The entropy change associated with the conversion of 1 k g of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapour are 4.2 k J K - and 2.0 k J K - 1 k g - 1 ; hea

Question

The entropy change associated with the conversion of $1 k g$ of ice at $273 K$ to water vapours at $383 K$ is:

(Specific heat of water liquid and water vapour are $4.2 k J K^{-}$ and $2.0 k J K^{- 1} k g^{- 1}$ ; heat of liquid fusion and vaporization of water are $334 k J k g^{- 1}$ and $2491 k J k g^{- 1}$ , respectively). ( $l o g$ $273 = 2.436, \log 373 = 2.572, \log 383 = 2.583$ )

TestHub · Accepted Answer

( i ) Phase change H 2 O ( l ) → H 2 O ( s ) So ( i i ) Temperature change phase same (fusion) H 2 O ( l ) → Δ S 2 Δ H 2 O ( l ) Δ S 2 = mS ln T 2 T 1 = 1 × 4 .2 ln 373 273 = 1 .31 ( i i i ) Phase change (vapourisation) Δ S 3 = Δ H v a p 373 = 2491 373 = 6.67 ( i v ) Temperature change phase same H 2 O ( g ) 373 K → Δ S 4 H 2 O ( g ) 383 K Δ S 4 = 2 l n 383 373 = 0.0529 Δ S total = Δ S 1 + Δ S 2 + Δ S 3 + Δ S 4 = 1 .223 + 1 .31 + 6 .67 + 0 .0529 = 9 .26 kJ / kg K

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