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

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

Explanation:

Δ = Δ+Δ+Δ+Δ

Δ = total entropy of the system in J/K

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

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

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

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

Therefore,

Δ = m**

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

Δ = 10*2.106* = 0.79 J/K

Δ = m*/273

m = 10 g; = 334 J/g

Δ = 10*334/273 = 12.23 J/K

Δ = m**

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

Δ = 10*4.218* = 14.83 J/K

Δ = m*/388

m = 10 g; = 2250 J/g

Δ = 10*2250/388 = 57.99 J/K

Therefore,

Δ = Δ+Δ+Δ+Δ = 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

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