Question

Calculate the entropy change when 1 kg of water is heated from `27^(@)C` to `200^(@)C` forming super heated steam under constant pressure.
Given `:` Specific heat ofwater `= 4180 J //kg .K`
Specific heat of steam `= 1670 + 0.49 T J // kg. K`
Latent heat of vaporisation `= 23 xx 10^(5) J // kg`

Answer 1

Entropy change for heatgin 1 kg of water from `27^(@)C` to `100^(@)C`
`Delta S = 2.303 n C_(p) log. ( T_(2))/( T_(1))`
`= 2.303xx ( 1000)/( 18) "moles"xx ( 4180xx18)/( 1000) J mol^(-1) xx log. ( 373K)/( 300K) `
`= 910.55J`
Entropy change for heating 1 kg of water at `100^(@)C` to 1 kg of steam at `100^(@)C`
`Delta S = ( DeltaH_(v))/( T) = ( 23 xx 10^(5)J)/( 373 ) = 6166.21J`
Entropy change for heating 1kg steam from `100^(@)C ( 373K) ` to `200^(@)(473K) `
`DeltaS=int_(373)^(473)(nC_(p) dT)/(T) = int_373^473((1670 +0.49T))/(T) = dT`
`=int_(373)^(473)((1670)/(T)+0.49)dT`
`=[1670ln T + 0.49 T]_(373)^(473)`
`= 1670 xx 2.303 ( log473- log 373) + 0.49 ( 473- 373)`
`=1670 xx 2.303 ( 0. 1032 ) + 49`
`= 396.91 +49 =445. 9 J`
`:. `Total entropychange `= 910.55 + 6166.21 +45.9 J = 7522.6 J`