How Does Melting Ice Affect Entropy in a Calorimeter System?

[laxplayer1189]

Homework Statement

What is the net change in entropy of the system from the time of mixing until the moment the ice completely melts? The heat of fusion of ice is .
Note that since the amount of ice is relatively small, the temperature of the water remains nearly constant throughout the process. Note also that the ice starts out at the melting point, and you are asked about the change in entropy by the time it just melts. In other words, you can assume that the temperature of the "ice water" remains constant as well.


In a well-insulated calorimeter, 1.0 of water at 20 is mixed with 1.0 of ice at 0.

Homework Equations

The Attempt at a Solution

 

[Durato]

Entropy is defined as dS = dQ/T (dS = change in entropy, dQ = change in heat, T = temperature in kelvin). Since the temperature is constant, all you have to do is take the total heat required to melt the ice and divide by the temperature in kelvin (i think, anyways!). Oh, and i was just wandering where the units are?

 

[Moetasim]

The net change in entropy in this system can be calculated by considering the entropy of each component at different stages of the process. Initially, the entropy of the water and ice are both at their respective ambient temperatures, which means they have a relatively low entropy. As the ice melts, the entropy of the system increases due to the increase in disorder and randomness as the solid ice turns into liquid water. This increase in entropy is accompanied by a decrease in the temperature of the ice water mixture, as the ice absorbs heat from the water to melt.

Once all the ice has melted, the temperature of the water will reach equilibrium with the ice water mixture and the entropy of the system will remain constant. This means that the net change in entropy is equal to the increase in entropy during the melting process.

To calculate this change in entropy, we can use the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature at which the heat is added. In this case, the heat added is the heat of fusion of ice, which is provided in the problem statement. Since the temperature of the ice water mixture remains constant throughout the process, we can use the initial temperature of 0 degrees Celsius to calculate the change in entropy.

Therefore, the net change in entropy is equal to the heat of fusion of ice divided by the initial temperature of the ice water mixture, or ΔS = (334 J/g) / (273 K) = 1.22 J/K. This means that the net change in entropy in this system is 1.22 J/K.