The latent heat of fusion of ice at 0° C is 80 cal / g. Entropy change (Δ S) accompanying the melting of 1 g of ice at 0° C would be (units: cal / g / K )

Question

The latent heat of fusion of ice at 0° C is 80 cal / g. Entropy change (Δ S) accompanying the melting of 1 g of ice at 0° C would be (units: cal / g / K ) (A) 273 (B) 8.0 (C) 0.0 (D) 0.293

Tardigrade Admin · Accepted Answer

Δ Sm e l=(Δ Hm e l/T)=(80/273) =0.293 cal / g / K

Q. The latent heat of fusion of ice at $0^{\circ} C$ is $80 c a l / g$ . Entropy change $(Δ S)$ accompanying the melting of $1 g$ of ice at $0^{\circ} C$ would be (units: $c a l / g / K$ )

273

8.0

0.0

0.293

$Δ S_{m e l} = \frac{Δ H _{m e l}}{T} = \frac{80}{273}$
$= 0.293 c a l / g / K$

Q. The latent heat of fusion of ice at 0∘C is 80cal/g. Entropy change (ΔS) accompanying the melting of 1g of ice at 0∘C would be (units: cal/g/K )

273

8.0

0.0

0.293

ΔSmel​=TΔHmel​​=27380​ =0.293cal/g/K

Q. The latent heat of fusion of ice at $0^{\circ} C$ is $80 c a l / g$ . Entropy change $(Δ S)$ accompanying the melting of $1 g$ of ice at $0^{\circ} C$ would be (units: $c a l / g / K$ )

$Δ S_{m e l} = \frac{Δ H _{m e l}}{T} = \frac{80}{273}$
$= 0.293 c a l / g / K$