Question

Calculate the change in entropy of the system when 10g of ice at -10C is converted into water vapor at 115C and at a constant pressure of 1bar. The constant pressure molar heat capacity of H2O(s) and H2O(l) is 75.291J/Kmol and that of H2O(g) is 33.58J/Kmo

Answer 1

Total change in entropy of the system is 85.84 J/K

Step-by-step explanation:

Δ
`s_(t)` = Δ
`s_(1)`+Δ
`s_(2)`+Δ
`s_(3)`+Δ
`s_(4)`

Δ
`s_(t)` = total entropy of the system in J/K

Δ
`s_(1)` = entropy due to warming ice from - 10°C (263 K) to 0°C (273 K) in J/K

Δ
`s_(2)` = entropy due to melting of ice at 0°C (273 K) in J/K

Δ
`s_(3)` = entropy due to warming liquid from 0°C (273 K) to 115°C (388 K) in J/K

Δ
`s_(4)` = entropy due to vaporizing liquid at 115°C (388 K) in J/K

Therefore,

Δ
`s_(1)` = m*
`c_(i)`*
`ln(273)/(263)`

m = mass = 10 g;
`c_(i)` = specific heat of ice = 2.106 J/(g*K)

Δ
`s_(1)` = 10*2.106*
`ln(273)/(263)` = 0.79 J/K

Δ
`s_(2)` = m*
`L_(f)`/273

m = 10 g;
`L_(f)` = 334 J/g

Δ
`s_(2)` = 10*334/273 = 12.23 J/K

Δ
`s_(3)` = m*
`c_(w)`*
`ln(388)/(273)`

m = mass = 10 g;
`c_(w)` = specific heat of water = 4.218 J/(g*K)

Δ
`s_(3)` = 10*4.218*
`ln(388)/(273)` = 14.83 J/K

Δ
`s_(4)` = m*
`L_(v)`/388

m = 10 g;
`L_(v)` = 2250 J/g

Δ
`s_(4)` = 10*2250/388 = 57.99 J/K

Therefore,

Δ
`s_(t)` = Δ
`s_(1)`+Δ
`s_(2)`+Δ
`s_(3)`+Δ
`s_(4)` = 0.79+12.23+14.83+57.99 = 85.84 J/K

Thus, the change in entropy of the system is 85.84 J/K