Re: Entropy Change Problem - Correction

["Polvani, Donald G." <donald_g_polvani@MD.NORTHGRUM.COM>]

This post corrects an error in my second post of yesterday.  See comments
below original message reproduced here.

> Thanks to Bob Sciamanda and David Bowman for correcting my thinking.  I now
> start with a spontaneous process with the surrounding gas at T_g which is
> less than the freezing temperature of the water(T_0) ,insulate the gas and
> the water from the surroundings, and set conditions up so the gas ends up
at
> T_0 just as the last bit of water freezes by suitable choice of the ratio
of
> the gas and water masses and original gas temperature (T_g).  I ask for the
> net entropy change at this equilibrium point.  Almost the same calculation
> as my original post now leads to the net entropy change dS_net = dS_gas +
> dS_ice = A[ln(1/x)/(1-x) - 1] where now A = mL_f/T_0 and x = T_g/T_0.>
> Now A > 0 and ln(1/x)/(1-x)- 1 > or = 0 for 0 < x < 1, so, indeed, dS_n>et
>
> or = 0.

The corrected analysis, for the same conditions, is:

dS_ice = -mL_f/T_0 and dS_gas = (mL_f/delta_T)ln(T_0/T_g)
where delta_T = T_0 - T_g

Letting A = mL_f/T_g and x = delta_T/T_g, we can put the net entropy change
in the form:
dS_net = dS_gas + dS_ice = Af(x) where f(x) = (1/x)ln(1+x) - 1/(1+x)

Therefore, dS_net is > 0 since A > 0 and f(x) > 0 for x >0

Brian Mcinnes suggested a much simpler solution, for a surrounding gas which
now acts as an infinite temperature reservoir.  This yields dS_net =
mL_f(1/T_g - 1/T_0) which must be positive for T_g < T_0.  This is certainly
a simpler aproach.  I guess it comes down to a question of taste.  Having a
process which ends in equilbrium is more "realistic" to me than an infinite
temperature reservoir whose temperature doesn't "increase" as thermal energy
is transferred to it by the water.

Don Polvani
Anne Arundel Community College