Entropy of Melting ice With Salt

Most recent answer: 03/09/2011

Q:
If ice melts at a temperature below 273K, say 263K, then how does the entropy during the process change? How do I calculate it if I know the heat of reaction at 263K, entropy change at 273K and other auxiliary info like heat capacities and densities of ice and water at 273K?
- Niharika (age 18)
India
A:
I assume that this ice must be exposed to some sort of solute- salts, sugar, ethanol, etc. That's why it would melt at the lower temperature. You're right to think that the entropy change wouldn't be quite the same as when ice melts to pure water at 273K.
If, as you suggest, the latent heat (L) of melting is known, you're in good shape. The entropy change is just given by:

ΔS=L/T.

You suggest knowing some  other quantities that could be useful for slight variants of the problem. Let's say you were freezing pure water instead of melting ice. Then maybe you could get it supercooled to 263K before it freezes. You want to know the (negative) entropy change on freezing. It's the same as the change on freezing at 273K, corrected for the different entropy changes of ice and liquid water between 263K and 273 K. You get those entropy changes by integrating their respective heat capacities divided by T from 263K to 273 K. I think it's a decent approximation to treat those heat capacities as independent of T in this range, so the correction would be:
(Cice-Cliquid)ln(273/263).

I can't think of any simple version of this type of problem where you need to know the densities.

Mike W.

(published on 03/09/2011)