# Entropy Problems

1. May 16, 2010

### salient

1. The problem statement, all variables and given/known data
80g of ice initially at zero degrees C is allowed to melt in 200g of water initially at 20 degrees C.

Find the energy change of the water, energy change of ice
Find entropy change in water, ice, and universe.

2. Relevant equations
Delta S = Q/T
Q = (Specific heat)(mass)(Delta T)
Lf = heat of fusion
Cw = specific heat of water (yes I know it's 4.184ish J/grams degrees C, but he doesn't like us using numbers).

3. The attempt at a solution
My physics instructor spent a half-hour defining entropy in about 18 different unique-to-him ways, and an hour and a half on random tangents that didn't really have much to do with entropy or even physics. No equations were given, no problems were demonstrated.

So I'm just going to attempt to solve this using methods I remember from high school chemistry about fifteen years ago, and I'm hoping maybe someone can tell me if I'm way off my rocker.

so Qice = (80)(Lf) + (80)(Cw)(Tf-0)
and Qwater = (200)(Cw)(Tf-20)

Since I know energy from the water is going into the ice:

(80)(Lf) + (80)(Cw)(Tf-0) = -(200)(Cw)(Tf-20)

Since the final temperature is the only real variable there, I can solve for that and come up with values for the Qs.

As far as entropy goes though, this isn't a constant temperature process. In the equation Delta S = Q/T, if indeed that is the equation I should be using, do I put the final temperature there below the Q change required to reach it? Would the entropy change of the universe be the sum of those two Delta Ss?

2. May 17, 2010

### ehild

I suppose that this whole process is adiabatic, there is no heat exchange with the environment, and there is no work done, or it can be ignored.
So you have three processes: Melting ice is isotherm, the temperature of ice stays 0 C°till it completely melts. You can get the entropy change during melting from the heat and temperature.
For the entropy change of the water, take into account that dQ=cmdT, so dS=dQ/T=cmdT/T. Integrate between the initial and final temperature to get the change of entropy. Do the same with both the warm water and with the melted ice.
The environment does not get any heat if the process was adiabatic, so its entropy does not change. So the entropy change of the universe is the sum of the three entropy terms: melting the ice, warming up the ice-water and cooling down the warm water.

ehild

3. May 17, 2010

### Andrew Mason

Yes.
Entropy is calculated along the reversible path between initial and final states.

$$\Delta S_{universe} = \Delta S_{ice} + \Delta S_{water} = \int_{273}^{T_f} dQ_{rev}/T + \int_{293}^{T_f} dQ_{rev}/T = \int_{273}^{T_f} (m_{ice}Lf +m_{ice}CdT) /T + \int_{293}^{T_f} m_{water}CdT/T$$

AM