Question

Calculate the entropy change when `1 kg` of water is heated from `27^(@)C` to `200^(@)C` forming supper heated steam under constant pressure. Given specific heat of water `= 4180 J Kg^(-1)K^(-1)`and specific heat of steam `= 1670 + 0.49 J kg^(-1)K^(-1)` and latent heat of vaporisation `=23 xx 10^(5) J kg^(-1)`.

Answer 1

`DeltaS = 2.303 xx C_(P) xx "log" (T_(2))/(T_(1))` [where `m` in `kg` and `C_(P)` in `J kg^(-1)]`
Entropy change for heating water from `27^(@)C` to `100^(@)C`.
`DeltaS = 2.303 xx (1000)/(18) xx (4180 xx 18)/(1000) "log"(373)/(300) = 910.55J`
Entropy change for heating `1 kg H_(2)O` to `1 kg` steam at `100^(@)C`.
`DeltaS = (DeltaH_(V))/(T) = (23 xx 10^(5))/(373) = 6166.21J`
Entropy change for heating `1 kg` steam from `373 to 473K, m` in `kg`.
`DeltaS = int_(373)^(473)(nC_(P).dT)/(T)=m int_(373)^(473)((1670+0.49T)dT)/(T)`
`=m int_(373)^(473) (1670 dT)/(R) +m int_(373)^(473) 0.49 dT`
`= m xx 1670 xx 2.3030 [log T]_(373)^(473) + m xx 0.49[T]_(373)^(473)`
`= 1 xx 1670 xx2.303 xx "log" (473)/(373) +1 xx 0.49 xx 100`
`= 396.73 +49 = 445.73J`
`:.` Total entropy change `= 910.55 +6166.21 +445.73`
`= 7522.50 J`