/(K*mol)

For 10 kg (10000/18 moles) of water, that ΔS works out to 1136 J/K

Since it takes about 540 calories (or 4.18*540 J) to evaporate 1 gram of water at atmospheric pressure and all of that energy goes into increasing its entropy, that change would add another 60,500 (10,000*(4.18*540/373) J/K

Consequently if the water ends up as steam. its total entropy change would 1136+60500 or 61700 J/degree.