If a 0.06247 kg ice cube initially at 0.00 C is allowed to melt in a Styrofoam cup holding 0.497 kg of water at 10.0 C , what is the total entropy change for this process?
You have two processes going on in this problem.
1.
The first is the melting of the ice cube, going from a solid state to a liquid state.
So this is the entropy of ice at zero degrees C going to water at zero degrees. It's the entropy of phase change. Be sure to use absolute temperature, Kelvin, and not C wen solving the equations.
Our equation is:
dS = dQ / T
We know T. But what about dQ. The only change happening in melting ice is a change of state. The value you'll want to look for is the enthalpy of fusion in units of kilograms.
2.
Once you've performed found the entropy of ice melting, we now move to our second problem.
Our mass of ice is now water at 0 C. It is going to mix with 0.497 kg of water that has an initial temperature of 10.0 C.
Now we have the benefit of being able to use heat capacities of liquid water and the fact that we know the initial temperature of each mass of water. The final factor will be the final temperature.
The formula:
dS=dQ/T
will be used again.
----
These are some hints. Let me know if I can help further.
Let me know if they help.
Billy W.